半导体物理与器件第四版课后习题答案
半导体物理与器件第四版课后习题集答案解析4
Chapter 44.1⎪⎪⎭⎫ ⎝⎛-=kTE N N n gc i exp 2υ ⎪⎪⎭⎫ ⎝⎛-⎪⎭⎫ ⎝⎛=kT E T N N g O cO exp 3003υwhere cO N and O N υ are the values at 300 K.(a) SiliconT (K) kT (eV)i n (cm 3-) 200 400 600 01727.0 03453.0 0518.041068.7⨯ 121038.2⨯ 141074.9⨯(b) Germanium (c) GaAsT (K) i n (cm 3-) i n (cm 3-)200 400 600101016.2⨯ 141060.8⨯ 161082.3⨯38.1 91028.3⨯ 121072.5⨯_______________________________________ 4.2Plot_______________________________________ 4.3(a) ⎪⎪⎭⎫⎝⎛-=kT E N N n g c i exp 2υ ()()()319192113001004.1108.2105⎪⎭⎫⎝⎛⨯⨯=⨯T()()⎥⎦⎤⎢⎣⎡-⨯3000259.012.1exp T()3382330010912.2105.2⎪⎭⎫⎝⎛⨯=⨯T()()()()⎥⎦⎤⎢⎣⎡-⨯T 0259.030012.1expBy trial and error, 5.367≅T K(b)()252122105.2105⨯=⨯=i n()()()()()⎥⎦⎤⎢⎣⎡-⎪⎭⎫⎝⎛⨯=T T 0259.030012.1exp 30010912.2338By trial and error, 5.417≅T K_______________________________________ 4.4At 200=T K, ()⎪⎭⎫⎝⎛=3002000259.0kT017267.0=eVAt 400=T K, ()⎪⎭⎫⎝⎛=3004000259.0kT034533.0=eV()()()()17222102210025.31040.11070.7200400⨯=⨯⨯=ii nn⎥⎦⎤⎢⎣⎡-⎥⎦⎤⎢⎣⎡-⨯⎪⎭⎫⎝⎛⎪⎭⎫ ⎝⎛=017267.0exp 034533.0exp 30020030040033g g E E⎥⎦⎤⎢⎣⎡-=034533.0017267.0exp 8g g E E()[]9578.289139.57exp 810025.317-=⨯g Eor()1714.38810025.3ln 9561.2817=⎪⎪⎭⎫⎝⎛⨯=g E or 318.1=g E eVNow ()32103004001070.7⎪⎭⎫⎝⎛=⨯o co N N υ⎪⎭⎫ ⎝⎛-⨯034533.0318.1exp ()()172110658.2370.210929.5-⨯=⨯o co N N υ so 371041.9⨯=o co N N υcm 6-_______________________________________ 4.5()()⎪⎭⎫ ⎝⎛-=⎪⎭⎫ ⎝⎛-⎪⎭⎫ ⎝⎛-=kT kT kT A n B n i i 20.0exp 90.0exp 10.1exp For 200=T K, 017267.0=kT eV For 300=T K, 0259.0=kT eV For 400=T K, 034533.0=kT eV (a) For 200=T K, ()()610325.9017267.020.0exp -⨯=⎪⎭⎫ ⎝⎛-=A n B n i i (b) For 300=T K, ()()41043.40259.020.0exp -⨯=⎪⎭⎫ ⎝⎛-=A n B n i i (c) For 400=T K, ()()31005.3034533.020.0exp -⨯=⎪⎭⎫ ⎝⎛-=A n B n i i _______________________________________ 4.6(a) ()⎥⎦⎤⎢⎣⎡---∝kT E E E E f g F c F c exp()⎥⎦⎤⎢⎣⎡---∝kT E E E E c c exp()⎥⎦⎤⎢⎣⎡--⨯kT E E F c exp Let x E E c =-Then ⎪⎭⎫⎝⎛-∝kT x x f g F c expTo find the maximum value: ()⎪⎭⎫⎝⎛-∝-kT x x dx f g d F c exp 212/10exp 12/1=⎪⎭⎫ ⎝⎛-⋅-kT x x kT which yields2212/12/1kTx kT x x =⇒= The maximum value occurs at2kTE E c +=(b)()()⎥⎦⎤⎢⎣⎡---∝-kT E E E E f g F F exp 1υυ()⎥⎦⎤⎢⎣⎡---∝kT E E E E υυexp()⎥⎦⎤⎢⎣⎡--⨯kT E E F υexp Let x E E =-υThen ()⎪⎭⎫ ⎝⎛-∝-kT x x f g F exp 1υTo find the maximum value()[]0exp 1=⎥⎦⎤⎢⎣⎡⎪⎭⎫ ⎝⎛-∝-kT x x dx d dx f g d F υ Same as part (a). Maximum occurs at2kTx =or2kTE E -=υ_______________________________________ 4.7()()()()⎥⎦⎤⎢⎣⎡---⎥⎦⎤⎢⎣⎡---=kT E E E E kT E E E E E n E n c c c c 221121exp expwherekT E E c 41+= and 22kTE E c += Then()()()⎥⎦⎤⎢⎣⎡--=kT E E kT kTE n E n 2121exp 24()5.3exp 22214exp 22-=⎥⎦⎤⎢⎣⎡⎪⎭⎫ ⎝⎛--=or()()0854.021=E n E n_______________________________________ 4.8Plot_______________________________________ 4.9Plot_______________________________________ 4.10⎪⎪⎭⎫ ⎝⎛=-**ln 43n p midgap Fi m m kT E E Silicon: o p m m 56.0*=, o n m m 08.1*=0128.0-=-midgap Fi E E eV Germanium: o pm m 37.0*=,o n m m 55.0*=0077.0-=-midgap Fi E E eV Gallium Arsenide: o p m m 48.0*=,o n m m 067.0*= 0382.0+=-midgap Fi E E eV _______________________________________ 4.11 ()⎪⎪⎭⎫⎝⎛=-c midgap Fi N N kT E E υln 21()()kT kT 4952.0108.21004.1ln 211919-=⎪⎪⎭⎫ ⎝⎛⨯⨯=T (K) kT (eV)(midgap Fi E E -)(eV) 200 400 600 01727.0 03453.0 0518.0 0086.0- 0171.0- 0257.0-_______________________________________ 4.12(a) ⎪⎪⎭⎫ ⎝⎛=-**ln 43n p midgap Fi m m kT E E ()⎪⎭⎫⎝⎛=21.170.0ln 0259.04363.10-⇒meV(b) ()⎪⎭⎫⎝⎛=-080.075.0ln 0259.043midgap Fi E E47.43+⇒meV_______________________________________ 4.13Let ()==K E g c constant Then()()dE E fE g n FE co c⎰∞=dE kT E E Kc E F⎰∞⎪⎪⎭⎫⎝⎛-+=exp 11()dE kT E E K cE F ⎰∞⎥⎦⎤⎢⎣⎡--≅exp Let kT E E c-=η so that ηd kT dE ⋅=We can write ()()c F c F E E E E E E -+-=-so that()()()η-⋅⎥⎦⎤⎢⎣⎡--=⎥⎦⎤⎢⎣⎡--e x p e x p e x p kT E E kT E E F c F The integral can then be written as()()ηηd kT E E kT K n F c o ⎰∞-⎥⎦⎤⎢⎣⎡--⋅⋅=0exp exp which becomes()⎥⎦⎤⎢⎣⎡--⋅⋅=kT E E kT K n F c o exp _______________________________________ 4.14Let ()()c c E E C E g -=1 for c E E ≥ Then()()dE E fE g n FE co c⎰∞=()dE kT E E E E C c E Fc ⎰∞⎪⎪⎭⎫⎝⎛-+-=exp 11()()dE kT E E E E C F E C c⎥⎦⎤⎢⎣⎡---≅⎰∞exp 1LetkTE E c-=η so that ηd kT dE ⋅= We can write()()F c c F E E E E E E -+-=-Then()⎥⎦⎤⎢⎣⎡--=kT E E C n F c o exp 1()()dE kT E E E E c E c c⎥⎦⎤⎢⎣⎡---⨯⎰∞exp or()⎥⎦⎤⎢⎣⎡--=kT E E C n F c o exp 1 ()()()[]()ηηηd kT kT -⨯⎰∞exp 0We find that()()()11exp exp 0+=---=-∞∞⎰ηηηηηdSo()()⎥⎦⎤⎢⎣⎡--=kT E E kT C n F c o exp 21 _______________________________________ 4.15We have ⎪⎪⎭⎫⎝⎛=∈*1m m a r o r o For germanium, 16=∈r , o m m 55.0*= Then()()()53.02955.01161=⎪⎭⎫⎝⎛=o a roroA r 4.151=The ionization energy can be written as ()6.132*⎪⎪⎭⎫⎝⎛∈∈⎪⎪⎭⎫ ⎝⎛=s o o m m E eV ()()029.06.131655.02=⇒=E eV_______________________________________ 4.16We have ⎪⎪⎭⎫⎝⎛=∈*1m m a r o r o For gallium arsenide, 1.13=∈r ,o m m 067.0*= Then()()oA r 10453.0067.011.131=⎪⎭⎫⎝⎛=The ionization energy is()()()6.131.13067.06.1322*=⎪⎪⎭⎫ ⎝⎛∈∈⎪⎪⎭⎫ ⎝⎛=s o o m m E or0053.0=E eV_______________________________________ 4.17(a) ⎪⎪⎭⎫⎝⎛=-o c F c n N kT E E ln()⎪⎪⎭⎫⎝⎛⨯⨯=1519107108.2ln 0259.02148.0=eV (b) ()F c g F E E E E E --=-υ90518.02148.012.1=-=eV(c) ()⎥⎦⎤⎢⎣⎡--=kT E E N p F o υυexp()⎥⎦⎤⎢⎣⎡-⨯=0259.090518.0exp 1004.119 31090.6⨯=cm 3- (d) Holes(e) ⎪⎪⎭⎫⎝⎛=-i o Fi F n n kT E E ln()⎪⎪⎭⎫⎝⎛⨯⨯=1015105.1107ln 0259.0338.0=eV_______________________________________ 4.18(a) ⎪⎪⎭⎫⎝⎛=-o F p N kT E E υυln()⎪⎪⎭⎫⎝⎛⨯⨯=16191021004.1ln 0259.0162.0=eV(b)()υE E E E E F g F c --=- 958.0162.012.1=-=eV(c) ()⎪⎭⎫⎝⎛-⨯=0259.0958.0exp 108.219o n31041.2⨯=cm 3-(d) ⎪⎪⎭⎫⎝⎛=-i o F Fi n p kT E E ln()⎪⎪⎭⎫⎝⎛⨯⨯=1016105.1102ln 0259.0365.0=eV_______________________________________ 4.19(a) ⎪⎪⎭⎫⎝⎛=-o c F c n N kT E E ln()⎪⎪⎭⎫⎝⎛⨯⨯=519102108.2ln 0259.08436.0=eV ()F c g F E E E E E --=-υ 8436.012.1-= 2764.0=-υE E F eV (b)()⎪⎭⎫⎝⎛-⨯=0259.027637.0exp 1004.119o p1410414.2⨯=cm 3-(c) p-type_______________________________________ 4.20(a) ()032375.03003750259.0=⎪⎭⎫⎝⎛=kT eV()⎥⎦⎤⎢⎣⎡-⎪⎭⎫ ⎝⎛⨯=032375.028.0exp 300375107.42/317o n 141015.1⨯=cm 3-()28.042.1-=--=-F c g F E E E E E υ 14.1=eV()⎥⎦⎤⎢⎣⎡-⎪⎭⎫⎝⎛⨯=032375.014.1exp 3003751072/318o p 31099.4⨯=cm 3-(b) ()⎪⎪⎭⎫⎝⎛⨯⨯=-14171015.1107.4ln 0259.0F c E E2154.0=eV()2154.042.1-=--=-F c g F E E E E E υ 2046.1=eV()⎥⎦⎤⎢⎣⎡-⨯=0259.02046.1exp 10718o p21042.4-⨯=cm 3-_______________________________________ 4.21(a) ()032375.03003750259.0=⎪⎭⎫⎝⎛=kT eV()⎥⎦⎤⎢⎣⎡-⎪⎭⎫⎝⎛⨯=032375.028.0exp 300375108.22/319o n 151086.6⨯= cm 3-()28.012.1-=--=-F c g F E E E E E υ 840.0=eV()⎥⎦⎤⎢⎣⎡-⎪⎭⎫⎝⎛⨯=032375.0840.0exp 3003751004.12/319o p 71084.7⨯=cm 3-(b) ⎪⎪⎭⎫⎝⎛=-o c F c n N kT E E ln()⎪⎪⎭⎫⎝⎛⨯⨯=151910862.6108.2ln 0259.02153.0=eV9047.02153.012.1=-=-υE E F eV()⎥⎦⎤⎢⎣⎡-⨯=0259.0904668.0exp 1004.119o p 31004.7⨯=cm 3-_______________________________________ 4.22 (a) p-type(b) 28.0412.14===-g F E E E υeV()⎥⎦⎤⎢⎣⎡--=kT E E N p F o υυexp ()⎥⎦⎤⎢⎣⎡-⨯=0259.028.0exp 1004.119141010.2⨯=cm 3- ()υE E E E E F g F c --=- 84.028.012.1=-=eV()⎥⎦⎤⎢⎣⎡--=kT E E N n F c c o exp ()⎥⎦⎤⎢⎣⎡-⨯=0259.084.0exp 108.219 51030.2⨯=cm 3-_______________________________________ 4.23(a) ⎥⎦⎤⎢⎣⎡-=kT E E n n Fi Fi o exp ()⎥⎦⎤⎢⎣⎡⨯=0259.022.0exp 105.110131033.7⨯=cm 3-⎥⎦⎤⎢⎣⎡-=kT E E n p F Fii o exp ()⎥⎦⎤⎢⎣⎡-⨯=0259.022.0exp 105.11061007.3⨯=cm 3-(b) ⎥⎦⎤⎢⎣⎡-=kT E E n n Fi F i o exp()⎥⎦⎤⎢⎣⎡⨯=0259.022.0exp 108.16 91080.8⨯=cm 3-⎥⎦⎤⎢⎣⎡-=kT E E n p F Fii o exp()⎥⎦⎤⎢⎣⎡-⨯=0259.022.0exp 108.16 21068.3⨯=cm 3-_______________________________________ 4.24(a) ⎪⎪⎭⎫⎝⎛=-o F p N kT E E υυln()⎪⎪⎭⎫⎝⎛⨯⨯=151********.1ln 0259.01979.0=eV(b)()υE E E E E F g F c --=- 92212.019788.012.1=-=eV(c) ()⎥⎦⎤⎢⎣⎡-⨯=0259.092212.0exp 108.219o n31066.9⨯=cm 3- (d) Holes(e) ⎪⎪⎭⎫⎝⎛=-i o F Fi n p kT E E ln()⎪⎪⎭⎫⎝⎛⨯⨯=1015105.1105ln 0259.03294.0=eV _______________________________________ 4.25()034533.03004000259.0=⎪⎭⎫⎝⎛=kT eV()2/3193004001004.1⎪⎭⎫⎝⎛⨯=υN1910601.1⨯=cm 3-()2/319300400108.2⎪⎭⎫⎝⎛⨯=c N19103109.4⨯=cm 3-()()1919210601.1103109.4⨯⨯=i n⎥⎦⎤⎢⎣⎡-⨯034533.012.1exp 24106702.5⨯=1210381.2⨯=⇒i n cm 3- (a)⎪⎪⎭⎫ ⎝⎛=-oF pN kT E E υυln ()⎪⎪⎭⎫⎝⎛⨯⨯=151910510601.1ln 034533.02787.0=eV (b) 84127.027873.012.1=-=-F c E E eV(c)()⎥⎦⎤⎢⎣⎡-⨯=034533.084127.0exp 103109.419o n910134.1⨯=cm 3- (d) Holes(e) ⎪⎪⎭⎫⎝⎛=-i o F Fi n p kT E E ln()⎪⎪⎭⎫⎝⎛⨯⨯=121510381.2105ln 034533.02642.0=eV _______________________________________ 4.26(a) ()⎥⎦⎤⎢⎣⎡-⨯=0259.025.0exp 10718o p141050.4⨯=cm 3-17.125.042.1=-=-F c E E eV()⎥⎦⎤⎢⎣⎡-⨯=0259.017.1exp 107.417o n21013.1-⨯=cm 3- (b) 034533.0=kT eV ()2/318300400107⎪⎭⎫ ⎝⎛⨯=υN1910078.1⨯=cm 3- ()2/317300400107.4⎪⎭⎫⎝⎛⨯=c N1710236.7⨯=cm 3- ⎪⎪⎭⎫⎝⎛=-o F p N kT E E υυln()⎪⎪⎭⎫⎝⎛⨯⨯=14191050.410078.1ln 034533.03482.0=eV072.13482.042.1=-=-F c E E eV()⎥⎦⎤⎢⎣⎡-⨯=034533.007177.1exp 10236.717o n 41040.2⨯=cm 3-_____________________________________ 4.27(a) ()⎥⎦⎤⎢⎣⎡-⨯=0259.025.0exp 1004.119o p141068.6⨯=cm 3-870.025.012.1=-=-F c E E eV()⎥⎦⎤⎢⎣⎡-⨯=0259.0870.0exp 108.219o n41023.7⨯=o n cm 3- (b)034533.0=kT eV()2/3193004001004.1⎪⎭⎫⎝⎛⨯=υN1910601.1⨯=cm 3- ()2/319300400108.2⎪⎭⎫ ⎝⎛⨯=c N1910311.4⨯=cm 3-⎪⎪⎭⎫⎝⎛=-o F p N kT E E υυln()⎪⎪⎭⎫⎝⎛⨯⨯=14191068.610601.1ln 034533.03482.0=eV7718.03482.012.1=-=-F c E E eV()⎥⎦⎤⎢⎣⎡-⨯=034533.077175.0exp 10311.419o n91049.8⨯=cm 3-_______________________________________ 4.28(a) ()F c o F N n ηπ2/12=For 2kT E E c F +=,5.02==-=kTkT kT E E c F F η Then ()0.12/1≅F F η()()0.1108.2219⨯=πo n191016.3⨯=cm 3-(b) ()F c o F N n ηπ2/12=()()0.1107.4217⨯=π171030.5⨯=cm 3-_______________________________________ 4.29()F o F N p ηπυ'=2/12()()FF ηπ'⨯=⨯2/119191004.12105So ()26.42/1='FF η We find kTE E FF-=≅'υη0.3()()0777.00259.00.3==-F E E υeV_______________________________________ 4.30(a) 44==-=kTkTkT E E c F F ηThen ()0.62/1≅F F η ()F c o F N n ηπ2/12=()()0.6108.2219⨯=π201090.1⨯=cm 3-(b) ()()0.6107.4217⨯=πo n181018.3⨯=cm 3-_______________________________________ 4.31For the electron concentration ()()()E f E g E n F c =The Boltzmann approximation applies, so ()()c nE E hm E n -=32/3*24π()⎥⎦⎤⎢⎣⎡--⨯kT E E F exp or()()()⎥⎦⎤⎢⎣⎡--=kT E E h m E n F c nexp 2432/3*π()⎥⎦⎤⎢⎣⎡---⨯kT E E kT E E kTc c exp DefinekTE E x c-= Then()()()x x K x n E n -=→exp To find maximum ()()x n E n →, set()()x x K dx x dn -⎢⎣⎡==-exp 2102/1 +()()⎥⎦⎤--x x exp 12/1or()⎥⎦⎤⎢⎣⎡--=-x x Kx 21exp 02/1which yieldskT E E kT E E x c c 2121+=⇒-==For the hole concentration ()()()[]E f E g E p F -=1υUsing the Boltzmann approximation ()()E E h m E p p-=υπ32/3*24()⎥⎦⎤⎢⎣⎡--⨯kT E E F exp or()()()⎥⎦⎤⎢⎣⎡--=kT E E h mE pF pυπexp 2432/3*()⎥⎦⎤⎢⎣⎡---⨯kT E E kT E E kTυυexp DefinekTEE x -='υThen()()x x K x p '-''='exp To find maximum value of ()()x p E p '→,set()0=''x d x dp Using the results from above,we find the maximum atkT E E 21-=υ_______________________________________4.32 (a) Silicon: We have()⎥⎦⎤⎢⎣⎡--=kT E E N n F c c o exp We can write()()F d d c F c E E E E E E -+-=- For045.0=-d c E E eV andkT E E F d 3=-eV we can write()⎥⎦⎤⎢⎣⎡--⨯=30259.0045.0exp 108.219o n()()737.4exp 108.219-⨯= or171045.2⨯=o n cm 3- We also have()⎥⎦⎤⎢⎣⎡--=kT E E N p F o υυexp Again, we can write()()υυE E E E E E a a F F -+-=- ForkT E E a F 3=- and045.0=-υE E a eV Then()⎥⎦⎤⎢⎣⎡--⨯=0259.0045.03exp 1004.119o p ()()737.4exp 1004.119-⨯= or161012.9⨯=o p cm 3- (b) GaAs: assume 0058.0=-d c E E eV Then()⎥⎦⎤⎢⎣⎡--⨯=30259.00058.0exp 107.417o n ()()224.3exp 107.417-⨯= or161087.1⨯=o n cm 3-Assume 0345.0=-υE E a eV Then()⎥⎦⎤⎢⎣⎡--⨯=30259.00345.0exp 10718o p ()()332.4exp 10718-⨯=or161020.9⨯=o p cm 3-_______________________________________ 4.33Plot_______________________________________ 4.34(a)151510310154⨯=-⨯=o p cm 3- ()415210105.7103105.1⨯=⨯⨯=o n cm 3-(b)16103⨯==d o N n cm 3-()316210105.7103105.1⨯=⨯⨯=o p cm 3-(c)10105.1⨯===i o o n p n cm 3-(d) ()()3191923003751004.1108.2⎪⎭⎫⎝⎛⨯⨯=in()()()()⎥⎦⎤⎢⎣⎡-⨯3750259.030012.1exp1110334.7⨯=⇒i n cm 3- 15104⨯==a o N p cm 3-()8152111034.110410334.7⨯=⨯⨯=on cm 3-(e) ()()3191923004501004.1108.2⎪⎭⎫⎝⎛⨯⨯=in()()()()⎥⎦⎤⎢⎣⎡-⨯4500259.030012.1exp1310722.1⨯=⇒i n cm 3-()2132141410722.1210210⨯+⎪⎪⎭⎫ ⎝⎛+=o n1410029.1⨯=cm 3-()12142131088.210029.110722.1⨯=⨯⨯=o p cm 3-_______________________________________ 4.35(a)151510104-⨯=-=d a o N N p15103⨯=cm 3-()3152621008.1103108.1-⨯=⨯⨯==o i o p n n cm 3-(b)16103⨯==d o N n cm 3-()416261008.1103108.1-⨯=⨯⨯=o p cm 3-(c)6108.1⨯===i o o n p n cm 3-(d) ()()318172300375100.7107.4⎪⎭⎫⎝⎛⨯⨯=in()()()()⎥⎦⎤⎢⎣⎡-⨯3750259.030042.1exp810580.7⨯=⇒i n cm 3- 15104⨯==a o N p cm 3-()215281044.110410580.7⨯=⨯⨯=on cm 3-(e) ()()318172300450100.7107.4⎪⎭⎫⎝⎛⨯⨯=in()()()()⎥⎦⎤⎢⎣⎡-⨯4500259.030042.1exp1010853.3⨯=⇒i n cm 3- 1410==d o N n cm 3-()7142101048.11010853.3⨯=⨯=op cm 3-_______________________________________ 4.36 (a) Ge: 13104.2⨯=i n cm 3-(i)2222i dd o n N N n +⎪⎪⎭⎫ ⎝⎛+=()21321515104.221022102⨯+⎪⎪⎭⎫ ⎝⎛⨯+⨯=or15102⨯=≅d o N n cm 3-()152132102104.2⨯⨯==o i o n n p111088.2⨯= cm 3- (ii)151610710⨯-=-≅d a o N N p 15103⨯=cm 3- ()152132103104.2⨯⨯==o i o p n n111092.1⨯=cm 3-(b) GaAs: 6108.1⨯=i n cm 3- (i)15102⨯=≅d o N n cm()315261062.1102108.1-⨯=⨯⨯=op cm 3-(ii)15103⨯=-≅d a o N N p cm 3-()315261008.1103108.1-⨯=⨯⨯=on cm 3-(c) The result implies that there is only one minority carrier in a volume of 310cm 3. _______________________________________ 4.37(a) For the donor level⎪⎪⎭⎫ ⎝⎛-+=kT E E N n F d d d exp 2111⎪⎭⎫ ⎝⎛+=0259.020.0exp 2111or41085.8-⨯=dd N n(b) We have()⎪⎪⎭⎫ ⎝⎛-+=kT E E E f F F exp 11Now()()F c c F E E E E E E -+-=- or245.0+=-kT E E FThen()⎪⎭⎫ ⎝⎛++=0259.0245.01exp 11E f For()51087.2-⨯=E f F_______________________________________ 4.38 (a) ⇒>d a N N p-type (b) Silicon:1313101105.2⨯-⨯=-=d a o N N p or13105.1⨯=o p cm 3- Then()7132102105.1105.1105.1⨯=⨯⨯==o i o p n n cm 3- Germanium:2222i da d a o n N N N N p +⎪⎪⎭⎫ ⎝⎛-+-=()21321313104.22105.12105.1⨯+⎪⎪⎭⎫⎝⎛⨯+⎪⎪⎭⎫ ⎝⎛⨯=or131026.3⨯=o p cm 3- Then()131321321076.110264.3104.2⨯=⨯⨯==o i o p n n cm 3-Gallium Arsenide:13105.1⨯=-=d a o N N p cm 3- and()216.0105.1108.113262=⨯⨯==o i o p n n cm 3- _______________________________________ 4.39 (a) ⇒>a d N N n-type(b)1515102.1102⨯-⨯=-≅a d o N N n14108⨯=cm 3-()51421021081.2108105.1⨯=⨯⨯==o i o n n p cm 3-(c)()d a ao N N N p -+'≅ 151515102102.1104⨯-⨯+'=⨯aN 15108.4⨯='⇒aN cm 3-()41521010625.5104105.1⨯=⨯⨯=on cm 3-_______________________________________ 4.40()155210210125.1102105.1⨯=⨯⨯==o i o p n n cm 3- ⇒>o o p n n-type_______________________________________ 4.41()()318192300250100.61004.1⎪⎭⎫⎝⎛⨯⨯=i n()()⎥⎦⎤⎢⎣⎡-⨯3002500259.066.0exp24108936.1⨯=1210376.1⨯=⇒i n cm 3- 2222414i o o i o i o n n n n p n n =⇒==i o n n 21=⇒So 111088.6⨯=o n cm 3-,Then 121075.2⨯=o p cm 3-2222i aa o n N N p +⎪⎪⎭⎫ ⎝⎛+= 212210752.2⎪⎪⎭⎫ ⎝⎛-⨯a N242108936.12⨯+⎪⎪⎭⎫ ⎝⎛=a N()21224210752.2105735.7⎪⎪⎭⎫ ⎝⎛+⨯-⨯aa N N242108936.12⨯+⎪⎪⎭⎫ ⎝⎛=aN so that 1210064.2⨯=a N cm 3-_______________________________________ 4.42Plot_______________________________________ 4.43Plot_______________________________________ 4.44Plot_______________________________________ 4.452222i ad a d o n N N N N n +⎪⎪⎭⎫ ⎝⎛-+-= 2102.1102101.1141414⨯-⨯=⨯2214142102.1102i n +⎪⎪⎭⎫⎝⎛⨯-⨯+()()221321314104104101.1i n +⨯=⨯-⨯22727106.1109.4i n +⨯=⨯ so 131074.5⨯=i n cm 3-1314272103101.1103.3⨯=⨯⨯==o i o n n p cm 3- _______________________________________ 4.46 (a) ⇒>d a N N p-typeMajority carriers are holes1616105.1103⨯-⨯=-=d a o N N p16105.1⨯=cm 3-Minority carriers are electrons()4162102105.1105.1105.1⨯=⨯⨯==o i o p n n cm 3- (b) Boron atoms must be addedd a ao N N N p -+'=161616105.1103105⨯-⨯+'=⨯aN So 16105.3⨯='aN cm 3-()316210105.4105105.1⨯=⨯⨯=on cm 3-_______________________________________ 4.47 (a) ⇒<<i o n p n-type (b) oi o o i o p n n n n p 22=⇒=on ()16421010125.1102105.1⨯=⨯⨯=cm 3-⇒electrons are majority carriers4102⨯=o p cm 3-⇒holes are minority carriers (c) a d o N N n -= 151610710125.1⨯-=⨯d N so 1610825.1⨯=d N cm 3-_______________________________________ 4.48⎪⎪⎭⎫⎝⎛=-i o F Fi n p kT E E ln For Germanium T (K) kT (eV)i n (cm 3-) 200400 60001727.0 03453.0 0518.0101016.2⨯ 141060.8⨯ 161082.3⨯2222i a a o n N N p +⎪⎪⎭⎫⎝⎛+=and 1510=a N cm 3- T (K) op (cm 3-)()F Fi E E -(eV)200400 60015100.1⨯151049.1⨯ 161087.3⨯1855.0 01898.0 000674.0_______________________________________ 4.49(a) ⎪⎪⎭⎫⎝⎛=-d c F c N N kT E E ln()⎪⎪⎭⎫⎝⎛⨯=d N 19108.2ln 0259.0 For 1410cm 3-, 3249.0=-F c E E eV 1510cm 3-, 2652.0=-F c E E eV 1610cm 3-, 2056.0=-F c E E eV 1710cm 3-, 1459.0=-F c E E eV(b) ⎪⎪⎭⎫⎝⎛=-i d Fi F n N kT E E ln()⎪⎪⎭⎫⎝⎛⨯=10105.1ln 0259.0d N For 1410cm 3-, 2280.0=-Fi F E E eV 1510cm 3-, 2877.0=-Fi F E E eV 1610cm 3-, 3473.0=-Fi F E E eV 1710cm 3-, 4070.0=-Fi F E E eV _______________________________________ 4.50 (a)2222i d d o n N N n +⎪⎪⎭⎫⎝⎛+= 151005.105.1⨯==d o N n cm 3- ()21515105.01005.1⨯-⨯()2215105.0i n +⨯=so 2821025.5⨯=i nNow()()3191923001004.1108.2⎪⎭⎫ ⎝⎛⨯⨯=T n i()()⎥⎦⎤⎢⎣⎡-⨯3000259.012.1exp T()3382830010912.21025.5⎪⎭⎫ ⎝⎛⨯=⨯T⎥⎦⎤⎢⎣⎡-⨯T 973.12972exp By trial and error, 5.536=T K (b) At 300=T K,⎪⎪⎭⎫⎝⎛=-o c F c n N kT E E ln()⎪⎪⎭⎫⎝⎛⨯=-151910108.2ln 0259.0F c E E2652.0=eV At 5.536=T K,()046318.03005.5360259.0=⎪⎭⎫⎝⎛=kT eV()2/3193005.536108.2⎪⎭⎫⎝⎛⨯=c N1910696.6⨯=cm 3-⎪⎪⎭⎫⎝⎛=-o c F c n N kT E E ln()⎪⎪⎭⎫⎝⎛⨯⨯=-15191005.110696.6ln 046318.0F c E E5124.0=eV then ()2472.0=-∆F c E E eV (c) Closer to the intrinsic energy level._______________________________________ 4.51⎪⎪⎭⎫⎝⎛=-i o F Fi n p kT E E ln At 200=T K, 017267.0=kT eV 400=T K, 034533.0=kT eV 600=T K, 0518.0=kT eVAt 200=T K,()()3191923002001004.1108.2⎪⎭⎫⎝⎛⨯⨯=in⎥⎦⎤⎢⎣⎡-⨯017267.012.1exp410638.7⨯=⇒i n cm 3- At 400=T K,()()3191923004001004.1108.2⎪⎭⎫⎝⎛⨯⨯=in⎥⎦⎤⎢⎣⎡-⨯034533.012.1exp 1210381.2⨯=⇒i n cm 3- At 600=T K,()()3191923006001004.1108.2⎪⎭⎫⎝⎛⨯⨯=in⎥⎦⎤⎢⎣⎡-⨯0518.012.1exp1410740.9⨯=⇒i n cm 3- At 200=T K and 400=T K, 15103⨯==a o N p cm 3- At 600=T K,2222i a a o n N N p +⎪⎪⎭⎫⎝⎛+=()2142151510740.921032103⨯+⎪⎪⎭⎫ ⎝⎛⨯+⨯=1510288.3⨯=cm 3-Then, 200=T K, 4212.0=-F Fi E E eV 400=T K, 2465.0=-F Fi E E eV600=T K, 0630.0=-F Fi E E eV_______________________________________ 4.52(a)()⎪⎪⎭⎫⎝⎛⨯=⎪⎪⎭⎫ ⎝⎛=-6108.1ln 0259.0ln a i a F Fi N n N kT E EFor 1410=a N cm 3-,4619.0=-F Fi E E eV1510=a N cm 3-,5215.0=-F Fi E E eV1610=a N cm 3-,5811.0=-F Fi E E eV1710=a N cm 3-,6408.0=-F Fi E E eV (b)()⎪⎪⎭⎫⎝⎛⨯=⎪⎪⎭⎫ ⎝⎛=-a a F N N N kT E E 18100.7ln 0259.0ln υυ For 1410=a N cm 3-,2889.0=-υE E F eV1510=a N cm 3-,2293.0=-υE E F eV1610=a N cm 3-,1697.0=-υE E F eV1710=a N cm 3-,1100.0=-υE E F eV_______________________________________ 4.53(a) ⎪⎪⎭⎫ ⎝⎛=-**ln 43n p midgap Fi m m kT E E ()()10ln 0259.043= or0447.0+=-midgap Fi E E eV(b) Impurity atoms to be added so 45.0=-F midgap E E eV (i) p-type, so add acceptor atoms (ii)4947.045.00447.0=+=-F Fi E E eV Then⎪⎪⎭⎫⎝⎛-=kT E E n p F Fi i o exp()⎪⎭⎫⎝⎛=0259.04947.0exp 105or131097.1⨯==a o N p cm 3-_______________________________________4.54()⎥⎦⎤⎢⎣⎡--=-=kT E E N N N n F c c a d o exp so()⎪⎭⎫⎝⎛-⨯+⨯=0259.0215.0exp 108.21051915d N15151095.6105⨯+⨯=or16102.1⨯=d N cm 3-_______________________________________ 4.55 (a) Silicon(i)⎪⎪⎭⎫⎝⎛=-d c F c N N kT E E ln()2188.0106108.2ln 0259.01519=⎪⎪⎭⎫⎝⎛⨯⨯=eV(ii)1929.00259.02188.0=-=-F c E E eV()⎥⎦⎤⎢⎣⎡--=kT E E N N F c c d exp()⎥⎦⎤⎢⎣⎡-⨯=0259.01929.0exp 108.2191610631.1⨯=d N cm 3-15106⨯+'=dN 1610031.1⨯='⇒dN cm 3- Additionaldonor atoms (b) GaAs(i)()⎪⎪⎭⎫⎝⎛⨯=-151710107.4ln 0259.0F c E E15936.0=eV(ii)13346.00259.015936.0=-=-F c E E eV()⎥⎦⎤⎢⎣⎡-⨯=0259.013346.0exp 107.417d N1510718.2⨯=cm 3-1510+'=dN 1510718.1⨯='⇒dN cm 3- Additionaldonor atoms _______________________________________4.56 (a) ⎪⎪⎭⎫ ⎝⎛=-a F Fi N N kT E E υln()⎪⎪⎭⎫⎝⎛⨯⨯=16191021004.1ln 0259.01620.0=eV(b) ⎪⎪⎭⎫⎝⎛=-d c Fi F N N kT E E ln ()1876.0102108.2ln 0259.01619=⎪⎪⎭⎫ ⎝⎛⨯⨯=eV (c) For part (a); 16102⨯=o p cm 3-()162102102105.1⨯⨯==o i o p n n410125.1⨯=cm 3-For part (b): 16102⨯=o n cm 3-()162102102105.1⨯⨯==o i o n n p410125.1⨯=cm 3-_______________________________________4.57⎥⎦⎤⎢⎣⎡-=kT E E n n Fi F i o exp ()⎥⎦⎤⎢⎣⎡⨯=0259.055.0exp 108.1615100.3⨯=cm 3-Add additional acceptor impuritiesa d o N N n -= a N -⨯=⨯151510710315104⨯=⇒a N cm 3-_______________________________________ 4.58(a) ⎪⎪⎭⎫⎝⎛=-i o F Fi n p kT E E ln ()3161.0105.1103ln 0259.01015=⎪⎪⎭⎫⎝⎛⨯⨯=eV (b) ⎪⎪⎭⎫⎝⎛=-i o Fi F n n kT E E ln ()3758.0105.1103ln 0259.01016=⎪⎪⎭⎫⎝⎛⨯⨯=eV(c) Fi F E E =(d) ⎪⎪⎭⎫⎝⎛=-i o F Fi n p kT E E ln ()⎪⎪⎭⎫ ⎝⎛⨯⨯⎪⎭⎫ ⎝⎛=111510334.7104ln 3003750259.0 2786.0=eV(e) ⎪⎪⎭⎫ ⎝⎛=-i o Fi F n n kT E E ln()⎪⎪⎭⎫ ⎝⎛⨯⨯⎪⎭⎫ ⎝⎛=131410722.110029.1ln 3004500259.0 06945.0=eV _______________________________________ 4.59 (a) ⎪⎪⎭⎫ ⎝⎛=-o F p N kT E E υυln()2009.0103100.7ln 0259.01518=⎪⎪⎭⎫ ⎝⎛⨯⨯=eV(b) ()⎪⎪⎭⎫ ⎝⎛⨯⨯=--4181008.1100.7ln 0259.0υE E F 360.1=eV(c) ()⎪⎪⎭⎫⎝⎛⨯⨯=-618108.1100.7ln 0259.0υE E F7508.0=eV(d) ()⎪⎭⎫⎝⎛=-3003750259.0υE E F()()⎥⎥⎦⎤⎢⎢⎣⎡⨯⨯⨯152/318104300375100.7ln 2526.0=eV(e) ()⎪⎭⎫⎝⎛=-3004500259.0υE E F()()⎥⎥⎦⎤⎢⎢⎣⎡⨯⨯⨯72/3181048.1300450100.7ln 068.1=eV_______________________________________ 4.60n-type⎪⎪⎭⎫⎝⎛=-i o Fi F n n kT E E ln()3504.0105.110125.1ln 0259.01016=⎪⎪⎭⎫⎝⎛⨯⨯=eV ______________________________________ 4.612222i aa o n N N p +⎪⎪⎭⎫ ⎝⎛+= 21051008.51515⨯=⨯22152105i n +⎪⎪⎭⎫ ⎝⎛⨯+()21515105.21008.5⨯-⨯()2215105.2i n +⨯=230301025.6106564.6i n +⨯=⨯29210064.4⨯=⇒i n⎥⎦⎤⎢⎣⎡-=kT E N N n g c i exp 2υ()030217.03003500259.0=⎪⎭⎫⎝⎛=kT eV()1921910633.1300350102.1⨯=⎪⎭⎫ ⎝⎛⨯=c N cm 3-()192191045.2300350108.1⨯=⎪⎭⎫ ⎝⎛⨯=υN cm 3- Now()()1919291045.210633.110064.4⨯⨯=⨯⎥⎦⎤⎢⎣⎡-⨯030217.0exp g ESo()()()⎥⎦⎤⎢⎣⎡⨯⨯⨯=29191910064.41045.210633.1ln 030217.0g E 6257.0=⇒g E eV_______________________________________ 4.62 (a) Replace Ga atoms ⇒Silicon acts as adonor()()1415105.310705.0⨯=⨯=d N cm 3-Replace As atoms ⇒Silicon acts as anacceptor()()15151065.610795.0⨯=⨯=a N cm 3-(b) ⇒>d a N N p-type(c) 1415105.31065.6⨯-⨯=-=d a o N N p 15103.6⨯=cm 3-()4152621014.5103.6108.1-⨯=⨯⨯==o i o p n n cm 3- (d)⎪⎪⎭⎫⎝⎛=-i o F Fi n p kT E E ln()5692.0108.1103.6ln 0259.0615=⎪⎪⎭⎫⎝⎛⨯⨯=eV_______________________________________。
半导体物理与器件第四版课后习题标准答案
半导体物理与器件第四版课后习题答案————————————————————————————————作者：————————————————————————————————日期：2______________________________________________________________________________________3Chapter 33.1If o a were to increase, the bandgap energy would decrease and the material would beginto behave less like a semiconductor and morelike a metal. If o a were to decrease, the bandgap energy would increase and thematerial would begin to behave more like an insulator._______________________________________ 3.2Schrodinger's wave equation is:()()()t x x V x t x m ,,2222ψ⋅+∂ψ∂-η()tt x j ∂ψ∂=,ηAssume the solution is of the form:()()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-=ψt E kx j x u t x ηexp , Region I: ()0=x V . Substituting theassumed solution into the wave equation, we obtain:()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-⎩⎨⎧∂∂-t E kx j x jku x m ηηexp 22 ()⎪⎭⎪⎬⎫⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-∂∂+t E kx j x x u ηexp()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-⋅⎪⎭⎫ ⎝⎛-=t E kx j x u jE j ηηηexp which becomes()()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-⎩⎨⎧-t E kx j x u jk m ηηexp 222()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-∂∂+t E kx j x x u jk ηexp 2()⎪⎭⎪⎬⎫⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-∂∂+t E kx j x x u ηexp 22()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-+=t E kx j x Eu ηexp This equation may be written as()()()()0222222=+∂∂+∂∂+-x u mE x x u x x u jk x u k ηSetting ()()x u x u 1= for region I, the equation becomes: ()()()()021221212=--+x u k dx x du jk dxx u d α where222ηmE=αQ.E.D.In Region II, ()O V x V =. Assume the same form of the solution:()()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-=ψt E kx j x u t x ηexp , Substituting into Schrodinger's wave equation, we find:()()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-⎩⎨⎧-t E kx j x u jk m ηηexp 222()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-∂∂+t E kx j x x u jk ηexp 2()⎪⎭⎪⎬⎫⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-∂∂+t E kx j x x u ηexp 22()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-+t E kx j x u V O ηexp()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-=t E kx j x Eu ηexp This equation can be written as:______________________________________________________________________________________4()()()2222xx u x x u jk x u k ∂∂+∂∂+- ()()02222=+-x u mEx u mV O ηη Setting ()()x u x u 2= for region II, this equation becomes()()dx x du jkdx x u d 22222+()022222=⎪⎪⎭⎫ ⎝⎛+--x u mV k O ηα where again222ηmE=αQ.E.D._______________________________________ 3.3We have ()()()()021221212=--+x u k dx x du jk dxx u d α Assume the solution is of the form: ()()[]x k j A x u -=αexp 1 ()[]x k j B +-+αexpThe first derivative is()()()[]x k j A k j dxx du --=ααexp 1 ()()[]x k j B k j +-+-ααexpand the second derivative becomes()()[]()[]x k j A k j dxx u d --=ααexp 2212()[]()[]x k j B k j +-++ααexp 2Substituting these equations into the differential equation, we find()()[]x k j A k ---ααexp 2()()[]x k j B k +-+-ααexp 2(){()[]x k j A k j jk --+ααexp 2()()[]}x k j B k j +-+-ααexp ()()[]{x k j A k ---ααexp 22 ()[]}0exp =+-+x k j B α Combining terms, we obtain()()()[]222222αααα----+--k k k k k ()[]x k j A -⨯αexp()()()[]222222αααα--++++-+k k k k k()[]0exp =+-⨯x k j B α We find that00= Q.E.D.For the differential equation in ()x u 2 and theproposed solution, the procedure is exactly the same as above._______________________________________ 3.4We have the solutions()()[]x k j A x u -=αexp 1()[]x k j B +-+αexp for a x <<0 and()()[]x k j C x u -=βexp 2()[]x k j D +-+βexp for 0<<-x b .The first boundary condition is ()()0021u u =which yields0=--+D C B AThe second boundary condition is201===x x dx dudx du which yields()()()C k B k A k --+--βαα ()0=++D k βThe third boundary condition is ()()b u a u -=21 which yields()[]()[]a k j B a k j A +-+-ααexp exp ()()[]b k j C --=βexp ()()[]b k j D -+-+βexp______________________________________________________________________________________5and can be written as()[]()[]a k j B a k j A +-+-ααexp exp ()[]b k j C ---βexp ()[]0exp =+-b k j D βThe fourth boundary condition isbx a x dx dudx du -===21 which yields()()[]a k j A k j --ααexp()()[]a k j B k j +-+-ααexp()()()[]b k j C k j ---=ββexp()()()[]b k j D k j -+-+-ββexp and can be written as()()[]a k j A k --ααexp()()[]a k j B k +-+-ααexp ()()[]b k j C k ----ββexp()()[]0exp =+++b k j D k ββ_______________________________________ 3.5(b) (i) First point: πα=aSecond point: By trial and error, πα729.1=a (ii) First point: πα2=aSecond point: By trial and error, πα617.2=a _______________________________________ 3.6(b) (i) First point: πα=aSecond point: By trial and error, πα515.1=a (ii) First point: πα2=aSecond point: By trial and error, πα375.2=a _______________________________________ 3.7ka a aaP cos cos sin =+'ααα Let y ka =, x a =α Theny x x xP cos cos sin =+'Consider dydof this function.()[]{}y x x x P dyd sin cos sin 1-=+⋅'- We find()()()⎭⎬⎫⎩⎨⎧⋅+⋅-'--dy dx x x dy dx x x P cos sin 112y dydx x sin sin -=-Theny x x x x x P dy dx sin sin cos sin 12-=⎭⎬⎫⎩⎨⎧-⎥⎦⎤⎢⎣⎡+-' For πn ka y ==,...,2,1,0=n 0sin =⇒y So that, in general,()()dkd ka d a d dy dxαα===0 And22ηmE=α SodkdEm mE dk d ⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛=-22/122221ηηα This implies thatdk dE dk d ==0α for an k π= _______________________________________ 3.8(a) πα=a 1π=⋅a E m o 212η______________________________________________________________________________________6()()()()2103123422221102.41011.9210054.12---⨯⨯⨯==ππa m E o η19104114.3-⨯=J From Problem 3.5 πα729.12=aπ729.1222=⋅a E m o η()()()()2103123422102.41011.9210054.1729.1---⨯⨯⨯=πE18100198.1-⨯=J12E E E -=∆1918104114.3100198.1--⨯-⨯= 19107868.6-⨯=Jor 24.4106.1107868.61919=⨯⨯=∆--E eV(b) πα23=aπ2223=⋅a E m o η()()()()2103123423102.41011.9210054.12---⨯⨯⨯=πE18103646.1-⨯=J From Problem 3.5, πα617.24=aπ617.2224=⋅a E m o η()()()()2103123424102.41011.9210054.1617.2---⨯⨯⨯=πE18103364.2-⨯=J34E E E -=∆1818103646.1103364.2--⨯-⨯= 1910718.9-⨯=Jor 07.6106.110718.91919=⨯⨯=∆--E eV_______________________________________ 3.9(a) At π=ka , πα=a 1π=⋅a E m o 212η()()()()2103123421102.41011.9210054.1---⨯⨯⨯=πE19104114.3-⨯=JAt 0=ka , By trial and error, πα859.0=a o()()()()210312342102.41011.9210054.1859.0---⨯⨯⨯=πo E19105172.2-⨯=J o E E E -=∆11919105172.2104114.3--⨯-⨯= 2010942.8-⨯=Jor 559.0106.110942.81920=⨯⨯=∆--E eV (b) At π2=ka , πα23=aπ2223=⋅a E m o η()()()()2103123423102.41011.9210054.12---⨯⨯⨯=πE18103646.1-⨯=J At π=ka . From Problem 3.5,πα729.12=aπ729.1222=⋅a E m o η()()()()2103123422102.41011.9210054.1729.1---⨯⨯⨯=πE18100198.1-⨯=J 23E E E -=∆1818100198.1103646.1--⨯-⨯= 19104474.3-⨯=Jor 15.2106.1104474.31919=⨯⨯=∆--E eV_____________________________________________________________________________________________________________________________73.10(a) πα=a 1π=⋅a E m o 212η()()()()2103123421102.41011.9210054.1---⨯⨯⨯=πE19104114.3-⨯=JFrom Problem 3.6, πα515.12=aπ515.1222=⋅a E m o η()()()()2103123422102.41011.9210054.1515.1---⨯⨯⨯=πE1910830.7-⨯=J 12E E E -=∆1919104114.310830.7--⨯-⨯= 19104186.4-⨯=Jor 76.2106.1104186.41919=⨯⨯=∆--E eV(b) πα23=aπ2223=⋅a E m o η()()()()2103123423102.41011.9210054.12---⨯⨯⨯=πE18103646.1-⨯=JFrom Problem 3.6, πα375.24=aπ375.2224=⋅a E m o η()()()()2103123424102.41011.9210054.1375.2---⨯⨯⨯=πE18109242.1-⨯=J 34E E E -=∆1818103646.1109242.1--⨯-⨯=1910597.5-⨯=Jor 50.3106.110597.51919=⨯⨯=∆--E eV _____________________________________3.11(a) At π=ka , πα=a 1π=⋅a E m o 212η()()()()2103123421102.41011.9210054.1---⨯⨯⨯=πE19104114.3-⨯=J At 0=ka , By trial and error, πα727.0=a o π727.022=⋅a E m o o η()()()()210312342102.41011.9210054.1727.0---⨯⨯⨯=πo E19108030.1-⨯=J o E E E -=∆11919108030.1104114.3--⨯-⨯= 19106084.1-⨯=Jor 005.1106.1106084.11919=⨯⨯=∆--E eV (b) At π2=ka , πα23=aπ2223=⋅a E m o η()()()()2103123423102.41011.9210054.12---⨯⨯⨯=πE18103646.1-⨯=JAt π=ka , From Problem 3.6, πα515.12=aπ515.1222=⋅a E m o η()()()()2103423422102.41011.9210054.1515.1---⨯⨯⨯=πE1910830.7-⨯=J 23E E E -=∆191810830.7103646.1--⨯-⨯=______________________________________________________________________________________81910816.5-⨯=Jor 635.3106.110816.51919=⨯⨯=∆--E eV_______________________________________ 3.12For 100=T K,()()⇒+⨯-=-1006361001073.4170.124g E164.1=g E eV200=T K, 147.1=g E eV 300=T K, 125.1=g E eV 400=T K, 097.1=g E eV 500=T K, 066.1=g E eV 600=T K, 032.1=g E eV _______________________________________ 3.13The effective mass is given by1222*1-⎪⎪⎭⎫ ⎝⎛⋅=dk E d m ηWe have()()B curve dkEd A curve dk E d 2222>so that ()()B curve m A curve m **<_______________________________________ 3.14The effective mass for a hole is given by1222*1-⎪⎪⎭⎫ ⎝⎛⋅=dk E d m p η We have that()()B curve dkEd A curve dk E d 2222> so that ()()B curve m A curve m p p **<_______________________________________ 3.15Points A,B: ⇒<0dkdEvelocity in -x direction Points C,D: ⇒>0dkdEvelocity in +x directionPoints A,D: ⇒<022dk Ednegative effective massPoints B,C: ⇒>022dkEd positive effective mass_______________________________________ 3.16For A: 2k C E i =At 101008.0+⨯=k m 1-, 05.0=E eV Or()()2119108106.105.0--⨯=⨯=E JSo ()2101211008.0108⨯=⨯-C3811025.1-⨯=⇒CNow ()()38234121025.1210054.12--*⨯⨯==C m η 311044.4-⨯=kgor o m m ⋅⨯⨯=--*31311011.9104437.4 o m m 488.0=*For B: 2k C E i =At 101008.0+⨯=k m 1-, 5.0=E eV Or ()()2019108106.15.0--⨯=⨯=E JSo ()2101201008.0108⨯=⨯-C 3711025.1-⨯=⇒CNow ()()37234121025.1210054.12--*⨯⨯==C m η321044.4-⨯=kgor o m m ⋅⨯⨯=--*31321011.9104437.4o m m 0488.0=*_____________________________________________________________________________________________________________________________93.17For A: 22k C E E -=-υ()()()2102191008.0106.1025.0⨯-=⨯--C3921025.6-⨯=⇒C ()()39234221025.6210054.12--*⨯⨯-=-=C m η31108873.8-⨯-=kgor o m m ⋅⨯⨯-=--*31311011.9108873.8o m m 976.0--=* For B: 22k C E E -=-υ()()()2102191008.0106.13.0⨯-=⨯--C382105.7-⨯=⇒C()()3823422105.7210054.12--*⨯⨯-=-=C m η3210406.7-⨯-=kgor o m m ⋅⨯⨯-=--*31321011.910406.7o m m 0813.0-=*_______________________________________ 3.18(a) (i) νh E =or ()()341910625.6106.142.1--⨯⨯==h E ν1410429.3⨯=Hz(ii) 141010429.3103⨯⨯===νλc E hc 51075.8-⨯=cm 875=nm(b) (i) ()()341910625.6106.112.1--⨯⨯==h E ν 1410705.2⨯=Hz(ii) 141010705.2103⨯⨯==νλc410109.1-⨯=cm 1109=nm _______________________________________ 3.19(c) Curve A: Effective mass is a constantCurve B: Effective mass is positive around 0=k , and is negativearound 2π±=k ._______________________________________ 3.20()[]O O k k E E E --=αcos 1 Then()()()[]O k k E dkdE---=ααsin 1()[]O k k E -+=ααsin 1 and()[]O k k E dkEd -=ααcos 2122Then221222*11ηηαE dk Ed m o k k =⋅== or212*αE m η=_______________________________________ 3.21(a) ()[]3/123/24l t dnm m m =*()()[]3/123/264.1082.04o o m m =o dnm m 56.0=*(b)oo l t cn m m m m m 64.11082.02123+=+=*oo m m 6098.039.24+=o cnm m 12.0=*_______________________________________ 3.22(a) ()()[]3/22/32/3lh hh dp m m m +=*______________________________________________________________________________________10()()[]3/22/32/3082.045.0o o m m +=[]o m ⋅+=3/202348.030187.0o dpm m 473.0=*(b) ()()()()2/12/12/32/3lh hh lh hh cpm m m m m ++=* ()()()()om ⋅++=2/12/12/32/3082.045.0082.045.0 o cpm m 34.0=*_______________________________________3.23For the 3-dimensional infinite potential well, ()0=x V when a x <<0, a y <<0, and a z <<0. In this region, the wave equation is:()()()222222,,,,,,z z y x y z y x x z y x ∂∂+∂∂+∂∂ψψψ()0,,22=+z y x mEψηUse separation of variables technique, so let ()()()()z Z y Y x X z y x =,,ψSubstituting into the wave equation, we have222222zZXY y Y XZ x X YZ ∂∂+∂∂+∂∂ 022=⋅+XYZ mEηDividing by XYZ , we obtain021*********=+∂∂⋅+∂∂⋅+∂∂⋅ηmE z Z Z y Y Y x X X Let01222222=+∂∂⇒-=∂∂⋅X k x X k x X X xx The solution is of the form: ()x k B x k A x X x x cos sin += Since ()0,,=z y x ψ at 0=x , then ()00=Xso that 0=B .Also, ()0,,=z y x ψ at a x =, so that()0=a X . Then πx x n a k = where...,3,2,1=x n Similarly, we have2221y k y Y Y -=∂∂⋅ and 2221z k zZ Z -=∂∂⋅From the boundary conditions, we find πy y n a k = and πz z n a k = where...,3,2,1=y n and ...,3,2,1=z n From the wave equation, we can write022222=+---ηmE k k k z y xThe energy can be written as()222222⎪⎭⎫ ⎝⎛++==a n n n m E E z y x n n n z y x πη _______________________________________ 3.24The total number of quantum states in the 3-dimensional potential well is given (in k-space) by()332a dk k dk k g T ⋅=ππ where222ηmEk =We can then writeηmEk 2=Taking the differential, we obtaindE Em dE E m dk ⋅⋅=⋅⋅⋅⋅=2112121ηηSubstituting these expressions into the densityof states function, we have()dE EmmE a dE E g T ⋅⋅⋅⎪⎭⎫ ⎝⎛=212233ηηππ Noting thatπ2h=ηthis density of states function can be simplified and written as______________________________________________________________________________________()()dE E m h a dE E g T ⋅⋅=2/33324πDividing by 3a will yield the density of states so that()()E hm E g ⋅=32/324π _______________________________________ 3.25For a one-dimensional infinite potential well,222222k an E m n ==*πη Distance between quantum states()()aa n a n k k n n πππ=⎪⎭⎫ ⎝⎛=⎪⎭⎫ ⎝⎛+=-+11Now()⎪⎭⎫ ⎝⎛⋅=a dkdk k g T π2NowE m k n*⋅=21ηdE Em dk n⋅⋅⋅=*2211η Then()dE Em a dE E g n T ⋅⋅⋅=*2212ηπDivide by the "volume" a , so()Em E g n *⋅=21πηSo()()()()()EE g 31341011.9067.0210054.11--⨯⋅⨯=π ()EE g 1810055.1⨯=m 3-J 1-_______________________________________ 3.26(a) Silicon, o nm m 08.1=*()()c nc E E h m E g -=*32/324π()dE E E h m g kTE E c nc c c⋅-=⎰+*232/324π()()kT E E c nc cE E h m 22/332/33224+*-⋅⋅=π()()2/332/323224kT h m n⋅⋅=*π ()()[]()()2/33342/33123210625.61011.908.124kT ⋅⋅⨯⨯=--π ()()2/355210953.7kT ⨯=(i) At 300=T K, 0259.0=kT eV ()()19106.10259.0-⨯=2110144.4-⨯=J Then()()[]2/3215510144.4210953.7-⨯⨯=c g25100.6⨯=m 3- or 19100.6⨯=c g cm 3-(ii) At 400=T K, ()⎪⎭⎫⎝⎛=3004000259.0kT034533.0=eV()()19106.1034533.0-⨯=21105253.5-⨯=J Then()()[]2/32155105253.5210953.7-⨯⨯=c g2510239.9⨯=m 3-or 191024.9⨯=c g cm 3-(b) GaAs, o nm m 067.0=*()()[]()()2/33342/33123210625.61011.9067.024kT g c ⋅⋅⨯⨯=--π ()()2/3542102288.1kT ⨯=______________________________________________________________________________________(i) At 300=T K, 2110144.4-⨯=kT J()()[]2/3215410144.42102288.1-⨯⨯=c g2310272.9⨯=m 3- or 171027.9⨯=c g cm 3-(ii) At 400=T K, 21105253.5-⨯=kT J()()[]2/32154105253.52102288.1-⨯⨯=c g2410427.1⨯=m 3- 181043.1⨯=c g cm 3-_______________________________________ 3.27(a) Silicon, o p m m 56.0=* ()()E E h mE g p-=*υυπ32/324()dE E E h m g E kTE p⋅-=⎰-*υυυυπ332/324()()υυυπE kTE pE E h m 32/332/33224-*-⎪⎭⎫ ⎝⎛-=()()[]2/332/333224kT h mp-⎪⎭⎫ ⎝⎛-=*π ()()[]()()2/33342/33133210625.61011.956.024kT ⎪⎭⎫ ⎝⎛⨯⨯=--π ()()2/355310969.2kT ⨯=(i)At 300=T K, 2110144.4-⨯=kT J ()()[]2/3215510144.4310969.2-⨯⨯=υg2510116.4⨯=m 3- or 191012.4⨯=υg cm 3-(ii)At 400=T K, 21105253.5-⨯=kT J ()()[]2/32155105253.5310969.2-⨯⨯=υg2510337.6⨯=m 3-or 191034.6⨯=υg cm 3- (b) GaAs, o p m m 48.0=*()()[]()()2/33342/33133210625.61011.948.024kT g ⎪⎭⎫ ⎝⎛⨯⨯=--πυ ()()2/3553103564.2kT ⨯=(i)At 300=T K, 2110144.4-⨯=kT J ()()[]2/3215510144.43103564.2-⨯⨯=υg2510266.3⨯=m 3- or 191027.3⨯=υg cm 3-(ii)At 400=T K, 21105253.5-⨯=kT J()()[]2/32155105253.53103564.2-⨯⨯=υg2510029.5⨯=m 3-or 191003.5⨯=υg cm 3-_______________________________________ 3.28(a) ()()c nc E E h m E g -=*32/324π()()[]()c E E -⨯⨯=--3342/33110625.61011.908.124πc E E -⨯=56101929.1 For c E E =; 0=c g1.0+=c E E eV; 4610509.1⨯=c g m 3-J 1-2.0+=c E E eV;4610134.2⨯=m 3-J 1- 3.0+=c E E eV; 4610614.2⨯=m 3-J 1- 4.0+=c E E eV; 4610018.3⨯=m 3-J 1-(b) ()E E hm g p-=*υυπ32/324()()[]()E E -⨯⨯=--υπ3342/33110625.61011.956.024E E -⨯=υ55104541.4______________________________________________________________________________________For υE E =; 0=υg 1.0-=υE E eV; 4510634.5⨯=υg m 3-J 1-2.0-=υE E eV;4510968.7⨯=m 3-J 1-3.0-=υE E eV; 4510758.9⨯=m 3-J 1-4.0-=υE E eV;4610127.1⨯=m 3-J 1-_______________________________________ 3.29(a) ()()68.256.008.12/32/32/3=⎪⎭⎫ ⎝⎛==**pnc m m g g υ(b) ()()0521.048.0067.02/32/32/3=⎪⎭⎫ ⎝⎛==**pncmm g g υ_______________________________________3.30 Plot_______________________________________ 3.31(a) ()()()!710!7!10!!!-=-=i i i i i N g N g W()()()()()()()()()()()()1201238910!3!7!78910===(b) (i) ()()()()()()()()12!10!101112!1012!10!12=-=i W 66= (ii)()()()()()()()()()()()()1234!8!89101112!812!8!12=-=i W 495=_______________________________________ 3.32()⎪⎪⎭⎫ ⎝⎛-+=kT E E E f F exp 11(a) kT E E F =-, ()()⇒+=1exp 11E f ()269.0=E f(b) kT E E F 5=-, ()()⇒+=5exp 11E f()31069.6-⨯=E f (c) kT E E F 10=-, ()()⇒+=10exp 11E f()51054.4-⨯=E f_______________________________________ 3.33()⎪⎪⎭⎫ ⎝⎛-+-=-kT E E E f F exp 1111or()⎪⎪⎭⎫⎝⎛-+=-kT E E E f F exp 111(a) kT E E F =-, ()269.01=-E f (b) kT E E F 5=-, ()31069.61-⨯=-E f (c) kT E E F 10=-, ()51054.41-⨯=-E f_______________________________________ 3.34(a) ()⎥⎦⎤⎢⎣⎡--≅kT E E f F F exp c E E =; 61032.90259.030.0exp -⨯=⎥⎦⎤⎢⎣⎡-=F f 2kT E c +; ()⎥⎦⎤⎢⎣⎡+-=0259.020259.030.0exp F f 61066.5-⨯=kT E c +; ()⎥⎦⎤⎢⎣⎡+-=0259.00259.030.0exp F f 61043.3-⨯=23kT E c +; ()()⎥⎦⎤⎢⎣⎡+-=0259.020259.0330.0exp F f 61008.2-⨯=kT E c 2+; ()()⎥⎦⎤⎢⎣⎡+-=0259.00259.0230.0exp F f 61026.1-⨯=______________________________________________________________________________________(b) ⎥⎦⎤⎢⎣⎡-+-=-kT E E f F F exp 1111()⎥⎦⎤⎢⎣⎡--≅kT E E F exp υE E =; ⎥⎦⎤⎢⎣⎡-=-0259.025.0exp 1F f 51043.6-⨯= 2kT E -υ; ()⎥⎦⎤⎢⎣⎡+-=-0259.020259.025.0exp 1F f 51090.3-⨯=kT E -υ; ()⎥⎦⎤⎢⎣⎡+-=-0259.00259.025.0exp 1F f 51036.2-⨯=23kT E -υ;()()⎥⎦⎤⎢⎣⎡+-=-0259.020259.0325.0exp 1F f 51043.1-⨯= kT E 2-υ;()()⎥⎦⎤⎢⎣⎡+-=-0259.00259.0225.0exp 1F f 61070.8-⨯=_______________________________________ 3.35()()⎥⎦⎤⎢⎣⎡-+-=⎥⎦⎤⎢⎣⎡--=kT E kT E kT E E f F c F F exp exp and()⎥⎦⎤⎢⎣⎡--=-kT E E f F F exp 1()()⎥⎦⎤⎢⎣⎡---=kT kT E E F υexp So ()⎥⎦⎤⎢⎣⎡-+-kT E kT E F c exp()⎥⎦⎤⎢⎣⎡+--=kT kT E E F υexp Then kT E E E kT E F F c +-=-+υOr midgap c F E E E E =+=2υ_______________________________________ 3.3622222man E n πη= For 6=n , Filled state()()()()()2103122234610121011.92610054.1---⨯⨯⨯=πE18105044.1-⨯=Jor 40.9106.1105044.119186=⨯⨯=--E eVFor 7=n , Empty state()()()()()2103122234710121011.92710054.1---⨯⨯⨯=πE1810048.2-⨯=Jor 8.12106.110048.219187=⨯⨯=--E eVTherefore 8.1240.9<<F E eV_______________________________________ 3.37(a) For a 3-D infinite potential well ()222222⎪⎭⎫⎝⎛++=a n n n mE z y x πη For 5 electrons, the 5thelectron occupies the quantum state 1,2,2===z y x n n n ; so()2222252⎪⎭⎫ ⎝⎛++=a n n n m E z y x πη()()()()()21031222223410121011.9212210054.1---⨯⨯++⨯=π 1910761.3-⨯=Jor 35.2106.110761.319195=⨯⨯=--E eV For the next quantum state, which is empty, the quantum state is 2,2,1===z y x n n n . This quantum state is at the same energy, so 35.2=F E eV(b) For 13 electrons, the 13th electronoccupies the quantum state______________________________________________________________________________________3,2,3===z y x n n n ; so ()()()()()2103122222341310121011.9232310054.1---⨯⨯++⨯=πE 1910194.9-⨯=Jor 746.5106.110194.9191913=⨯⨯=--E eVThe 14th electron would occupy the quantum state 3,3,2===z y x n n n . This state is at the same energy, so 746.5=F E eV_______________________________________ 3.38The probability of a state at E E E F ∆+=1 being occupied is()⎪⎭⎫ ⎝⎛∆+=⎪⎪⎭⎫ ⎝⎛-+=kT E kT E E E f F exp 11exp 11111 The probability of a state at E E E F ∆-=2being empty is()⎪⎪⎭⎫ ⎝⎛-+-=-kT E E E f F 222exp 1111⎪⎭⎫ ⎝⎛∆-+⎪⎭⎫ ⎝⎛∆-=⎪⎭⎫ ⎝⎛∆-+-=kT E kT E kT E exp 1exp exp 111 or()⎪⎭⎫ ⎝⎛∆+=-kT E E f exp 11122 so ()()22111E f E f -= Q.E.D._______________________________________ 3.39(a) At energy 1E , we want01.0exp 11exp 11exp 1111=⎪⎪⎭⎫ ⎝⎛-+⎪⎪⎭⎫ ⎝⎛-+-⎪⎪⎭⎫ ⎝⎛-kT E E kT E E kT E E F F FThis expression can be written as01.01exp exp 111=-⎪⎪⎭⎫ ⎝⎛-⎪⎪⎭⎫ ⎝⎛-+kT E E kT E E F For()⎪⎪⎭⎫⎝⎛-=kT E E F 1exp 01.01Then()100ln 1kT E E F += orkT E E F 6.41+= (b)At kT E E F 6.4+=,()()6.4exp 11exp 1111+=⎪⎪⎭⎫ ⎝⎛-+=kT E E E f F which yields()01.000990.01≅=E f_______________________________________ 3.40 (a)()()⎥⎦⎤⎢⎣⎡--=⎥⎦⎤⎢⎣⎡--=0259.050.580.5exp exp kT E E f F F 61032.9-⨯=(b) ()060433.03007000259.0=⎪⎭⎫⎝⎛=kT eV31098.6060433.030.0exp -⨯=⎥⎦⎤⎢⎣⎡-=F f (c) ()⎥⎦⎤⎢⎣⎡--≅-kT E E f F F exp 1 ⎥⎦⎤⎢⎣⎡-=kT 25.0exp 02.0______________________________________________________________________________________or 5002.0125.0exp ==⎥⎦⎤⎢⎣⎡+kT ()50ln 25.0=kTor()()⎪⎭⎫⎝⎛===3000259.0063906.050ln 25.0T kTwhich yields 740=T K_______________________________________ 3.41 (a)()00304.00259.00.715.7exp 11=⎪⎭⎫ ⎝⎛-+=E for 0.304%(b) At 1000=T K, 08633.0=kT eV Then()1496.008633.00.715.7exp 11=⎪⎭⎫ ⎝⎛-+=E for 14.96%(c) ()997.00259.00.785.6exp 11=⎪⎭⎫⎝⎛-+=E for 99.7% (d)At F E E =, ()21=E f for alltemperatures_______________________________________ 3.42(a) For 1E E =()()⎥⎦⎤⎢⎣⎡--≅⎪⎪⎭⎫ ⎝⎛-+=kT E E kTE E E fF F11exp exp 11Then()611032.90259.030.0exp -⨯=⎪⎭⎫ ⎝⎛-=E fFor 2E E =,82.030.012.12=-=-E E F eV Then()⎪⎭⎫ ⎝⎛-+-=-0259.082.0exp 1111E for()⎥⎦⎤⎢⎣⎡⎪⎭⎫ ⎝⎛---≅-0259.082.0exp 111E f141078.10259.082.0exp -⨯=⎪⎭⎫ ⎝⎛-=(b) For 4.02=-E E F eV, 72.01=-F E E eV At 1E E =,()()⎪⎭⎫⎝⎛-=⎥⎦⎤⎢⎣⎡--=0259.072.0exp exp 1kT E E E f F or()131045.8-⨯=E f At 2E E =,()()⎥⎦⎤⎢⎣⎡--=-kT E E E f F 2exp 1⎪⎭⎫ ⎝⎛-=0259.04.0expor()71096.11-⨯=-E f_______________________________________ 3.43(a) At 1E E =()()⎪⎭⎫⎝⎛-=⎥⎦⎤⎢⎣⎡--=0259.030.0exp exp 1kT E E E f F or()61032.9-⨯=E f At 2E E =, 12.13.042.12=-=-E E F eV So()()⎥⎦⎤⎢⎣⎡--=-kT E E E f F 2exp 1⎪⎭⎫ ⎝⎛-=0259.012.1exp______________________________________________________________________________________or()191066.11-⨯=-E f(b) For 4.02=-E E F ,02.11=-F E E eV At 1E E =,()()⎪⎭⎫⎝⎛-=⎥⎦⎤⎢⎣⎡--=0259.002.1exp exp 1kT E E E f F or()181088.7-⨯=E f At 2E E =,()()⎥⎦⎤⎢⎣⎡--=-kT E E E f F 2exp 1⎪⎭⎫ ⎝⎛-=0259.04.0expor ()71096.11-⨯=-E f_______________________________________ 3.44()1exp 1-⎥⎦⎤⎢⎣⎡⎪⎪⎭⎫ ⎝⎛-+=kTE E E f Fso()()2exp 11-⎥⎦⎤⎢⎣⎡⎪⎪⎭⎫ ⎝⎛-+-=kT E E dE E df F⎪⎪⎭⎫ ⎝⎛-⎪⎭⎫⎝⎛⨯kT E E kT F exp 1 or()2exp 1exp 1⎥⎦⎤⎢⎣⎡⎪⎪⎭⎫ ⎝⎛-+⎪⎪⎭⎫ ⎝⎛-⎪⎭⎫⎝⎛-=kT E E kT E E kT dE E df F F (a) At 0=T K, For()00exp =⇒=∞-⇒<dE dfE E F()0exp =⇒+∞=∞+⇒>dEdfE E FAt -∞=⇒=dEdfE E F(b) At 300=T K, 0259.0=kT eVFor F E E <<,0=dE dfFor F E E >>, 0=dEdfAt F E E =,()()65.91110259.012-=+⎪⎭⎫ ⎝⎛-=dE df (eV)1-(c) At 500=T K, 04317.0=kT eVFor F E E <<, 0=dE dfFor F E E >>, 0=dEdfAt F E E =, ()()79.511104317.012-=+⎪⎭⎫ ⎝⎛-=dE df (eV)1- _______________________________________ 3.45(a) At midgap E E =,()⎪⎪⎭⎫ ⎝⎛+=⎪⎪⎭⎫ ⎝⎛-+=kTE kTE E E f gF2exp 11exp 11Si: 12.1=g E eV, ()()⎥⎦⎤⎢⎣⎡+=0259.0212.1exp 11E for()101007.4-⨯=E fGe: 66.0=g E eV______________________________________________________________________________________()()⎥⎦⎤⎢⎣⎡+=0259.0266.0exp 11E for()61093.2-⨯=E f GaAs: 42.1=g E eV ()()⎥⎦⎤⎢⎣⎡+=0259.0242.1exp 11E for()121024.1-⨯=E f(b) Using the results of Problem 3.38, the answers to part (b) are exactly the same as those given in part (a)._______________________________________ 3.46(a) ()⎥⎦⎤⎢⎣⎡--=kT E E f F F exp ⎥⎦⎤⎢⎣⎡-=-kT 60.0exp 108 or ()810ln 60.0+=kT()032572.010ln 60.08==kT eV()⎪⎭⎫⎝⎛=3000259.0032572.0Tso 377=T K(b) ⎥⎦⎤⎢⎣⎡-=-kT 60.0exp 106()610ln 60.0+=kT()043429.010ln 60.06==kT ()⎪⎭⎫⎝⎛=3000259.0043429.0Tor 503=T K_______________________________________ 3.47(a) At 200=T K,()017267.03002000259.0=⎪⎭⎫⎝⎛=kT eV⎪⎪⎭⎫ ⎝⎛-+==kTE E f FF exp 1105.019105.01exp =-=⎪⎪⎭⎫⎝⎛-kT E E F()()()19ln 017267.019ln ==-kT E E F 05084.0=eV By symmetry, for 95.0=F f , 05084.0-=-F E E eVThen ()1017.005084.02==∆E eV (b) 400=T K, 034533.0=kT eV For 05.0=F f , from part (a),()()()19ln 034533.019ln ==-kT E E F 10168.0=eV Then ()2034.010168.02==∆E eV_______________________________________。
半导体物理与器件第四课后习题答案3.doc
Chapter 33.1If o a were to increase, the bandgap energy would decrease and the material would begin to behave less like a semiconductor and more like a metal. If o a were to decrease, the bandgap energy would increase and thematerial would begin to behave more like an insulator._______________________________________ 3.2Schrodinger's wave equation is:()()()t x x V xt x m ,,2222ψ⋅+∂ψ∂- ()tt x j ∂ψ∂=, Assume the solution is of the form:()()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-=ψt E kx j x u t x exp , Region I: ()0=x V . Substituting theassumed solution into the wave equation, we obtain:()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-⎩⎨⎧∂∂-t E kx j x jku x m exp 22 ()⎪⎭⎪⎬⎫⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-∂∂+t E kx j x x u exp ()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-⋅⎪⎭⎫ ⎝⎛-=t E kx j x u jE j exp which becomes()()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-⎩⎨⎧-t E kx j x u jk m exp 222 ()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-∂∂+t E kx j x x u jkexp 2 ()⎪⎭⎪⎬⎫⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-∂∂+t E kx j x x u exp 22 ()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-+=t E kx j x Eu exp This equation may be written as()()()()0222222=+∂∂+∂∂+-x u mE x x u x x u jk x u kSetting ()()x u x u 1= for region I, the equation becomes:()()()()021221212=--+x u k dx x du jk dxx u d α where222mE=αIn Region II, ()O V x V =. Assume the same form of the solution:()()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-=ψt E kx j x u t x exp , Substituting into Schrodinger's wave equation, we find:()()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-⎩⎨⎧-t E kx j x u jk m exp 222 ()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-∂∂+t E kx j x x u jkexp 2 ()⎪⎭⎪⎬⎫⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-∂∂+t E kx j x x u exp 22 ()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-+t E kx j x u V O exp ()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-=t E kx j x Eu exp This equation can be written as:()()()2222x x u x x u jk x u k ∂∂+∂∂+- ()()02222=+-x u mEx u mV OSetting ()()x u x u 2= for region II, this equation becomes()()dx x du jk dxx u d 22222+ ()022222=⎪⎪⎭⎫ ⎝⎛+--x u mV k O α where again222mE=α_______________________________________3.3We have()()()()021221212=--+x u k dx x du jk dxx u d α Assume the solution is of the form: ()()[]x k j A x u -=αexp 1()[]x k j B +-+αexp The first derivative is()()()[]x k j A k j dxx du --=ααexp 1 ()()[]x k j B k j +-+-ααexp and the second derivative becomes()()[]()[]x k j A k j dxx u d --=ααexp 2212 ()[]()[]x k j B k j +-++ααexp 2Substituting these equations into the differential equation, we find()()[]x k j A k ---ααexp 2()()[]x k j B k +-+-ααexp 2(){()[]x k j A k j jk --+ααexp 2()()[]}x k j B k j +-+-ααexp ()()[]{x k j A k ---ααexp 22 ()[]}0exp =+-+x k j B α Combining terms, we obtain()()()[]222222αααα----+--k k k k k ()[]x k j A -⨯αexp()()()[]222222αααα--++++-+k k k k k ()[]0exp =+-⨯x k j B α We find that 00=For the differential equation in ()x u 2 and the proposed solution, the procedure is exactly the same as above._______________________________________ 3.4We have the solutions ()()[]x k j A x u -=αexp 1()[]x k j B +-+αexp for a x <<0 and()()[]x k j C x u -=βexp 2()[]x k j D +-+βexp for 0<<-x b .The first boundary condition is ()()0021u u =which yields0=--+D C B AThe second boundary condition is201===x x dx dudx du which yields()()()C k B k A k --+--βαα()0=++D k β The third boundary condition is ()()b u a u -=21 which yields()[]()[]a k j B a k j A +-+-ααexp exp ()()[]b k j C --=βexp()()[]b k j D -+-+βexp and can be written as()[]()[]a k j B a k j A +-+-ααexp exp ()[]b k j C ---βexp()[]0exp =+-b k j D β The fourth boundary condition isbx a x dx dudx du -===21 which yields()()[]a k j A k j --ααexp()()[]a k j B k j +-+-ααexp ()()()[]b k j C k j ---=ββexp()()()[]b k j D k j -+-+-ββexp and can be written as ()()[]a k j A k --ααexp()()[]a k j B k +-+-ααexp()()[]b k j C k ----ββexp()()[]0exp =+++b k j D k ββ_______________________________________ 3.5(b) (i) First point: πα=aSecond point: By trial and error, πα729.1=a (ii) First point: πα2=aSecond point: By trial and error, πα617.2=a_______________________________________3.6(b) (i) First point: πα=aSecond point: By trial and error, πα515.1=a (ii) First point: πα2=aSecond point: By trial and error, πα375.2=a_______________________________________ 3.7ka a aaP cos cos sin =+'αααLet y ka =, x a =α Theny x x xP cos cos sin =+'Consider dy dof this function.()[]{}y x x x P dyd sin cos sin 1-=+⋅'- We find()()()⎭⎬⎫⎩⎨⎧⋅+⋅-'--dy dx x x dy dx x x P cos sin 112y dydxx sin sin -=- Theny x x x x x P dy dx sin sin cos sin 12-=⎭⎬⎫⎩⎨⎧-⎥⎦⎤⎢⎣⎡+-'For πn ka y ==, ...,2,1,0=n 0sin =⇒y So that, in general,()()dk d ka d a d dy dxαα===0 And22 mE=αSodk dEm mE dk d ⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛=-22/122221 α This implies thatdk dE dk d ==0α for an k π= _______________________________________ 3.8(a) πα=a 1π=⋅a E m o 212()()()()2103123422221102.41011.9210054.12---⨯⨯⨯==ππa m E o19104114.3-⨯=J From Problem 3.5 πα729.12=aπ729.1222=⋅a E m o()()()()2103123422102.41011.9210054.1729.1---⨯⨯⨯=πE18100198.1-⨯=J 12E E E -=∆1918104114.3100198.1--⨯-⨯= 19107868.6-⨯=Jor 24.4106.1107868.61919=⨯⨯=∆--E eV(b) πα23=aπ2223=⋅a E m o()()()()2103123423102.41011.9210054.12---⨯⨯⨯=πE18103646.1-⨯=J From Problem 3.5, πα617.24=aπ617.2224=⋅a E m o()()()()2103123424102.41011.9210054.1617.2---⨯⨯⨯=πE18103364.2-⨯=J 34E E E -=∆1818103646.1103364.2--⨯-⨯= 1910718.9-⨯=Jor 07.6106.110718.91919=⨯⨯=∆--E eV_______________________________________3.9(a) At π=ka , πα=a 1π=⋅a E m o 212()()()()2103123421102.41011.9210054.1---⨯⨯⨯=πE19104114.3-⨯=JAt 0=ka , By trial and error, πα859.0=a o ()()()()210312342102.41011.9210054.1859.0---⨯⨯⨯=πoE19105172.2-⨯=J o E E E -=∆11919105172.2104114.3--⨯-⨯= 2010942.8-⨯=Jor 559.0106.110942.81920=⨯⨯=∆--E eV (b) At π2=ka , πα23=aπ2223=⋅a E m o()()()()2103123423102.41011.9210054.12---⨯⨯⨯=πE18103646.1-⨯=JAt π=ka . From Problem 3.5, πα729.12=aπ729.1222=⋅a E m o()()()()2103123422102.41011.9210054.1729.1---⨯⨯⨯=πE18100198.1-⨯=J23E E E -=∆1818100198.1103646.1--⨯-⨯= 19104474.3-⨯=Jor 15.2106.1104474.31919=⨯⨯=∆--E eV_______________________________________3.10(a) πα=a 1π=⋅a E m o 212()()()()2103123421102.41011.9210054.1---⨯⨯⨯=πE19104114.3-⨯=JFrom Problem 3.6, πα515.12=aπ515.1222=⋅a E m o()()()()2103123422102.41011.9210054.1515.1---⨯⨯⨯=πE1910830.7-⨯=J 12E E E -=∆1919104114.310830.7--⨯-⨯= 19104186.4-⨯=Jor 76.2106.1104186.41919=⨯⨯=∆--E eV (b) πα23=aπ2223=⋅a E m o()()()()2103123423102.41011.9210054.12---⨯⨯⨯=πE18103646.1-⨯=JFrom Problem 3.6, πα375.24=aπ375.2224=⋅a E m o()()()()2103123424102.41011.9210054.1375.2---⨯⨯⨯=πE18109242.1-⨯=J 34E E E -=∆1818103646.1109242.1--⨯-⨯= 1910597.5-⨯=Jor 50.3106.110597.51919=⨯⨯=∆--E eV_____________________________________3.11(a) At π=ka , πα=a 1π=⋅a E m o 212()()()()2103123421102.41011.9210054.1---⨯⨯⨯=πE19104114.3-⨯=JAt 0=ka , By trial and error, πα727.0=a oπ727.022=⋅a E m o o()()()()210312342102.41011.9210054.1727.0---⨯⨯⨯=πo E19108030.1-⨯=Jo E E E -=∆11919108030.1104114.3--⨯-⨯= 19106084.1-⨯=Jor 005.1106.1106084.11919=⨯⨯=∆--E eV (b) At π2=ka , πα23=aπ2223=⋅a E m o()()()()2103123423102.41011.9210054.12---⨯⨯⨯=πE18103646.1-⨯=JAt π=ka , From Problem 3.6,πα515.12=aπ515.1222=⋅a E m o()()()()2103423422102.41011.9210054.1515.1---⨯⨯⨯=πE1910830.7-⨯=J23E E E -=∆191810830.7103646.1--⨯-⨯= 1910816.5-⨯=Jor 635.3106.110816.51919=⨯⨯=∆--E eV_______________________________________3.12For 100=T K, ()()⇒+⨯-=-1006361001073.4170.124gE164.1=g E eV200=T K, 147.1=g E eV 300=T K, 125.1=g E eV 400=T K, 097.1=g E eV 500=T K, 066.1=g E eV 600=T K, 032.1=g E eV_______________________________________3.13The effective mass is given by1222*1-⎪⎪⎭⎫⎝⎛⋅=dk E d mWe have()()B curve dkE d A curve dk E d 2222> so that ()()B curve m A curve m **<_______________________________________ 3.14The effective mass for a hole is given by1222*1-⎪⎪⎭⎫ ⎝⎛⋅=dk E d m p We have that()()B curve dkEd A curve dk E d 2222> so that ()()B curve m A curve m p p **<_______________________________________ 3.15Points A,B: ⇒<0dk dEvelocity in -x directionPoints C,D: ⇒>0dk dEvelocity in +x directionPoints A,D: ⇒<022dk Ednegative effective massPoints B,C: ⇒>022dkEd positive effective mass _______________________________________3.16For A: 2k C E i =At 101008.0+⨯=k m 1-, 05.0=E eV Or ()()2119108106.105.0--⨯=⨯=E J So ()2101211008.0108⨯=⨯-C3811025.1-⨯=⇒CNow ()()38234121025.1210054.12--*⨯⨯==C m 311044.4-⨯=kgor o m m ⋅⨯⨯=--*31311011.9104437.4o m m 488.0=* For B: 2k C E i =At 101008.0+⨯=k m 1-, 5.0=E eV Or ()()2019108106.15.0--⨯=⨯=E JSo ()2101201008.0108⨯=⨯-C 3711025.1-⨯=⇒CNow ()()37234121025.1210054.12--*⨯⨯==C m 321044.4-⨯=kg or o m m ⋅⨯⨯=--*31321011.9104437.4o m m 0488.0=*_______________________________________ 3.17For A: 22k C E E -=-υ()()()2102191008.0106.1025.0⨯-=⨯--C 3921025.6-⨯=⇒C()()39234221025.6210054.12--*⨯⨯-=-=C m31108873.8-⨯-=kgor o m m ⋅⨯⨯-=--*31311011.9108873.8o m m 976.0--=* For B: 22k C E E -=-υ()()()2102191008.0106.13.0⨯-=⨯--C 382105.7-⨯=⇒C()()3823422105.7210054.12--*⨯⨯-=-=C m3210406.7-⨯-=kgor o m m ⋅⨯⨯-=--*31321011.910406.7o m m 0813.0-=*_______________________________________ 3.18(a) (i) νh E =or ()()341910625.6106.142.1--⨯⨯==h E ν1410429.3⨯=Hz(ii) 141010429.3103⨯⨯===νλc E hc 51075.8-⨯=cm 875=nm(b) (i) ()()341910625.6106.112.1--⨯⨯==h E ν1410705.2⨯=Hz(ii) 141010705.2103⨯⨯==νλc410109.1-⨯=cm 1109=nm_______________________________________ 3.19(c) Curve A: Effective mass is a constantCurve B: Effective mass is positive around 0=k , and is negativearound 2π±=k . _______________________________________ 3.20()[]O O k k E E E --=αcos 1 Then()()()[]O k k E dkdE ---=ααsin 1()[]O k k E -+=ααsin 1 and()[]O k k E dk E d -=ααcos 2122Then221222*11 αE dk Ed m o k k =⋅== or212*αE m =_______________________________________ 3.21(a) ()[]3/123/24lt dn m m m =*()()[]3/123/264.1082.04oom m =o dn m m 56.0=*(b)o o l t cnm m m m m 64.11082.02123+=+=*oo m m 6098.039.24+=o cn m m 12.0=*_______________________________________ 3.22(a) ()()[]3/22/32/3lh hh dp m m m +=*()()[]3/22/32/3082.045.0o om m +=[]o m ⋅+=3/202348.030187.0o dp m m 473.0=*(b) ()()()()2/12/12/32/3lh hh lh hh cpm m m m m ++=*()()()()om ⋅++=2/12/12/32/3082.045.0082.045.0 o cp m m 34.0=*_______________________________________ 3.23For the 3-dimensional infinite potential well, ()0=x V when a x <<0, a y <<0, and a z <<0. In this region, the wave equation is:()()()222222,,,,,,z z y x y z y x x z y x ∂∂+∂∂+∂∂ψψψ()0,,22=+z y x mEψ Use separation of variables technique, so let ()()()()z Z y Y x X z y x =,,ψSubstituting into the wave equation, we have222222zZXY y Y XZ x X YZ ∂∂+∂∂+∂∂ 022=⋅+XYZ mEDividing by XYZ , we obtain021*********=+∂∂⋅+∂∂⋅+∂∂⋅ mEz Z Z y Y Y x X XLet01222222=+∂∂⇒-=∂∂⋅X k x X k x X X xx The solution is of the form: ()x k B x k A x X x x cos sin +=Since ()0,,=z y x ψ at 0=x , then ()00=X so that 0=B .Also, ()0,,=z y x ψ at a x =, so that ()0=a X . Then πx x n a k = where ...,3,2,1=x n Similarly, we have2221y k y Y Y -=∂∂⋅ and 2221z k zZ Z -=∂∂⋅From the boundary conditions, we find πy y n a k = and πz z n a k =where...,3,2,1=y n and ...,3,2,1=z n From the wave equation, we can write022222=+---mE k k k z y xThe energy can be written as()222222⎪⎭⎫⎝⎛++==a n n n m E E z y x n n n z y x π _______________________________________ 3.24The total number of quantum states in the 3-dimensional potential well is given (in k-space) by()332a dk k dk k g T ⋅=ππ where222 mEk =We can then writemEk 2=Taking the differential, we obtaindE Em dE E m dk ⋅⋅=⋅⋅⋅⋅=2112121 Substituting these expressions into the density of states function, we have()dE E mmE a dE E g T ⋅⋅⋅⎪⎭⎫ ⎝⎛=212233 ππ Noting thatπ2h=this density of states function can be simplified and written as()()dE E m h a dE E g T ⋅⋅=2/33324π Dividing by 3a will yield the density of states so that()()E h m E g ⋅=32/324π _______________________________________ 3.25For a one-dimensional infinite potential well,222222k a n E m n ==*π Distance between quantum states()()aa n a n k k n n πππ=⎪⎭⎫ ⎝⎛=⎪⎭⎫ ⎝⎛+=-+11Now()⎪⎭⎫ ⎝⎛⋅=a dkdk k g T π2NowE m k n *⋅=21dE Em dk n⋅⋅⋅=*2211 Then()dE Em a dE E g n T ⋅⋅⋅=*2212 π Divide by the "volume" a , so ()Em E g n *⋅=21πSo()()()()()EE g 31341011.9067.0210054.11--⨯⋅⨯=π ()EE g 1810055.1⨯=m 3-J 1-_______________________________________ 3.26(a) Silicon, o n m m 08.1=*()()c nc E E h m E g -=*32/324π()dE E E h m g kTE E c nc c c⋅-=⎰+*232/324π()()kT E E c nc cE E h m 22/332/33224+*-⋅⋅=π()()2/332/323224kT hm n⋅⋅=*π ()()[]()()2/33342/33123210625.61011.908.124kT ⋅⋅⨯⨯=--π ()()2/355210953.7kT ⨯=(i) At 300=T K, 0259.0=kT eV()()19106.10259.0-⨯= 2110144.4-⨯=J Then ()()[]2/3215510144.4210953.7-⨯⨯=c g25100.6⨯=m 3-or 19100.6⨯=c g cm 3-(ii) At 400=T K, ()⎪⎭⎫⎝⎛=3004000259.0kT034533.0=eV()()19106.1034533.0-⨯= 21105253.5-⨯=J Then()()[]2/32155105253.5210953.7-⨯⨯=c g2510239.9⨯=m 3- or 191024.9⨯=c g cm 3-(b) GaAs, o nm m 067.0=*()()[]()()2/33342/33123210625.61011.9067.024kT g c ⋅⋅⨯⨯=--π ()()2/3542102288.1kT ⨯=(i) At 300=T K, 2110144.4-⨯=kT J ()()[]2/3215410144.42102288.1-⨯⨯=c g2310272.9⨯=m 3- or 171027.9⨯=c g cm 3-(ii) At 400=T K, 21105253.5-⨯=kT J ()()[]2/32154105253.52102288.1-⨯⨯=c g2410427.1⨯=m 3-181043.1⨯=c g cm 3-_______________________________________ 3.27(a) Silicon, o p m m 56.0=* ()()E E h mE g p-=*υυπ32/324()dE E E h mg E kTE p⋅-=⎰-*υυυυπ332/324()()υυυπE kTE pE E hm 32/332/33224-*-⎪⎭⎫ ⎝⎛-=()()[]2/332/333224kT hmp-⎪⎭⎫ ⎝⎛-=*π ()()[]()()2/33342/33133210625.61011.956.024kT ⎪⎭⎫ ⎝⎛⨯⨯=--π ()()2/355310969.2kT ⨯=(i)At 300=T K, 2110144.4-⨯=kT J ()()[]2/3215510144.4310969.2-⨯⨯=υg2510116.4⨯=m3-or 191012.4⨯=υg cm 3- (ii)At 400=T K, 21105253.5-⨯=kT J()()[]2/32155105253.5310969.2-⨯⨯=υg2510337.6⨯=m3-or 191034.6⨯=υg cm 3- (b) GaAs, o p m m 48.0=*()()[]()()2/33342/33133210625.61011.948.024kT g ⎪⎭⎫ ⎝⎛⨯⨯=--πυ ()()2/3553103564.2kT ⨯=(i)At 300=T K, 2110144.4-⨯=kT J()()[]2/3215510144.43103564.2-⨯⨯=υg2510266.3⨯=m 3- or 191027.3⨯=υg cm 3-(ii)At 400=T K, 21105253.5-⨯=kT J()()[]2/32155105253.53103564.2-⨯⨯=υg2510029.5⨯=m 3-or 191003.5⨯=υg cm 3-_______________________________________ 3.28(a) ()()c nc E E h m E g -=*32/324π()()[]()c E E -⨯⨯=--3342/33110625.61011.908.124πc E E -⨯=56101929.1 For c E E =; 0=c g1.0+=c E E eV; 4610509.1⨯=c g m 3-J 1-2.0+=c E E eV; 4610134.2⨯=m 3-J 1-3.0+=c E E eV; 4610614.2⨯=m 3-J 1- 4.0+=c E E eV; 4610018.3⨯=m 3-J 1- (b) ()E E h m g p-=*υυπ32/324()()[]()E E -⨯⨯=--υπ3342/33110625.61011.956.024E E -⨯=υ55104541.4 For υE E =; 0=υg1.0-=υE E eV; 4510634.5⨯=υg m 3-J 1-2.0-=υE E eV; 4510968.7⨯=m 3-J 1-3.0-=υE E eV; 4510758.9⨯=m 3-J 1-4.0-=υE E eV; 4610127.1⨯=m 3-J 1-_______________________________________ 3.29(a) ()()68.256.008.12/32/32/3=⎪⎭⎫ ⎝⎛==**pnc m m g g υ(b) ()()0521.048.0067.02/32/32/3=⎪⎭⎫ ⎝⎛==**pncmm g g υ_______________________________________3.30 Plot_______________________________________ 3.31(a) ()()()!710!7!10!!!-=-=i i i i i N g N g W()()()()()()()()()()()()1201238910!3!7!78910===(b) (i) ()()()()()()()()12!10!101112!1012!10!12=-=i W 66=(ii) ()()()()()()()()()()()()1234!8!89101112!812!8!12=-=i W 495=_______________________________________ 3.32()⎪⎪⎭⎫ ⎝⎛-+=kT E E E f F exp 11(a) kT E E F =-, ()()⇒+=1exp 11E f()269.0=E f (b) kT E E F 5=-, ()()⇒+=5exp 11E f()31069.6-⨯=E f(c) kT E E F 10=-, ()()⇒+=10exp 11E f ()51054.4-⨯=E f_______________________________________ 3.33()⎪⎪⎭⎫ ⎝⎛-+-=-kT E E E f F exp 1111or()⎪⎪⎭⎫ ⎝⎛-+=-kT E E E f F exp 111(a) kT E E F =-, ()269.01=-E f (b) kT E E F 5=-, ()31069.61-⨯=-E f(c) kT E E F 10=-, ()51054.41-⨯=-E f_______________________________________ 3.34(a) ()⎥⎦⎤⎢⎣⎡--≅kT E E f F F exp c E E =; 61032.90259.030.0exp -⨯=⎥⎦⎤⎢⎣⎡-=F f 2kT E c +; ()⎥⎦⎤⎢⎣⎡+-=0259.020259.030.0exp F f 61066.5-⨯=kT E c +; ()⎥⎦⎤⎢⎣⎡+-=0259.00259.030.0exp F f 61043.3-⨯=23kT E c +; ()()⎥⎦⎤⎢⎣⎡+-=0259.020259.0330.0exp F f 61008.2-⨯=kT E c 2+; ()()⎥⎦⎤⎢⎣⎡+-=0259.00259.0230.0exp F f 61026.1-⨯=(b) ⎥⎦⎤⎢⎣⎡-+-=-kT E E f F F exp 1111()⎥⎦⎤⎢⎣⎡--≅kT E E F exp υE E =; ⎥⎦⎤⎢⎣⎡-=-0259.025.0exp 1F f 51043.6-⨯= 2kT E -υ; ()⎥⎦⎤⎢⎣⎡+-=-0259.020259.025.0exp 1F f 51090.3-⨯=kT E -υ; ()⎥⎦⎤⎢⎣⎡+-=-0259.00259.025.0exp 1F f 51036.2-⨯=23kTE -υ; ()()⎥⎦⎤⎢⎣⎡+-=-0259.020259.0325.0exp 1F f 51043.1-⨯= kT E 2-υ;()()⎥⎦⎤⎢⎣⎡+-=-0259.00259.0225.0exp 1F f 61070.8-⨯=_______________________________________3.35()()⎥⎦⎤⎢⎣⎡-+-=⎥⎦⎤⎢⎣⎡--=kT E kT E kT E E f F c F F exp exp and()⎥⎦⎤⎢⎣⎡--=-kT E E f F F exp 1 ()()⎥⎦⎤⎢⎣⎡---=kT kT E E F υexp So ()⎥⎦⎤⎢⎣⎡-+-kT E kT E F c exp ()⎥⎦⎤⎢⎣⎡+--=kT kT E E F υexp Then kT E E E kT E F F c +-=-+υOr midgap c F E E E E =+=2υ_______________________________________ 3.3622222ma n E n π =For 6=n , Filled state()()()()()2103122234610121011.92610054.1---⨯⨯⨯=πE18105044.1-⨯=Jor 40.9106.1105044.119186=⨯⨯=--E eV For 7=n , Empty state ()()()()()2103122234710121011.92710054.1---⨯⨯⨯=πE1810048.2-⨯=Jor 8.12106.110048.219187=⨯⨯=--E eV Therefore 8.1240.9<<F E eV_______________________________________ 3.37(a) For a 3-D infinite potential well()222222⎪⎭⎫ ⎝⎛++=a n n n mE z y x π For 5 electrons, the 5th electron occupies the quantum state 1,2,2===z y x n n n ; so()2222252⎪⎭⎫ ⎝⎛++=a n n n m E z y x π()()()()()21031222223410121011.9212210054.1---⨯⨯++⨯=π1910761.3-⨯=Jor 35.2106.110761.319195=⨯⨯=--E eV For the next quantum state, which is empty, the quantum state is 2,2,1===z y x n n n . This quantum state is at the same energy, so 35.2=F E eV(b) For 13 electrons, the 13th electronoccupies the quantum state 3,2,3===z y x n n n ; so ()()()()()2103122222341310121011.9232310054.1---⨯⨯++⨯=πE 1910194.9-⨯=Jor 746.5106.110194.9191913=⨯⨯=--E eVThe 14th electron would occupy the quantum state 3,3,2===z y x n n n . This state is at the same energy, so 746.5=F E eV_______________________________________ 3.38The probability of a state at E E E F ∆+=1 being occupied is()⎪⎭⎫ ⎝⎛∆+=⎪⎪⎭⎫ ⎝⎛-+=kT E kT E E E f F exp 11exp 11111 The probability of a state at E E E F ∆-=2being empty is()⎪⎪⎭⎫ ⎝⎛-+-=-kT E E E f F 222exp 1111⎪⎭⎫ ⎝⎛∆-+⎪⎭⎫ ⎝⎛∆-=⎪⎭⎫ ⎝⎛∆-+-=kT E kT E kT E exp 1exp exp 111or()⎪⎭⎫ ⎝⎛∆+=-kT E E f exp 11122so ()()22111E f E f -=_______________________________________3.39(a) At energy 1E , we want01.0exp 11exp 11exp 1111=⎪⎪⎭⎫ ⎝⎛-+⎪⎪⎭⎫ ⎝⎛-+-⎪⎪⎭⎫ ⎝⎛-kT E E kT E E kT E E F F FThis expression can be written as01.01exp exp 111=-⎪⎪⎭⎫ ⎝⎛-⎪⎪⎭⎫ ⎝⎛-+kT E E kT E E F F or()⎪⎪⎭⎫⎝⎛-=kT E E F 1exp 01.01Then()100ln 1kT E E F += orkT E E F 6.41+= (b)At kT E E F 6.4+=, ()()6.4exp 11exp 1111+=⎪⎪⎭⎫ ⎝⎛-+=kT E E E f F which yields()01.000990.01≅=E f_______________________________________ 3.40 (a)()()⎥⎦⎤⎢⎣⎡--=⎥⎦⎤⎢⎣⎡--=0259.050.580.5exp exp kT E E f F F 61032.9-⨯=(b) ()060433.03007000259.0=⎪⎭⎫⎝⎛=kT eV31098.6060433.030.0exp -⨯=⎥⎦⎤⎢⎣⎡-=F f (c) ()⎥⎦⎤⎢⎣⎡--≅-kT E E f F F exp 1 ⎥⎦⎤⎢⎣⎡-=kT 25.0exp 02.0or 5002.0125.0exp ==⎥⎦⎤⎢⎣⎡+kT ()50ln 25.0=kTor()()⎪⎭⎫⎝⎛===3000259.0063906.050ln 25.0T kT which yields 740=T K_______________________________________ 3.41 (a)()00304.00259.00.715.7exp 11=⎪⎭⎫ ⎝⎛-+=E for 0.304%(b) At 1000=T K, 08633.0=kT eV Then()1496.008633.00.715.7exp 11=⎪⎭⎫ ⎝⎛-+=E for 14.96%(c) ()997.00259.00.785.6exp 11=⎪⎭⎫ ⎝⎛-+=E for 99.7% (d)At F E E =, ()21=E f for all temperatures_______________________________________ 3.42(a) For 1E E =()()⎥⎦⎤⎢⎣⎡--≅⎪⎪⎭⎫ ⎝⎛-+=kT E E kTE E E fF F11exp exp 11Then()611032.90259.030.0exp -⨯=⎪⎭⎫ ⎝⎛-=E fFor 2E E =, 82.030.012.12=-=-E E F eV Then()⎪⎭⎫ ⎝⎛-+-=-0259.082.0exp 1111E for()⎥⎦⎤⎢⎣⎡⎪⎭⎫ ⎝⎛---≅-0259.082.0exp 111E f141078.10259.082.0exp -⨯=⎪⎭⎫ ⎝⎛-=(b) For 4.02=-E E F eV,72.01=-F E E eVAt 1E E =,()()⎪⎭⎫⎝⎛-=⎥⎦⎤⎢⎣⎡--=0259.072.0exp exp 1kT E E E f F or()131045.8-⨯=E f At 2E E =,()()⎥⎦⎤⎢⎣⎡--=-kT E E E f F 2exp 1 ⎪⎭⎫ ⎝⎛-=0259.04.0expor()71096.11-⨯=-E f_______________________________________ 3.43(a) At 1E E =()()⎪⎭⎫⎝⎛-=⎥⎦⎤⎢⎣⎡--=0259.030.0exp exp 1kT E E E f F or()61032.9-⨯=E fAt 2E E =, 12.13.042.12=-=-E E F eV So()()⎥⎦⎤⎢⎣⎡--=-kT E E E f F 2exp 1 ⎪⎭⎫ ⎝⎛-=0259.012.1expor()191066.11-⨯=-E f (b) For 4.02=-E E F ,02.11=-F E E eV At 1E E =,()()⎪⎭⎫⎝⎛-=⎥⎦⎤⎢⎣⎡--=0259.002.1exp exp 1kT E E E f F or()181088.7-⨯=E f At 2E E =,()()⎥⎦⎤⎢⎣⎡--=-kT E E E f F 2exp 1 ⎪⎭⎫ ⎝⎛-=0259.04.0expor ()71096.11-⨯=-E f_______________________________________ 3.44()1exp 1-⎥⎦⎤⎢⎣⎡⎪⎪⎭⎫ ⎝⎛-+=kTE E E f Fso()()2exp 11-⎥⎦⎤⎢⎣⎡⎪⎪⎭⎫ ⎝⎛-+-=kT E E dE E df F⎪⎪⎭⎫ ⎝⎛-⎪⎭⎫⎝⎛⨯kT E E kT F exp 1or()2exp 1exp 1⎥⎦⎤⎢⎣⎡⎪⎪⎭⎫ ⎝⎛-+⎪⎪⎭⎫ ⎝⎛-⎪⎭⎫⎝⎛-=kT E E kT E E kT dE E df F F (a) At 0=T K, For()00exp =⇒=∞-⇒<dE dfE E F()0exp =⇒+∞=∞+⇒>dEdfE E FAt -∞=⇒=dEdfE E F(b) At 300=T K, 0259.0=kT eVFor F E E <<, 0=dE dfFor F E E >>, 0=dEdfAt F E E =,()()65.91110259.012-=+⎪⎭⎫ ⎝⎛-=dE df (eV)1-(c) At 500=T K, 04317.0=kT eVFor F E E <<, 0=dE dfFor F E E >>, 0=dEdfAt F E E =,()()79.511104317.012-=+⎪⎭⎫ ⎝⎛-=dE df (eV)1- _______________________________________ 3.45(a) At midgap E E =,()⎪⎪⎭⎫⎝⎛+=⎪⎪⎭⎫ ⎝⎛-+=kT E kTE E E f g F2exp 11exp 11Si: 12.1=g E eV, ()()⎥⎦⎤⎢⎣⎡+=0259.0212.1exp 11E for()101007.4-⨯=E fGe: 66.0=g E eV ()()⎥⎦⎤⎢⎣⎡+=0259.0266.0exp 11E for()61093.2-⨯=E f GaAs: 42.1=g E eV ()()⎥⎦⎤⎢⎣⎡+=0259.0242.1exp 11E for()121024.1-⨯=E f(b) Using the results of Problem 3.38, the answers to part (b) are exactly the same as those given in part (a)._______________________________________3.46(a) ()⎥⎦⎤⎢⎣⎡--=kT E E f F F exp ⎥⎦⎤⎢⎣⎡-=-kT 60.0exp 108or()810ln 60.0+=kT()032572.010ln 60.08==kT eV ()⎪⎭⎫⎝⎛=3000259.0032572.0Tso 377=T K(b) ⎥⎦⎤⎢⎣⎡-=-kT 60.0exp 106()610ln 60.0+=kT()043429.010ln 60.06==kT ()⎪⎭⎫⎝⎛=3000259.0043429.0Tor 503=T K_______________________________________ 3.47(a) At 200=T K,()017267.03002000259.0=⎪⎭⎫⎝⎛=kT eV⎪⎪⎭⎫ ⎝⎛-+==kT E E f F F exp 1105.019105.01exp =-=⎪⎪⎭⎫ ⎝⎛-kT E E F()()()19ln 017267.019ln ==-kT E E F 05084.0=eV By symmetry, for 95.0=F f , 05084.0-=-F E E eVThen ()1017.005084.02==∆E eV (b) 400=T K, 034533.0=kT eV For 05.0=F f , from part (a),()()()19ln 034533.019ln ==-kT E E F 10168.0=eVThen ()2034.010168.02==∆E eV _______________________________________。
半导体物理与器件第四版课后习题答案3
Chapter 33.1If o a were to increase, the bandgap energy would decrease and the material would begin to behave less like a semiconductor and more like a metal. If o a were to decrease, the bandgap energy would increase and thematerial would begin to behave more like an insulator._______________________________________ 3.2Schrodinger's wave equation is:()()()t x x V x t x m ,,2222ψ⋅+∂ψ∂- ()tt x j ∂ψ∂=, Assume the solution is of the form:()()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-=ψt E kx j x u t x exp , Region I: ()0=x V . Substituting theassumed solution into the wave equation, we obtain:()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-⎩⎨⎧∂∂-t E kx j x jku x m exp 22 ()⎪⎭⎪⎬⎫⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-∂∂+t E kx j x x u exp ()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-⋅⎪⎭⎫ ⎝⎛-=t E kx j x u jE j exp which becomes()()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-⎩⎨⎧-t E kx j x u jk m exp 222 ()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-∂∂+t E kx j x x u jkexp 2 ()⎪⎭⎪⎬⎫⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-∂∂+t E kx j x x u exp 22 ()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-+=t E kx j x Eu exp This equation may be written as()()()()0222222=+∂∂+∂∂+-x u mE x x u x x u jk x u kSetting ()()x u x u 1= for region I, the equation becomes:()()()()021221212=--+x u k dx x du jk dxx u d α where222mE=α Q.E.D.In Region II, ()O V x V =. Assume the same form of the solution:()()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-=ψt E kx j x u t x exp , Substituting into Schrodinger's wave equation, we find:()()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-⎩⎨⎧-t E kx j x u jk m exp 222 ()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-∂∂+t E kx j x x u jkexp 2 ()⎪⎭⎪⎬⎫⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-∂∂+t E kx j x x u exp 22 ()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-+t E kx j x u V O exp ()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-=t E kx j x Eu exp This equation can be written as:()()()2222x x u x x u jk x u k ∂∂+∂∂+- ()()02222=+-x u mEx u mV OSetting ()()x u x u 2= for region II, this equation becomes()()dx x du jkdx x u d 22222+ ()022222=⎪⎪⎭⎫ ⎝⎛+--x u mV k O αwhere again222mE=α Q.E.D._______________________________________3.3We have()()()()021221212=--+x u k dx x du jk dxx u d α Assume the solution is of the form: ()()[]x k j A x u -=αexp 1()[]x k j B +-+αexp The first derivative is()()()[]x k j A k j dxx du --=ααexp 1 ()()[]x k j B k j +-+-ααexp and the second derivative becomes()()[]()[]x k j A k j dx x u d --=ααexp 2212()[]()[]x k j B k j +-++ααexp 2Substituting these equations into the differential equation, we find()()[]x k j A k ---ααexp 2()()[]x k j B k +-+-ααexp 2(){()[]x k j A k j jk --+ααexp 2()()[]}x k j B k j +-+-ααexp ()()[]{x k j A k ---ααexp 22 ()[]}0exp =+-+x k j B α Combining terms, we obtain()()()[]222222αααα----+--k k k k k ()[]x k j A -⨯αexp()()()[]222222αααα--++++-+k k k k k ()[]0exp =+-⨯x k j B α We find that00= Q.E.D. For the differential equation in ()x u 2 and the proposed solution, the procedure is exactly the same as above._______________________________________ 3.4We have the solutions ()()[]x k j A x u -=αexp 1 ()[]x k j B +-+αexp for a x <<0 and()()[]x k j C x u -=βexp 2 ()[]x k j D +-+βexpfor 0<<-x b .The first boundary condition is()()0021u u =which yields0=--+D C B AThe second boundary condition is201===x x dx du dx du which yields()()()C k B k A k --+--βαα ()0=++D k β The third boundary condition is ()()b u a u -=21 which yields()[]()[]a k j B a k j A +-+-ααexp exp ()()[]b k j C --=βexp()()[]b k j D -+-+βexp and can be written as()[]()[]a k j B a k j A +-+-ααexp exp ()[]b k j C ---βexp()[]0exp =+-b k j D β The fourth boundary condition isbx a x dx dudx du -===21 which yields()()[]a k j A k j --ααexp ()()[]a k j B k j +-+-ααexp ()()()[]b k j C k j ---=ββexp()()()[]b k j D k j -+-+-ββexp and can be written as ()()[]a k j A k --ααexp ()()[]a k j B k +-+-ααexp ()()[]b k j C k ----ββexp()()[]0exp =+++b k j D k ββ_______________________________________ 3.5(b) (i) First point: πα=aSecond point: By trial and error, πα729.1=a (ii) First point: πα2=aSecond point: By trial and error, πα617.2=a_______________________________________3.6(b) (i) First point: πα=aSecond point: By trial and error, πα515.1=a (ii) First point: πα2=aSecond point: By trial and error, πα375.2=a_______________________________________ 3.7ka a aaP cos cos sin =+'αααLet y ka =, x a =α Theny x x xP cos cos sin =+'Consider dydof this function.()[]{}y x x x P dy d sin cos sin 1-=+⋅'- We find()()()⎭⎬⎫⎩⎨⎧⋅+⋅-'--dy dx x x dy dx x x P cos sin 112y dydxx sin sin -=-Theny x x x x x P dy dx sin sin cos sin 12-=⎭⎬⎫⎩⎨⎧-⎥⎦⎤⎢⎣⎡+-'For πn ka y ==, ...,2,1,0=n 0sin =⇒y So that, in general,()()dk d ka d a d dy dxαα===0 And 22 mE=α Sodk dEm mE dk d ⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛=-22/122221 α This implies thatdk dE dk d ==0α for an k π= _______________________________________3.8(a) πα=a 1π=⋅a E m o 212 ()()()()2103123422221102.41011.9210054.12---⨯⨯⨯==ππa m E o19104114.3-⨯=J From Problem 3.5 πα729.12=aπ729.1222=⋅a E m o()()()()2103123422102.41011.9210054.1729.1---⨯⨯⨯=πE18100198.1-⨯=J 12E E E -=∆1918104114.3100198.1--⨯-⨯= 19107868.6-⨯=Jor 24.4106.1107868.61919=⨯⨯=∆--E eV (b) πα23=aπ2223=⋅a E m o()()()()2103123423102.41011.9210054.12---⨯⨯⨯=πE18103646.1-⨯=J From Problem 3.5, πα617.24=aπ617.2224=⋅a E m o()()()()2103123424102.41011.9210054.1617.2---⨯⨯⨯=πE18103364.2-⨯=J 34E E E -=∆1818103646.1103364.2--⨯-⨯= 1910718.9-⨯=Jor 07.6106.110718.91919=⨯⨯=∆--E eV_______________________________________3.9(a) At π=ka , πα=a 1π=⋅a E m o 212()()()()2103123421102.41011.9210054.1---⨯⨯⨯=πE19104114.3-⨯=JAt 0=ka , By trial and error, πα859.0=a o ()()()()210312342102.41011.9210054.1859.0---⨯⨯⨯=πoE19105172.2-⨯=J o E E E -=∆11919105172.2104114.3--⨯-⨯= 2010942.8-⨯=Jor 559.0106.110942.81920=⨯⨯=∆--E eV(b) At π2=ka , πα23=a π2223=⋅a E m o()()()()2103123423102.41011.9210054.12---⨯⨯⨯=πE18103646.1-⨯=JAt π=ka . From Problem 3.5, πα729.12=aπ729.1222=⋅a E m o()()()()2103123422102.41011.9210054.1729.1---⨯⨯⨯=πE18100198.1-⨯=J 23E E E -=∆1818100198.1103646.1--⨯-⨯= 19104474.3-⨯=Jor 15.2106.1104474.31919=⨯⨯=∆--E eV _______________________________________3.10(a) πα=a 1π=⋅a E m o 212()()()()2103123421102.41011.9210054.1---⨯⨯⨯=πE19104114.3-⨯=JFrom Problem 3.6, πα515.12=aπ515.1222=⋅a E m o()()()()2103123422102.41011.9210054.1515.1---⨯⨯⨯=πE1910830.7-⨯=J 12E E E -=∆1919104114.310830.7--⨯-⨯= 19104186.4-⨯=Jor 76.2106.1104186.41919=⨯⨯=∆--E eV(b) πα23=aπ2223=⋅a E m o()()()()2103123423102.41011.9210054.12---⨯⨯⨯=πE18103646.1-⨯=JFrom Problem 3.6, πα375.24=aπ375.2224=⋅a E m o()()()()2103123424102.41011.9210054.1375.2---⨯⨯⨯=πE18109242.1-⨯=J 34E E E -=∆1818103646.1109242.1--⨯-⨯= 1910597.5-⨯=Jor 50.3106.110597.51919=⨯⨯=∆--E eV_____________________________________3.11(a) At π=ka , πα=a 1π=⋅a E m o 212()()()()2103123421102.41011.9210054.1---⨯⨯⨯=πE19104114.3-⨯=JAt 0=ka , By trial and error, πα727.0=a oπ727.022=⋅a E m o o()()()()210312342102.41011.9210054.1727.0---⨯⨯⨯=πo E19108030.1-⨯=J o E E E -=∆11919108030.1104114.3--⨯-⨯= 19106084.1-⨯=Jor 005.1106.1106084.11919=⨯⨯=∆--E eV(b) At π2=ka , πα23=aπ2223=⋅a E m o()()()()2103123423102.41011.9210054.12---⨯⨯⨯=πE18103646.1-⨯=JAt π=ka , From Problem 3.6, πα515.12=aπ515.1222=⋅a E m o()()()()2103423422102.41011.9210054.1515.1---⨯⨯⨯=πE1910830.7-⨯=J23E E E -=∆191810830.7103646.1--⨯-⨯= 1910816.5-⨯=Jor 635.3106.110816.51919=⨯⨯=∆--E eV _______________________________________3.12For 100=T K, ()()⇒+⨯-=-1006361001073.4170.124gE164.1=g E eV200=T K, 147.1=g E eV 300=T K, 125.1=g E eV 400=T K, 097.1=g E eV 500=T K, 066.1=g E eV 600=T K, 032.1=g E eV_______________________________________ 3.13The effective mass is given by1222*1-⎪⎪⎭⎫ ⎝⎛⋅=dk E d mWe have()()B curve dkE d A curve dk E d 2222>so that ()()B curve m A curve m **<_______________________________________ 3.14The effective mass for a hole is given by1222*1-⎪⎪⎭⎫ ⎝⎛⋅=dk E d m p We have that()()B curve dkEd A curve dk E d 2222> so that ()()B curve m A curve m p p **<_______________________________________ 3.15Points A,B: ⇒<0dk dEvelocity in -x directionPoints C,D: ⇒>0dk dEvelocity in +x directionPoints A,D: ⇒<022dk Ednegative effective massPoints B,C: ⇒>022dk Edpositive effective mass_______________________________________ 3.16For A: 2k C E i =At 101008.0+⨯=k m 1-, 05.0=E eV Or ()()2119108106.105.0--⨯=⨯=E J So ()2101211008.0108⨯=⨯-C3811025.1-⨯=⇒CNow ()()38234121025.1210054.12--*⨯⨯==C m 311044.4-⨯=kgor o m m ⋅⨯⨯=--*31311011.9104437.4 o m m 488.0=*For B: 2k C E i =At 101008.0+⨯=k m 1-, 5.0=E eV Or ()()2019108106.15.0--⨯=⨯=E JSo ()2101201008.0108⨯=⨯-C3711025.1-⨯=⇒CNow ()()37234121025.1210054.12--*⨯⨯==C m 321044.4-⨯=kgor o m m ⋅⨯⨯=--*31321011.9104437.4o m m 0488.0=*_______________________________________ 3.17For A: 22k C E E -=-υ()()()2102191008.0106.1025.0⨯-=⨯--C 3921025.6-⨯=⇒C ()()39234221025.6210054.12--*⨯⨯-=-=C m31108873.8-⨯-=kgor o m m ⋅⨯⨯-=--*31311011.9108873.8o m m 976.0--=* For B: 22k C E E -=-υ()()()2102191008.0106.13.0⨯-=⨯--C 382105.7-⨯=⇒C()()3823422105.7210054.12--*⨯⨯-=-=C m3210406.7-⨯-=kgor o m m ⋅⨯⨯-=--*31321011.910406.7o m m 0813.0-=*_______________________________________ 3.18(a) (i) νh E =or ()()341910625.6106.142.1--⨯⨯==h E ν 1410429.3⨯=Hz(ii) 141010429.3103⨯⨯===νλc E hc 51075.8-⨯=cm 875=nm(b) (i) ()()341910625.6106.112.1--⨯⨯==h E ν1410705.2⨯=Hz(ii) 141010705.2103⨯⨯==νλc410109.1-⨯=cm 1109=nm_______________________________________ 3.19(c) Curve A: Effective mass is a constantCurve B: Effective mass is positive around 0=k , and is negativearound 2π±=k ._______________________________________ 3.20()[]O O k k E E E --=αcos 1 Then()()()[]O k k E dkdE ---=ααsin 1()[]O k k E -+=ααsin 1 and()[]O k k E dk E d -=ααcos 2122Then221222*11αE dk Ed m o k k =⋅== or212*αE m = _______________________________________ 3.21(a) ()[]3/123/24l t dnm m m =*()()[]3/123/264.1082.04oo m m =o dnm m 56.0=*(b)oo l t cn m m m m m 64.11082.02123+=+=*oo m m 6098.039.24+=o cnm m 12.0=*_______________________________________ 3.22(a) ()()[]3/22/32/3lhhh dp m m m +=*()()[]3/22/32/3082.045.0oom m +=[]o m ⋅+=3/202348.030187.0o dpm m 473.0=*(b) ()()()()2/12/12/32/3lh hh lh hh cpm m m m m ++=*()()()()o m ⋅++=2/12/12/32/3082.045.0082.045.0 o cpm m 34.0=*_______________________________________3.23For the 3-dimensional infinite potential well, ()0=x V when a x <<0, a y <<0, and a z <<0. In this region, the wave equation is:()()()222222,,,,,,z z y x y z y x x z y x ∂∂+∂∂+∂∂ψψψ()0,,22=+z y x mEψUse separation of variables technique, so let ()()()()z Z y Y x X z y x =,,ψSubstituting into the wave equation, we have222222zZXY y Y XZ x X YZ ∂∂+∂∂+∂∂022=⋅+XYZ mEDividing by XYZ , we obtain021*********=+∂∂⋅+∂∂⋅+∂∂⋅ mEz Z Z y Y Y x X XLet01222222=+∂∂⇒-=∂∂⋅X k x X k x X X x x The solution is of the form: ()x k B x k A x X x x cos sin +=Since ()0,,=z y x ψ at 0=x , then ()00=X so that 0=B .Also, ()0,,=z y x ψ at a x =, so that ()0=a X . Then πx x n a k = where ...,3,2,1=x n Similarly, we have2221y k y Y Y -=∂∂⋅ and 2221z k zZ Z -=∂∂⋅From the boundary conditions, we find πy y n a k = and πz z n a k = where...,3,2,1=y n and ...,3,2,1=z n From the wave equation, we can write022222=+---mE k k k z y xThe energy can be written as()222222⎪⎭⎫ ⎝⎛++==a n n n m E E z y x n n n z y x π _______________________________________ 3.24The total number of quantum states in the 3-dimensional potential well is given (in k-space) by()332a dk k dk k g T ⋅=ππ where222 mEk =We can then writemEk 2=Taking the differential, we obtaindE Em dE E m dk ⋅⋅=⋅⋅⋅⋅=2112121 Substituting these expressions into the density of states function, we have()dE EmmE a dE E g T ⋅⋅⋅⎪⎭⎫ ⎝⎛=212233 ππ Noting thatπ2h=this density of states function can be simplified and written as()()dE E m h a dE E g T ⋅⋅=2/33324πDividing by 3a will yield the density of states so that()()E hm E g ⋅=32/324π _______________________________________ 3.25For a one-dimensional infinite potential well,222222k an E m n ==*π Distance between quantum states()()aa n a n k k n n πππ=⎪⎭⎫ ⎝⎛=⎪⎭⎫ ⎝⎛+=-+11Now()⎪⎭⎫ ⎝⎛⋅=a dkdk k g T π2NowE m k n*⋅=21dE Em dk n⋅⋅⋅=*2211Then()dE Em a dE E g n T ⋅⋅⋅=*2212 π Divide by the "volume" a , so ()Em E g n *⋅=21πSo()()()()()EE g 31341011.9067.0210054.11--⨯⋅⨯=π ()EE g 1810055.1⨯=m 3-J 1-_______________________________________ 3.26(a) Silicon, o nm m 08.1=*()()c nc E E h m E g -=*32/324π()dE E E h m g kTE E c nc c c⋅-=⎰+*232/324π()()kT E E c nc cE E h m 22/332/33224+*-⋅⋅=π()()2/332/323224kT h m n⋅⋅=*π ()()[]()()2/33342/33123210625.61011.908.124kT ⋅⋅⨯⨯=--π ()()2/355210953.7kT ⨯=(i) At 300=T K, 0259.0=kT eV()()19106.10259.0-⨯=2110144.4-⨯=J Then ()()[]2/3215510144.4210953.7-⨯⨯=c g25100.6⨯=m 3- or 19100.6⨯=c g cm 3-(ii) At 400=T K, ()⎪⎭⎫⎝⎛=3004000259.0kT034533.0=eV()()19106.1034533.0-⨯=21105253.5-⨯=J Then()()[]2/32155105253.5210953.7-⨯⨯=c g2510239.9⨯=m 3- or 191024.9⨯=c g cm 3-(b) GaAs, o nm m 067.0=*()()[]()()2/33342/33123210625.61011.9067.024kT g c ⋅⋅⨯⨯=--π()()2/3542102288.1kT ⨯=(i) At 300=T K, 2110144.4-⨯=kT J()()[]2/3215410144.42102288.1-⨯⨯=c g2310272.9⨯=m3-or 171027.9⨯=c g cm 3- (ii) At 400=T K, 21105253.5-⨯=kT J()()[]2/32154105253.52102288.1-⨯⨯=c g2410427.1⨯=m 3-181043.1⨯=c g cm 3-_______________________________________ 3.27(a) Silicon, o p m m 56.0=* ()()E E h mE g p-=*υυπ32/324()dE E E hm g E kTE p⋅-=⎰-*υυυυπ332/324()()υυυπE kTE pE E h m32/332/33224-*-⎪⎭⎫ ⎝⎛-=()()[]2/332/333224kT h m p-⎪⎭⎫ ⎝⎛-=*π ()()[]()()2/33342/33133210625.61011.956.024kT ⎪⎭⎫ ⎝⎛⨯⨯=--π ()()2/355310969.2kT ⨯=(i)At 300=T K, 2110144.4-⨯=kT J ()()[]2/3215510144.4310969.2-⨯⨯=υg2510116.4⨯=m 3- or 191012.4⨯=υg cm 3-(ii)At 400=T K, 21105253.5-⨯=kT J ()()[]2/32155105253.5310969.2-⨯⨯=υg2510337.6⨯=m 3-or 191034.6⨯=υg cm 3- (b) GaAs, o p m m 48.0=*()()[]()()2/33342/33133210625.61011.948.024kT g ⎪⎭⎫ ⎝⎛⨯⨯=--πυ ()()2/3553103564.2kT ⨯=(i)At 300=T K, 2110144.4-⨯=kT J()()[]2/3215510144.43103564.2-⨯⨯=υg2510266.3⨯=m 3- or 191027.3⨯=υg cm 3-(ii)At 400=T K, 21105253.5-⨯=kT J()()[]2/32155105253.53103564.2-⨯⨯=υg2510029.5⨯=m 3-or 191003.5⨯=υg cm 3-_______________________________________ 3.28(a) ()()c nc E E h m E g -=*32/324π()()[]()c E E -⨯⨯=--3342/33110625.61011.908.124πc E E -⨯=56101929.1 For c E E =; 0=c g1.0+=c E E eV; 4610509.1⨯=c g m 3-J 1-2.0+=c E E eV; 4610134.2⨯=m 3-J 1-3.0+=c E E eV; 4610614.2⨯=m 3-J 1- 4.0+=c E E eV; 4610018.3⨯=m 3-J 1- (b) ()E E hm g p-=*υυπ32/324()()[]()E E -⨯⨯=--υπ3342/33110625.61011.956.024E E -⨯=υ55104541.4 For υE E =; 0=υg1.0-=υE E eV; 4510634.5⨯=υg m 3-J 1-2.0-=υE E eV; 4510968.7⨯=m 3-J 1-3.0-=υE E eV; 4510758.9⨯=m 3-J 1-4.0-=υE E eV; 4610127.1⨯=m 3-J 1- _______________________________________ 3.29(a) ()()68.256.008.12/32/32/3=⎪⎭⎫ ⎝⎛==**pnc m m g g υ(b) ()()0521.048.0067.02/32/32/3=⎪⎭⎫ ⎝⎛==**pnc m m g g υ_______________________________________ 3.30 Plot_______________________________________ 3.31(a) ()()()!710!7!10!!!-=-=i i i i i N g N g W ()()()()()()()()()()()()1201238910!3!7!78910===(b) (i) ()()()()()()()()12!10!101112!1012!10!12=-=i W 66=(ii) ()()()()()()()()()()()()1234!8!89101112!812!8!12=-=i W 495=_______________________________________ 3.32()⎪⎪⎭⎫ ⎝⎛-+=kT E E E f F exp 11(a) kT E E F =-, ()()⇒+=1exp 11E f()269.0=E f (b) kT E E F 5=-, ()()⇒+=5exp 11E f()31069.6-⨯=E f (c) kT E E F 10=-, ()()⇒+=10exp 11E f()51054.4-⨯=E f_______________________________________ 3.33()⎪⎪⎭⎫ ⎝⎛-+-=-kT E E E f F exp 1111or()⎪⎪⎭⎫ ⎝⎛-+=-kT E E E f F exp 111(a) kT E E F =-, ()269.01=-E f (b) kT E E F 5=-, ()31069.61-⨯=-E f(c) kT E E F 10=-, ()51054.41-⨯=-E f _______________________________________ 3.34(a) ()⎥⎦⎤⎢⎣⎡--≅kT E E f F F exp c E E =; 61032.90259.030.0exp -⨯=⎥⎦⎤⎢⎣⎡-=F f 2kT E c +; ()⎥⎦⎤⎢⎣⎡+-=0259.020259.030.0exp F f 61066.5-⨯=kT E c +; ()⎥⎦⎤⎢⎣⎡+-=0259.00259.030.0exp F f 61043.3-⨯=23kT E c +; ()()⎥⎦⎤⎢⎣⎡+-=0259.020259.0330.0exp F f 61008.2-⨯=kT E c 2+; ()()⎥⎦⎤⎢⎣⎡+-=0259.00259.0230.0exp F f 61026.1-⨯=(b) ⎥⎦⎤⎢⎣⎡-+-=-kT E E f F F exp 1111()⎥⎦⎤⎢⎣⎡--≅kT E E F exp υE E =; ⎥⎦⎤⎢⎣⎡-=-0259.025.0exp 1F f 51043.6-⨯= 2kT E -υ; ()⎥⎦⎤⎢⎣⎡+-=-0259.020259.025.0exp 1F f 51090.3-⨯=kT E -υ; ()⎥⎦⎤⎢⎣⎡+-=-0259.00259.025.0exp 1F f 51036.2-⨯=23kT E -υ;()()⎥⎦⎤⎢⎣⎡+-=-0259.020259.0325.0exp 1F f 51043.1-⨯= kT E 2-υ;()()⎥⎦⎤⎢⎣⎡+-=-0259.00259.0225.0exp 1F f 61070.8-⨯=_______________________________________ 3.35()()⎥⎦⎤⎢⎣⎡-+-=⎥⎦⎤⎢⎣⎡--=kT E kT E kT E E f F c F F exp exp and()⎥⎦⎤⎢⎣⎡--=-kT E E f F F exp 1 ()()⎥⎦⎤⎢⎣⎡---=kT kT E E F υexp So ()⎥⎦⎤⎢⎣⎡-+-kT E kT E F c exp ()⎥⎦⎤⎢⎣⎡+--=kT kT E E F υexp Then kT E E E kT E F F c +-=-+υOr midgap cF E E E E =+=2υ_______________________________________ 3.3622222ma n E n π =For 6=n , Filled state ()()()()()2103122234610121011.92610054.1---⨯⨯⨯=πE18105044.1-⨯=Jor 40.9106.1105044.119186=⨯⨯=--E eV For 7=n , Empty state()()()()()2103122234710121011.92710054.1---⨯⨯⨯=πE1810048.2-⨯=Jor 8.12106.110048.219187=⨯⨯=--E eVTherefore 8.1240.9<<F E eV_______________________________________ 3.37(a) For a 3-D infinite potential well()222222⎪⎭⎫⎝⎛++=a n n n mE z y x π For 5 electrons, the 5th electron occupies the quantum state 1,2,2===z y x n n n ; so()2222252⎪⎭⎫ ⎝⎛++=a n n n m E z y x π()()()()()21031222223410121011.9212210054.1---⨯⨯++⨯=π1910761.3-⨯=Jor 35.2106.110761.319195=⨯⨯=--E eVFor the next quantum state, which is empty, the quantum state is 2,2,1===z y x n n n . This quantum state is at the same energy, so 35.2=F E eV(b) For 13 electrons, the 13th electronoccupies the quantum state 3,2,3===z y x n n n ; so ()()()()()2103122222341310121011.9232310054.1---⨯⨯++⨯=πE 1910194.9-⨯=Jor 746.5106.110194.9191913=⨯⨯=--E eVThe 14th electron would occupy the quantum state 3,3,2===z y x n n n . This state is at the same energy, so 746.5=F E eV_______________________________________ 3.38The probability of a state at E E E F ∆+=1 being occupied is()⎪⎭⎫ ⎝⎛∆+=⎪⎪⎭⎫ ⎝⎛-+=kT E kT E E E f F exp 11exp 11111 The probability of a state at E E E F ∆-=2being empty is()⎪⎪⎭⎫ ⎝⎛-+-=-kT E E E f F 222exp 1111⎪⎭⎫ ⎝⎛∆-+⎪⎭⎫ ⎝⎛∆-=⎪⎭⎫ ⎝⎛∆-+-=kT E kT E kT E exp 1exp exp 111or()⎪⎭⎫⎝⎛∆+=-kT E E f exp 11122so ()()22111E f E f -= Q.E.D. _______________________________________ 3.39(a) At energy 1E , we want01.0exp 11exp 11exp 1111=⎪⎪⎭⎫ ⎝⎛-+⎪⎪⎭⎫ ⎝⎛-+-⎪⎪⎭⎫ ⎝⎛-kT E E kT E E kT E E F F F This expression can be written as01.01exp exp 111=-⎪⎪⎭⎫ ⎝⎛-⎪⎪⎭⎫ ⎝⎛-+kT E E kT E E F For()⎪⎪⎭⎫⎝⎛-=kT E E F 1exp 01.01Then()100ln 1kT E E F += orkT E E F 6.41+= (b)At kT E E F 6.4+=, ()()6.4exp 11exp 1111+=⎪⎪⎭⎫ ⎝⎛-+=kT E E E f F which yields()01.000990.01≅=E f_______________________________________ 3.40 (a)()()⎥⎦⎤⎢⎣⎡--=⎥⎦⎤⎢⎣⎡--=0259.050.580.5exp exp kT E E f F F 61032.9-⨯=(b) ()060433.03007000259.0=⎪⎭⎫⎝⎛=kT eV31098.6060433.030.0exp -⨯=⎥⎦⎤⎢⎣⎡-=F f (c) ()⎥⎦⎤⎢⎣⎡--≅-kT E E f F F exp 1 ⎥⎦⎤⎢⎣⎡-=kT 25.0exp 02.0or 5002.0125.0exp ==⎥⎦⎤⎢⎣⎡+kT ()50ln 25.0=kTor()()⎪⎭⎫⎝⎛===3000259.0063906.050ln 25.0T kT which yields 740=T K_______________________________________ 3.41 (a)()00304.00259.00.715.7exp 11=⎪⎭⎫ ⎝⎛-+=E for 0.304%(b) At 1000=T K, 08633.0=kT eV Then()1496.008633.00.715.7exp 11=⎪⎭⎫ ⎝⎛-+=E for 14.96%(c) ()997.00259.00.785.6exp 11=⎪⎭⎫⎝⎛-+=E for 99.7% (d)At F E E =, ()21=E f for all temperatures_______________________________________ 3.42(a) For 1E E =()()⎥⎦⎤⎢⎣⎡--≅⎪⎪⎭⎫ ⎝⎛-+=kT E E kTE E E fF F11exp exp 11Then()611032.90259.030.0exp -⨯=⎪⎭⎫ ⎝⎛-=E fFor 2E E =, 82.030.012.12=-=-E E F eV Then()⎪⎭⎫ ⎝⎛-+-=-0259.082.0exp 1111E for()⎥⎦⎤⎢⎣⎡⎪⎭⎫ ⎝⎛---≅-0259.082.0exp 111E f141078.10259.082.0exp -⨯=⎪⎭⎫ ⎝⎛-=(b) For 4.02=-E E F eV,72.01=-F E E eV At 1E E =,()()⎪⎭⎫⎝⎛-=⎥⎦⎤⎢⎣⎡--=0259.072.0exp exp 1kT E E E f F or()131045.8-⨯=E f At 2E E =,()()⎥⎦⎤⎢⎣⎡--=-kT E E E f F 2exp 1 ⎪⎭⎫ ⎝⎛-=0259.04.0expor()71096.11-⨯=-E f_______________________________________ 3.43(a) At 1E E = ()()⎪⎭⎫⎝⎛-=⎥⎦⎤⎢⎣⎡--=0259.030.0exp exp 1kT E E E f F or()61032.9-⨯=E fAt 2E E =, 12.13.042.12=-=-E E F eV So()()⎥⎦⎤⎢⎣⎡--=-kT E E E f F 2exp 1 ⎪⎭⎫⎝⎛-=0259.012.1expor()191066.11-⨯=-E f(b) For 4.02=-E E F ,02.11=-F E E eV At 1E E =,()()⎪⎭⎫⎝⎛-=⎥⎦⎤⎢⎣⎡--=0259.002.1exp exp 1kT E E E f F or()181088.7-⨯=E f At 2E E =,()()⎥⎦⎤⎢⎣⎡--=-kT E E E f F 2exp 1 ⎪⎭⎫ ⎝⎛-=0259.04.0expor ()71096.11-⨯=-E f_______________________________________ 3.44()1exp 1-⎥⎦⎤⎢⎣⎡⎪⎪⎭⎫ ⎝⎛-+=kTE E E f Fso()()2exp 11-⎥⎦⎤⎢⎣⎡⎪⎪⎭⎫ ⎝⎛-+-=kT E E dE E df F⎪⎪⎭⎫ ⎝⎛-⎪⎭⎫⎝⎛⨯kT E E kT F exp 1 or()2exp 1exp 1⎥⎦⎤⎢⎣⎡⎪⎪⎭⎫ ⎝⎛-+⎪⎪⎭⎫ ⎝⎛-⎪⎭⎫⎝⎛-=kT E E kT E E kT dE E df F F (a) At 0=T K, For()00exp =⇒=∞-⇒<dE dfE E F()0exp =⇒+∞=∞+⇒>dEdfE E FAt -∞=⇒=dEdfE E F(b) At 300=T K, 0259.0=kT eVFor F E E <<, 0=dEdfFor F E E >>, 0=dEdfAt F E E =,()()65.91110259.012-=+⎪⎭⎫ ⎝⎛-=dE df (eV)1-(c) At 500=T K, 04317.0=kT eVFor F E E <<, 0=dE dfFor F E E >>, 0=dEdfAt F E E =, ()()79.511104317.012-=+⎪⎭⎫ ⎝⎛-=dE df (eV)1- _______________________________________ 3.45(a) At midgap E E =,()⎪⎪⎭⎫ ⎝⎛+=⎪⎪⎭⎫ ⎝⎛-+=kTE kTE E E f gF2exp 11exp 11Si: 12.1=g E eV, ()()⎥⎦⎤⎢⎣⎡+=0259.0212.1exp 11E for()101007.4-⨯=E fGe: 66.0=g E eV ()()⎥⎦⎤⎢⎣⎡+=0259.0266.0exp 11E for()61093.2-⨯=E fGaAs: 42.1=g E eV ()()⎥⎦⎤⎢⎣⎡+=0259.0242.1exp 11E for()121024.1-⨯=E f(b) Using the results of Problem 3.38, the answers to part (b) are exactly the same as those given in part (a)._______________________________________ 3.46(a) ()⎥⎦⎤⎢⎣⎡--=kT E E f F F exp ⎥⎦⎤⎢⎣⎡-=-kT 60.0exp 108 or ()810ln 60.0+=kT()032572.010ln 60.08==kT eV()⎪⎭⎫⎝⎛=3000259.0032572.0Tso 377=T K(b) ⎥⎦⎤⎢⎣⎡-=-kT 60.0exp 106()610ln 60.0+=kT()043429.010ln 60.06==kT()⎪⎭⎫⎝⎛=3000259.0043429.0Tor 503=T K_______________________________________ 3.47(a) At 200=T K,()017267.03002000259.0=⎪⎭⎫⎝⎛=kT eV⎪⎪⎭⎫ ⎝⎛-+==kTE E f FF exp 1105.019105.01exp =-=⎪⎪⎭⎫⎝⎛-kT E E F()()()19ln 017267.019ln ==-kT E E F 05084.0=eV By symmetry, for 95.0=F f , 05084.0-=-F E E eVThen ()1017.005084.02==∆E eV (b) 400=T K, 034533.0=kT eV For 05.0=F f , from part (a),()()()19ln 034533.019ln ==-kT E E F 10168.0=eVThen ()2034.010168.02==∆E eV_______________________________________。
半导体物理与器件第四版课后习题答案2
Chapter 22.1Sketch_______________________________________ 2.2Sketch_______________________________________ 2.3Sketch_______________________________________2.4From Problem 2.2, phase t xωλπ-=2= constant Then⎪⎭⎫ ⎝⎛+==⇒=-⋅πλωυωλπ2,02p dt dx dt dx From Problem 2.3, phase t xωλπ+=2 = constantThen ⎪⎭⎫⎝⎛-==⇒=+⋅πλωυωλπ2,02p dt dx dt dx _______________________________________2.5E hchc h E =⇒==λλν Gold: 90.4=E eV ()()19106.190.4-⨯= JSo,()()()()51910341054.2106.190.410310625.6---⨯=⨯⨯⨯=λcm orμλ254.0=mCesium: 90.1=E eV ()()19106.190.1-⨯=JSo,()()()()51910341054.6106.190.110310625.6---⨯=⨯⨯⨯=λcm orμλ654.0=m_______________________________________ 2.6(a) 9341055010625.6--⨯⨯==λh p2710205.1-⨯=kg-m/s331271032.11011.9102045.1⨯=⨯⨯==--m p υm/s or 51032.1⨯=υcm/s (b) 9341044010625.6--⨯⨯==λh p 2710506.1-⨯=kg-m/s331271065.11011.9105057.1⨯=⨯⨯==--m p υm/sor 51065.1⨯=υcm/s (c) Yes_______________________________________ 2.7(a) (i)()()()1931106.12.11011.922--⨯⨯==mE p 2510915.5-⨯=kg-m/s 925341012.110915.510625.6---⨯=⨯⨯==p h λm or oA 2.11=λ(ii)()()()1931106.1121011.92--⨯⨯=p 241087.1-⨯=kg-m/s1024341054.3108704.110625.6---⨯=⨯⨯=λmor oA 54.3=λ(iii) ()()()1931106.11201011.92--⨯⨯=p 2410915.5-⨯=kg-m/s1024341012.110915.510625.6---⨯=⨯⨯=λm or oA 12.1=λ (b)()()()1927106.12.11067.12--⨯⨯=p2310532.2-⨯=kg-m/s1123341062.210532.210625.6---⨯=⨯⨯=λm or oA 262.0=λ_______________________________________ 2.8()03885.00259.02323=⎪⎭⎫⎝⎛==kT E avg eVNowavg avg mE p 2=()()()1931106.103885.01011.92--⨯⨯= or2510064.1-⨯=avg p kg-m/sNow9253410225.610064.110625.6---⨯=⨯⨯==p h λm oroA 25.62=λ_______________________________________ 2.9pp p hch E λν==Nowmp E ee 22= and221⎪⎪⎭⎫ ⎝⎛=⇒=ee e e h m E hp λλ Set e p E E = and e p λλ10=Then22102121⎪⎪⎭⎫ ⎝⎛=⎪⎪⎭⎫ ⎝⎛=p ep h m hm hcλλλ which yieldsmchp 2100=λ100221002mc mc h hc hc E E p p =⋅===λ ()()1001031011.922831⨯⨯=-151064.1-⨯=J 25.10=keV _______________________________________ 2.10(a) 1034108510625.6--⨯⨯==λh p2610794.7-⨯= kg-m/s431261056.81011.910794.7⨯=⨯⨯==--m p υm/s or 61056.8⨯=υcm/s()()243121056.81011.92121⨯⨯==-υm E211033.3-⨯=Jor 219211008.2106.110334.3---⨯=⨯⨯=E eV (b) ()()23311081011.921⨯⨯=-E2310915.2-⨯=Jor 419231082.1106.110915.2---⨯=⨯⨯=E eV ()()3311081011.9⨯⨯==-υm p2710288.7-⨯=kg-m/s827351009.910288.710625.6---⨯-⨯⨯==p h λmor oA 909=λ_______________________________________ 2.11(a) ()()1083410110310625.6--⨯⨯⨯===λνhc h E 151099.1-⨯=J Now1915106.11099.1--⨯⨯==⇒⋅=e E V V e E41024.1⨯=V V 4.12=kV (b)()()15311099.11011.922--⨯⨯==mE p231002.6-⨯=kg-m/s Then1123341010.11002.610625.6---⨯=⨯⨯==p h λm oroA 11.0=λ_______________________________________ 2.126341010054.1--⨯=∆=∆x p η 2810054.1-⨯=kg-m/s_______________________________________ 2.13(a) (i) η=∆∆x p26103410783.8101210054.1---⨯=⨯⨯=∆p kg-m/s (ii)p m p dp d p dp dE E ∆⋅⎪⎪⎭⎫⎝⎛=∆⋅=∆22 mpp p m p ∆=∆⋅=22 Now mE p 2=()()()1931106.1161092--⨯⨯= 2410147.2-⨯=kg-m/sso ()()31262410910783.8101466.2---⨯⨯⨯=∆E1910095.2-⨯=Jor 31.1106.110095.21919=⨯⨯=∆--E eV(b) (i) 2610783.8-⨯=∆p kg-m/s(ii)()()()1928106.1161052--⨯⨯=p 231006.5-⨯=kg-m/s()()28262310510783.81006.5---⨯⨯⨯=∆E2110888.8-⨯=Jor 219211055.5106.110888.8---⨯=⨯⨯=∆E eV _______________________________________ 2.143223410054.11010054.1---⨯=⨯=∆=∆x p ηkg-m/s150010054.132-⨯=∆=∆⇒=m p m p υυ 36107-⨯=∆υm/s_______________________________________ 2.15(a) η=∆∆t E()()1619341023.8106.18.010054.1---⨯=⨯⨯=∆t s (b) 1034105.110054.1--⨯⨯=∆=∆x p η 251003.7-⨯=kg-m/s_______________________________________ 2.16(a) If ()t x ,1ψ and ()t x ,2ψ are solutionstoSchrodinger's wave equation, then()()()()t t x j t x x V x t x m ∂ψ∂=ψ+∂ψ∂⋅-,,,2112122ηη and()()()()t t x j t x x V xt x m ∂ψ∂=ψ+∂ψ∂⋅-,,,2222222ηη Adding the two equations, we obtain()()[]t x t x x m ,,221222ψ+ψ∂∂⋅-η ()()()[]t x t x x V ,,21ψ+ψ+()()[]t x t x tj ,,21ψ+ψ∂∂=ηwhich is Schrodinger's wave equation. So ()()t x t x ,,21ψ+ψ is also a solution.(b) If ()()t x t x ,,21ψ⋅ψ were a solution toSchrodinger's wave equation, then we could write []()[]21212222ψ⋅ψ+ψ⋅ψ∂∂⋅-x V x m η[]21ψ⋅ψ∂∂=tj ηwhich can be written as⎥⎦⎤⎢⎣⎡∂ψ∂⋅∂ψ∂+∂ψ∂ψ+∂ψ∂ψ-x x x x m 2121222221222η()[]⎥⎦⎤⎢⎣⎡∂ψ∂ψ+∂ψ∂ψ=ψ⋅ψ+t t j x V 122121ηDividing by 21ψ⋅ψ, we find⎥⎦⎤⎢⎣⎡∂ψ∂∂ψ∂ψψ+∂ψ∂⋅ψ+∂ψ∂⋅ψ-x x x xm21212121222222112η ()⎥⎦⎤⎢⎣⎡∂ψ∂ψ+∂ψ∂ψ=+t t j x V 112211ηSince 1ψ is a solution, then ()t j x V x m ∂ψ∂⋅ψ⋅=+∂ψ∂⋅ψ⋅-1121212112ηηSubtracting these last two equations, we have⎥⎦⎤⎢⎣⎡∂ψ∂∂ψ∂ψψ+∂ψ∂⋅ψ-x x x m 212122222212ηt j ∂ψ∂⋅ψ⋅=221η Since 2ψ is also a solution, we have()t j x V x m ∂ψ∂⋅ψ⋅=+∂ψ∂⋅ψ⋅-2222222112ηη Subtracting these last two equations, we obtain()02221212=-∂ψ∂⋅∂ψ∂⋅ψψ⋅-x V xx m ηThis equation is not necessarily valid, which means that 21ψψ is, in general, not a solutionto Schrodinger's wave equation._______________________________________ 2.1712cos 2312=⎪⎭⎫⎝⎛⎰+-dx x A π()12sin 2312=⎥⎦⎤⎢⎣⎡++-ππx x A121232=⎥⎦⎤⎢⎣⎡⎪⎭⎫ ⎝⎛--Aso 212=Aor 21=A_______________________________________ 2.18()1cos 22/12/12=⎰+-dx x n Aπ()142sin 22/12/12=⎥⎦⎤⎢⎣⎡++-ππn x n x A ⎪⎭⎫⎝⎛==⎥⎦⎤⎢⎣⎡⎪⎭⎫ ⎝⎛--211414122A Aor 2=A_______________________________________2.19Note that10*=ψ⋅ψ⎰∞dxFunction has been normalized. (a) Nowdx a x a P oa o o 24exp 2⎰⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛-=dx a x a o a o o ⎰⎪⎪⎭⎫⎝⎛-=42exp 2402exp 22o a o ooa x a a ⎪⎪⎭⎫ ⎝⎛-⎪⎪⎭⎫⎝⎛-=or()⎪⎭⎫ ⎝⎛--=⎥⎥⎦⎤⎢⎢⎣⎡-⎪⎪⎭⎫⎝⎛--=21exp 1142exp 1ooa a P which yields 393.0=P (b)dx a x a P o oa a o o 224exp 2⎰⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛-=dx a x a o oa a o o⎰⎪⎪⎭⎫⎝⎛-=242exp 2242exp 22o oa a o o o a x a a ⎪⎪⎭⎫ ⎝⎛-⎪⎪⎭⎫⎝⎛-=or()()⎥⎦⎤⎢⎣⎡⎪⎭⎫ ⎝⎛----=21exp 1exp 1Pwhich yields239.0=P (c)dx a x a P oa o o 20exp 2⎰⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛-= dx a x a oa o o⎰⎪⎪⎭⎫⎝⎛-=2exp 2o a o o oa x a a 02exp 22⎪⎪⎭⎫ ⎝⎛-⎪⎪⎭⎫⎝⎛-=()()[]12exp 1---= which yields 865.0=P_______________________________________ 2.20()dx x P 2⎰=ψ(a)dx x a a ⎪⎭⎫⎝⎛⎪⎭⎫ ⎝⎛⎰2cos 224/0π 4/042sin 22a a a x x a ⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛+⎪⎭⎫ ⎝⎛=ππ⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛+⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛=a a a ππ42sin 242 ()()⎥⎦⎤⎢⎣⎡+⎪⎭⎫ ⎝⎛=π4182a a a or 409.0=P(b) dx a x a P a a ⎪⎭⎫⎝⎛⎪⎭⎫ ⎝⎛=⎰π2/4/2cos 2 2/4/42sin 22a a a a x x a ⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛+⎪⎭⎫ ⎝⎛=ππ()⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛--⎪⎭⎫ ⎝⎛+⎪⎭⎫ ⎝⎛=a a a a a ππππ42sin 84sin 42 ⎥⎦⎤⎢⎣⎡--+=π41810412or 0908.0=P(c) dx a x a P a a ⎪⎭⎫⎝⎛⎪⎭⎫ ⎝⎛=⎰+-π22/2/cos 2 2/2/42sin 22a a a a x x a +-⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛+⎪⎭⎫ ⎝⎛=ππ()()⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡⎪⎭⎫ ⎝⎛--⎪⎭⎫ ⎝⎛--⎪⎭⎫ ⎝⎛+⎪⎭⎫ ⎝⎛=a a a a a ππππ4sin 44sin 42 or 1=P_______________________________________2.21(a) dx a x a P a ⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛=⎰π2sin 224/04/0244sin 22a a a x x a ⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-⎪⎭⎫ ⎝⎛=ππ()⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡⎪⎭⎫ ⎝⎛-⎪⎭⎫ ⎝⎛=a a a ππ8sin 82or 25.0=P(b) dx a x a P a a ⎪⎭⎫⎝⎛⎪⎭⎫ ⎝⎛=⎰π2sin 222/4/ 2/4/244sin 22a a a a x x a ⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-⎪⎭⎫ ⎝⎛=ππ()()⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡⎪⎭⎫ ⎝⎛+⎪⎭⎫ ⎝⎛-⎪⎭⎫ ⎝⎛-⎪⎭⎫ ⎝⎛=a a a a a ππππ8sin 882sin 42 or 25.0=P(c) dx a x a P a a ⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛=⎰+-π2sin 222/2/ 2/2/244sin 22a a a a x x a +-⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-⎪⎭⎫ ⎝⎛=ππ()()⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡⎪⎭⎫ ⎝⎛-+⎪⎭⎫ ⎝⎛--⎪⎭⎫ ⎝⎛-⎪⎭⎫ ⎝⎛=a a a a a ππππ82sin 482sin 42 or 1=P_______________________________________ 2.22(a) (i) 481210108108=⨯⨯==k p ωυm/s or 610=p υcm/s9810854.710822-⨯=⨯==ππλk mor oA 54.78=λ (ii)()()431101011.9-⨯==υm p271011.9-⨯=kg-m/s()()24312101011.92121-⨯==υm E2310555.4-⨯=Jor 419231085.2106.110555.4---⨯=⨯⨯=E eV (b) (i) 491310105.1105.1-=⨯-⨯==k p ωυm/s or 610-=p υcm/s991019.4105.122-⨯=⨯==ππλk m or oA 9.41=λ(ii) 271011.9-⨯-=p kg-m/s41085.2-⨯=E eV_______________________________________ 2.23(a) ()()t kx j Ae t x ω+-=ψ, (b) ()()21921106.1025.0υm E =⨯=- ()2311011.921υ-⨯= so41037.9⨯=υm/s 61037.9⨯=cm/sFor electron traveling in x -direction, 61037.9⨯-=υcm/s()()4311037.91011.9⨯-⨯==-υm p2610537.8-⨯-=kg-m/s926341076.710537.810625.6---⨯=⨯⨯==p h λm8910097.81076.722⨯=⨯==-πλπk m 1- ()()481037.910097.8⨯⨯=⋅=υωkor 1310586.7⨯=ωrad/s_______________________________________ 2.24(a) ()()4311051011.9⨯⨯==-υm p2610555.4-⨯=kg-m/s8263410454.110555.410625.6---⨯=⨯⨯==p h λm881032.410454.122⨯=⨯==-πλπk m 1- ()()481051032.4⨯⨯==υωk131016.2⨯=rad/s (b) ()()631101011.9-⨯=p251011.9-⨯=kg-m/s1025341027.71011.910625.6---⨯=⨯⨯=λm 9101064.810272.72⨯=⨯=-πk m 1-()()15691064.8101064.8⨯=⨯=ωrad/s_______________________________________ 2.25()()()2103122342222210751011.9210054.12---⨯⨯⨯==ππn ma n E n η()212100698.1-⨯=n E nJor()19212106.1100698.1--⨯⨯=n E n or ()3210686.6-⨯=n E n eV Then311069.6-⨯=E eV221067.2-⨯=E eV231002.6-⨯=E eV_______________________________________ 2.26(a) ()()()2103122342222210101011.9210054.12---⨯⨯⨯==ππn ma n E n η ()20210018.6-⨯=n Jor()()3761.0106.110018.6219202n n E n =⨯⨯=--eV Then376.01=E eV 504.12=E eV385.33=E eV(b) Ehc ∆=λ ()()19106.1504.1385.3-⨯-=∆E191001.3-⨯=J()()198341001.310310625.6--⨯⨯⨯=λ710604.6-⨯=m or 4.660=λnm_______________________________________ 2.27(a) 22222ma n E n πη=()()()223223423102.11015210054.11015----⨯⨯⨯=⨯πn()622310538.21015--⨯=⨯n or 2910688.7⨯=n (b) 151≅+n E mJ (c) No_______________________________________ 2.28For a neutron and 1=n :()()()2142722342221101066.1210054.12---⨯⨯==ππma E η13103025.3-⨯=Jor6191311006.2106.1103025.3⨯=⨯⨯=--E eV For an electron in the same potential well: ()()()2143122341101011.9210054.1---⨯⨯=πE10100177.6-⨯=J or9191011076.3106.1100177.6⨯=⨯⨯=--E eV _______________________________________ 2.29Schrodinger's time-independent wave equation()()()()02222=-+∂∂x x V E mx x ψψηWe know that()0=x ψ for 2a x ≥ and 2ax -≤We have()0=x V for 22a x a +<<-so in this region()()02222=+∂∂x mEx x ψψηThe solution is of the form ()kx B kx A x sin cos +=ψ where22ηmEk =Boundary conditions: ()0=x ψ at 2,2a x a x +=-=First mode solution:()x k A x 111cos =ψ where222112ma E a k ηππ=⇒=Second mode solution: ()x k B x 222sin =ψ where22222242ma E a k ηππ=⇒= Third mode solution: ()x k A x 333cos =ψ where22233293maE a k ηππ=⇒= Fourth mode solution: ()x k B x 444sin =ψ where222442164ma E a k ηππ=⇒= _______________________________________2.30The 3-D time-independent wave equation incartesian coordinates for ()0,,=z y x V is:()()()222222,,,,,,zz y x y z y x x z y x ∂∂+∂∂+∂∂ψψψ ()0,,22=+z y x mEψηUse separation of variables, so let ()()()()z Z y Y x X z y x =,,ψSubstituting into the wave equation, we obtain222222zZXY y Y XZ x X YZ ∂∂+∂∂+∂∂022=+XYZ mEηDividing by XYZ and letting222ηmEk =, wefind(1)01112222222=+∂∂⋅+∂∂⋅+∂∂⋅k zZZ y Y Y x X X We may set01222222=+∂∂⇒-=∂∂⋅X k x X k x X X xx Solution is of the form()()()x k B x k A x X x x cos sin += Boundary conditions: ()000=⇒=B X and ()an k a x X x x π=⇒==0 where ....3,2,1=x n Similarly, let2221y k y Y Y -=∂∂⋅ and 2221z k zZ Z -=∂∂⋅Applying the boundary conditions, we findan k y y π=, ....3,2,1=y nan k z z π=, ...3,2,1=z n From Equation (1) above, we have02222=+---k k k k z y xor222222ηmEk k k k z y x ==++ so that()2222222z y x n n n n n n maE E z y x ++=→πη _______________________________________ 2.31 (a)()()()0,2,,22222=⋅+∂∂+∂∂y x mEy y x x y x ψψψηSolution is of the form:()y k x k A y x y x sin sin ,⋅=ψWe find()y k x k Ak x y x y x x sin cos ,⋅=∂∂ψ ()y k x k Ak xy x y x x sin sin ,222⋅-=∂∂ψ()y k x k Ak y y x y x y cos sin ,⋅=∂∂ψ()y k x k Ak yy x y x y sin sin ,222⋅-=∂∂ψSubstituting into the original equation, we find:(1) 02222=+--ηmE k k y xFrom the boundary conditions, 0sin =a k A x , where oA a 40= So an k x x π=, ...,3,2,1=x n Also 0sin =b k A y , where oA b 20= So bn k y y π=, ...,3,2,1=y n Substituting into Eq. (1) above⎪⎪⎭⎫ ⎝⎛+=22222222b n a n m E y x n n y x ππη (b)Energy is quantized - similar to 1-D result. There can be more than one quantum stateper given energy - different than 1-D result._______________________________________ 2.32(a) Derivation of energy levels exactly thesame as in the text(b) ()21222222n n maE -=∆πη For 1,212==n n Then22223maE πη=∆ (i) For oA a 4= ()()()2102722341041067.1210054.13---⨯⨯⨯=∆πE2210155.6-⨯=Jor 319221085.3106.110155.6---⨯=⨯⨯=∆E eV(ii) For 5.0=a cm()()()22272234105.01067.1210054.13---⨯⨯⨯=∆πE3610939.3-⨯=J or1719361046.2106.110939.3---⨯=⨯⨯=∆E eV _______________________________________ 2.33(a) For region II, 0>x()()()0222222=-+∂∂x V E mx x O ψψηGeneral form of the solution is()()()x jk B x jk A x 22222exp exp -+=ψ where()O V E mk -=222ηTerm with 2B represents incident wave andterm with 2A represents reflected wave. Region I, 0<x()()0212212=+∂∂x mEx x ψψηGeneral form of the solution is()()()x jk B x jk A x 11111exp exp -+=ψ where212ηmEk =Term involving 1B represents thetransmitted wave and the term involving 1A represents reflected wave: but if a particle is transmitted into region I, it will not be reflected so that 01=A . Then()()x jk B x 111exp -=ψ()()()x jk B x jk A x 22222exp exp -+=ψ (b)Boundary conditions: (1) ()()0021===x x ψψ(2) 0201==∂∂=∂∂x x x x ψψ Applying the boundary conditions to the solutions, we find221B A B +=112222B k B k A k -=-Combining these two equations, we find212122B k k k k A ⋅⎪⎪⎭⎫⎝⎛+-=212212B k k k B ⋅⎪⎪⎭⎫⎝⎛+=The reflection coefficient is21212*22*22⎪⎪⎭⎫ ⎝⎛+-==k k k k B B A A R The transmission coefficient is()2212141k k k k T R T +=⇒-= _______________________________________2.34()()x k A x 222exp -=ψ ()()x k A A x P 2*2222exp -==ψwhere ()222ηE V m k o -=()()()34193110054.1106.18.25.31011.92---⨯⨯-⨯=9210286.4⨯=k m 1-(a) For 101055-⨯==oA x m ()x k P 22exp -=()()[]109105102859.42exp -⨯⨯-= 0138.0=(b) For 10101515-⨯==oA x m()()[]1091015102859.42exp -⨯⨯-=P61061.2-⨯= (c) For 10104040-⨯==oA x m()()[]1091040102859.42exp -⨯⨯-=P151029.1-⨯=_______________________________________ 2.35()a k V E V E T o o 22exp 116-⎪⎪⎭⎫⎝⎛-⎪⎪⎭⎫ ⎝⎛≅ where ()222ηE V m k o -=()()()34193110054.1106.11.00.11011.92---⨯⨯-⨯=or 2k 910860.4⨯=m 1-(a) For 10104-⨯=a m()()[]1091041085976.42exp 0.11.010.11.016-⨯⨯-⎪⎭⎫ ⎝⎛-⎪⎭⎫ ⎝⎛≅T 0295.0=(b) For 101012-⨯=a m()()[]10910121085976.42exp 0.11.010.11.016-⨯⨯-⎪⎭⎫ ⎝⎛-⎪⎭⎫ ⎝⎛≅T 51024.1-⨯= (c) υe N J t =, where t N is the density of transmitted electrons. 1.0=E eV 20106.1-⨯=J()23121011.92121υυ-⨯==m510874.1⨯=⇒υm/s 710874.1⨯=cm/s()()719310874.1106.1102.1⨯⨯=⨯--t N810002.4⨯=t N electrons/cm 3 Density of incident electrons, 10810357.10295.010002.4⨯=⨯=i N cm 3- _______________________________________2.36 ()a k V E V E T O O 22exp 116-⎪⎪⎭⎫⎝⎛-⎪⎪⎭⎫ ⎝⎛≅ (a) For ()o m m 067.0= ()222ηE V m k O -= ()()()()()2/1234193110054.1106.12.08.01011.9067.02⎪⎭⎪⎬⎫⎪⎩⎪⎨⎧⨯⨯-⨯=---or9210027.1⨯=k m 1- Then⎪⎭⎫⎝⎛-⎪⎭⎫ ⎝⎛=8.02.018.02.016T()()[]109101510027.12exp -⨯⨯-⨯ or138.0=T (b) For ()o m m 08.1=2k =()()()()()2/1234193110054.1106.12.08.01011.908.12⎪⎭⎪⎬⎫⎪⎩⎪⎨⎧⨯⨯-⨯--- or 9210124.4⨯=k m 1- Then ⎪⎭⎫⎝⎛-⎪⎭⎫ ⎝⎛=8.02.018.02.016T ()()[]109101510124.42exp -⨯⨯-⨯ or 51027.1-⨯=T _______________________________________2.37()a k V E V E T o o 22exp 116-⎪⎪⎭⎫ ⎝⎛-⎪⎪⎭⎫ ⎝⎛≅ where ()222ηE V m k o -=()()()341962710054.1106.1101121067.12---⨯⨯⨯⨯-⨯=1410274.7⨯=m 1- (a)()()[]14141010274.72exp 121112116-⨯-⎪⎭⎫ ⎝⎛-⎪⎭⎫ ⎝⎛≅T []548.14exp 222.1-=710875.5-⨯= (b)()()710875.510-⨯=T()[]a 1410274.72exp 222.1⨯-=()⎪⎭⎫⎝⎛⨯=⨯-61410875.5222.1ln 10274.72a or 1410842.0-⨯=a m_______________________________________ 2.38Region I ()0<x , 0=V ; Region II ()a x <<0, O V V = Region III ()a x >, 0=V (a) Region I:()()()x jk B x jk A x 11111exp exp -+=ψ(incident) (reflected)where 212ηmEk =Region II:()()()x k B x k A x 22222exp exp -+=ψ where ()222ηE V m k O -=Region III:()()()x jk B x jk A x 13133exp exp -+=ψ (b)In Region III, the 3B term represents areflected wave. However, once a particleis transmitted into Region III, there will not be a reflected wave so that 03=B .(c) Boundary conditions:At 0=x : ⇒=21ψψ2211B A B A +=+ ⇒=dxd dx d 21ψψ 22221111B k A k B jk A jk -=-At a x =: ⇒=32ψψ()()a k B a k A 2222exp exp -+()a jk A 13exp =⇒=dx d dx d 32ψψ ()()a k B k a k A k 222222exp exp --()a jk A jk 131exp = The transmission coefficient is defined as *11*33A A A A T =so from the boundary conditions, we wantto solve for 3A in terms of 1A . Solvingfor 1A in terms of 3A , we find(){()()[]a k a k k k k k jA A 2221222131exp exp 4---+=()()[]}a k a k k jk 2221exp exp 2-+-()a jk 1exp ⨯We then find()(){()[a k k k k k A A A A 22122221*33*11exp 4-=()]22exp a k --()()[]}2222221exp exp 4a k a k k k -++ We have()222ηE V m k O -= If we assume that E V O >>, then a k 2 will be large so that ()()a k a k 22exp exp ->>We can then write ()(){()[]222122221*33*11exp 4a k k k k k A A A A -= ()[]}222221exp 4a k k k + which becomes()()()a k k k k k A A A A 22122221*33*112exp 4+=Substituting the expressions for 1k and2k , we find222212ηO mV k k =+and()⎥⎦⎤⎢⎣⎡⎥⎦⎤⎢⎣⎡-=22222122ηηmE E V m k k O ()()E E V m O -⎪⎭⎫⎝⎛=222η ()()E V E V m O O ⎪⎪⎭⎫ ⎝⎛-⎪⎭⎫⎝⎛=1222η Then()()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫⎝⎛-⎪⎭⎫ ⎝⎛⎪⎪⎭⎫ ⎝⎛=E V E V m a k mV A A A A O O O 12162exp 222222*33*11ηη()a k V E V E A A O O 2*332exp 116-⎪⎪⎭⎫⎝⎛-⎪⎪⎭⎫ ⎝⎛=Finally,()a k V E V E A A A A T O O 2*11*332exp 116-⎪⎪⎭⎫⎝⎛-⎪⎪⎭⎫ ⎝⎛== _____________________________________2.39 Region I: 0=V ()()⇒=+∂∂0212212x mE x x ψψη()()()x jk B x jk A x 11111exp exp -+=ψ incident reflectedwhere212ηmEk = Region II: 1V V =()()()⇒=-+∂∂02221222x V E m xx ψψη()()()x jk B x jk A x 22222exp exp -+=ψtransmitted reflected where ()2122ηV E m k -=Region III: 2V V =()()()⇒=-+∂∂02322232x V E m x x ψψη()()x jk A x 333exp =ψtransmitted where()2232ηV E m k -=There is no reflected wave in Region III. The transmission coefficient is defined as:*11*3313*11*3313A A A A k k A A A A T ⋅=⋅=υυ From the boundary conditions, solve for 3Ain terms of 1A . The boundary conditions are:At 0=x : ⇒=21ψψ 2211B A B A +=+ ⇒∂∂=∂∂x x 21ψψ22221111B k A k B k A k -=- At a x =: ⇒=32ψψ ()()a jk B a jk A 2222exp exp -+()a jk A 33exp =⇒∂∂=∂∂xx 32ψψ()()a jk B k a jk A k 222222exp exp --()a jk A k 333exp =But ⇒=πn a k 22()()1exp exp 22=-=a jk a jk Then, eliminating 1B , 2A , 2B fromtheboundary condition equations, we find ()()23131231211344k k k k k k k k k T +=+⋅= _______________________________________ 2.40(a) Region I: Since E V O >, we can write()()()0212212=--∂∂x E V m x x O ψψη Region II: 0=V , so ()()0222222=+∂∂x mEx x ψψηRegion III: 03=⇒∞→ψVThe general solutions can be written, keeping in mind that 1ψ must remain finite for 0<x , as()()x k B x 111exp =ψ()()()x k B x k A x 22222cos sin +=ψ ()03=x ψ where ()212ηE V m k O -=and222ηmE k =(b) Boundary conditionsAt 0=x : ⇒=21ψψ21B B = 221121A k B k xx =⇒∂∂=∂∂ψψ At a x =: ⇒=32ψψ ()()0cos sin 2222=+a k B a k A or()a k A B 222tan -= (c)12122211B k k A A k B k ⎪⎪⎭⎫⎝⎛=⇒=and since 21B B =, then 2212B k k A ⎪⎪⎭⎫⎝⎛=From ()a k A B 222tan -=, we can write ()a k B k k B 22212tan ⎪⎪⎭⎫⎝⎛-= or()a k k k 221tan 1⎪⎪⎭⎫⎝⎛-=This equation can be written as ⎥⎥⎦⎤⎢⎢⎣⎡⋅⋅--=a mE E E V O 22tan 1η or⎥⎥⎦⎤⎢⎢⎣⎡⋅-=-a mE E V EO 22tan η This last equation is valid only forspecific values of the total energy E . The energy levels are quantized._______________________________________ 2.41()222424n e m E o o n η∈-=π(J)()222324n e m o o η∈-=π(eV)()()()[]()22342123193110054.121085.84106.11011.9n----⨯⨯⨯⨯-=πor258.13n E n -=(eV) 58.1311-=⇒=E n eV 395.322-=⇒=E n eV 51.133-=⇒=E n eV 849.044-=⇒=E n eV_______________________________________ 2.42We have⎪⎪⎭⎫⎝⎛-⎪⎪⎭⎫⎝⎛⋅=o oa r a exp 112/3100πψ and*10010024ψψπr P =⎪⎪⎭⎫ ⎝⎛-⎪⎪⎭⎫⎝⎛⋅⋅=o o a r a r 2exp 11432ππ or()⎪⎪⎭⎫ ⎝⎛-⋅=o o a r r a P 2exp 423 To find the maximum probability()0=dr r dP ()()⎪⎪⎭⎫ ⎝⎛-⎪⎩⎪⎨⎧⎪⎪⎭⎫ ⎝⎛-=o o o a r r a a 2exp 2423⎪⎭⎪⎬⎫⎪⎪⎭⎫ ⎝⎛-+o a r r 2exp 2 which giveso oa r a r =⇒+-=10 or o a r = is the radius that gives the greatest probability. _______________________________________2.43 100ψ is independent of θ and φ, sothe waveequation in spherical coordinates reducesto ()()021222=-+⎪⎭⎫ ⎝⎛∂∂∂∂⋅ψψr V E m r r r r oη where ()ra m r e r V o o o 224η-=∈-=π For ⎪⎪⎭⎫ ⎝⎛-⎪⎪⎭⎫⎝⎛⋅=o o a r a exp 112/3100πψ Then⎪⎪⎭⎫ ⎝⎛-⎪⎪⎭⎫ ⎝⎛-⎪⎪⎭⎫ ⎝⎛⋅=∂∂o o o a r a a r exp 1112/3100πψ so⎪⎪⎭⎫⎝⎛-⎪⎪⎭⎫ ⎝⎛⋅-=∂∂o o a r r a r rexp 1122/51002πψ We then obtain2/5100211⎪⎪⎭⎫⎝⎛⋅-=⎪⎪⎭⎫ ⎝⎛∂∂∂∂o a r r r πψ ⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛-⎪⎪⎭⎫ ⎝⎛-⎪⎪⎭⎫ ⎝⎛-⨯o o o a r a r a r r exp exp 22Substituting into the wave equation, we have⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛--⎪⎪⎭⎫ ⎝⎛-⎪⎪⎭⎫ ⎝⎛⋅-o o o o a r a r a r r a r exp exp 21122/52π ⎥⎦⎤⎢⎣⎡++r a m E m o o o 222ηη0exp 112/3=⎪⎪⎭⎫ ⎝⎛-⎪⎪⎭⎫ ⎝⎛⋅⎪⎪⎭⎫ ⎝⎛⨯o o a r a π where ()222241224o o o o a m e m E E ηη-=∈-==π Then the above equation becomes ⎪⎩⎪⎨⎧⎥⎦⎤⎢⎣⎡--⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛-⎪⎪⎭⎫ ⎝⎛⋅o o o o a r r a r a r a 222/321exp 11π 022222=⎪⎭⎪⎬⎫⎪⎪⎭⎫ ⎝⎛+-+r a m a m m o o o o o ηηη or ⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛-⎪⎪⎭⎫ ⎝⎛⋅o o a r a exp 112/3π 0211222=⎪⎭⎪⎬⎫⎪⎩⎪⎨⎧⎪⎪⎭⎫ ⎝⎛+-++-⨯r a a a r a o o o ois which gives 0 = 0 and shows that100 indeed a solution to the wave equation. _______________________________________2.44All elements are from the Group I column ofthe periodic table. All have one valenceelectron in the outer shell._______________________________________。
半导体物理与器件第四版课后习题答案(供参考).doc
Chapter 44.1n i 2E gN c N expkTT 3E gexpN cO N O300kTwhere N cO and N Oare the values at 300 K.(a) SiliconT (K) kT (eV) n i (cm 3) 200 0.01727 7.68 104 400 0.03453 2.38 1210 6000.05189.74 1014(c) GaAs(b) GermaniumT (K)n i (cm 3 ) n i (cm 3 ) 200 2.16 10101.38 4008.60 1014 3.28 109 6003.82 10165.72 1012_______________________________________ 4.2Plot_______________________________________4.3(a) n i 2 N c NexpE gkT31121919T5 2.8 1.04 101010300exp1.120.0259 T 300T 32.5 10 232.912 10 38300exp1.12 3000.0259 TBy trial and error, T 367.5 K(b)n i25 10 1222.5 10 2532.912 10 38T exp 1.12 300300 0.0259 TBy trial and error,T 417.5 K _______________________________________4.4At T200 K, kT0.02592003000. 017267eVAt T400 K, kT0.02594003000. 034533eVn i 2400 7.70 101023.025 10 17n i 2 2001.40 10 2 23400expE g3000.0345333200Egexp300 0.017267E gE g8 exp0.0345330.0172673.025 10178 exp E g 57 .9139 28.9578orE g 28.9561ln 3.025 1017 38.17148 or E g 1.318 eVNow7.70 1010N co N o340023001.318 exp0.03453321N co N o 2.370 175.929 10 2.658 10so N co N o 9.41 10 37 cm 6_______________________________________4.5exp 1.10n i kT 0.20Bexpn i A 0.90 kTexp kTFor T 200 K, kT 0.017267 eVFor T 300 K, kT 0.0259 eVFor T 400 K, kT 0.034533 eV(a) For T 200K,n i B exp 0.20 9.325 10 6n i A 0.017267(b) For T 300K,n i Bexp 0.204.43 10 4n i A 0.0259 (c) For T 400K,n i Bexp 0.203.05 10 3n i A 0.034533_______________________________________ 4.6(a) g c f FE E FE E c expkTThen g c f F x expxkTTo find the maximum value:d g c f F 1 x1 / 2 exp xdx 2 kT1 x1 /2 exp x 0kT kTwhich yields1/ 21 x kT2x1/ 2 x 2kTThe maximum value occurs atEkTE c2(b)g 1 f FE F EE E expkTE EE E expkTexpE F EkTLet E E xThen g 1 f F x expxkTTo find the maximum valued g 1 f F d xdx dxx expkTSame as part (a). Maximum occurs atxkT2E E c exp E E ckTorkTE E2E c EF expkTLet E E c x _______________________________________ 4.7E1 E c exp E1 E cn E1 kTn E2E2 E c exp E2 E c kTwhereE1 E c 4kT and E 2 E c kT 2Thenn E1 4kTexp E1 E2n E2 kT kT22 2 exp 4 12 exp 3.522orn E10.0854n E 2_______________________________________ 4.8Plot_______________________________________4.9Plot_______________________________________ 4.10E Fi E midgap 3kT ln m*pm n* 4Silicon: m*p 0.56 m o , m n* 1.08m oE Fi E midgap 0.0128 eVGermanium: m*p 0. 37m o ,*0.55m om nE Fi E midgap 0 .0077 eVGallium Arsenide: m*p 0.48m o ,m n* 0.067m oE Fi E midgap 0 .0382 eV_______________________________________ 4.11E Fi E midgap 1 kT ln N2 N c1kT ln 1.04 1019 0.4952 kT2 2.8 1019T (K) kT (eV) ( E Fi E midgap )(eV) 200 0.01727 0.0086 400 0.03453 0.0171 600 0.0518 0.0257_______________________________________4.12(a) E Fi E midgapm*p3 kT ln4 m n*3 0.0259 ln0.704 1.2110.63 meV(b) E Fi E midgap 3 0.0259 ln0.754 0.08043.47 meV_______________________________________4.13Let g c E K constantThenn o g c E f F E dEE cK1dEE E FEc 1 expkTK expE E FdEkTE cLetE E cso that dE kT dkTWe can writeE EF E c E F E E cso thatE E Fexp E c E FexpexpkTkTThe integral can then be written asn o K kT exp E c E Fexp d kTwhich becomesn o K kTE c EF expkT_______________________________________4.14Let g c E C1E E c for E E cThenn o g c E f F E dEE cC1 E E cdEE c 1exp E EF kTC1 EE E FdE E C expE ckTLetE E cdE kT dso thatkTWe can writeE EF E E c E c E FThenE c E Fn o C1 expkTE E cE E cdE expE ckT orn oE c EF C1 expkTkT exp kT d 0We find thatexp d exp 1 1So2 E c E Fn o C1 kT expkT_______________________________________4.15r1 m oWe have rm*a oFor germanium, r 16 , m* 0.55m oThenr1 16 1 a o 29 0.530.55oror1 15.4 AThe ionization energy can be written asm*2E o 13.6 eVm o s0.552 13.6 E 0.029 eV16_______________________________________ 4.16We have r1 m orm*a oFor gallium arsenide, r 13.1 , *m0.067 m o1or1 13.1 104 A0.530.067The ionization energy ism*20.067E o 13.6 13.6m o s 13.1 2orE0.0053 eV_______________________________________4.17Nc(a) E c E F kT ln2.8 10190.0259 ln 157 100.2148 eV(b) E F E E g E c E F1.12 0.2148 0.90518eV(c) p o NE F E expkT1.04 19 0.9051810 exp0.02596.90 103cm 3(d) Holesn o(e) E F E Fi kT lnn i710 150.0259 ln1.5 10100.338 eV_______________________________________4.18N(a) E F E kT lnp o190.0259 ln 1.0410210160.162 eV(b) E c E F E g E F E1.12 0.162 0.958 eV(c) n o 2.8 19 0.95810 exp0.02592.41 103cm3p o(d) E Fi E F kT lnn i2 10 160.0259 ln 101.5 100.365eV_______________________________________4.19Nc(a) E c E F kT ln0.0259 ln 2.810192 1050.8436 eVE F E E g E c E F1.12 0.8436E F E 0.2764 eV(b) p o 1.04 1019 exp 0.276370.02592.414 1014cm3(c)p-type_______________________________________4.20(a) kT3750.032375 eV0.02593003 / 2n o 4.7 10 17 375 exp 0.28300 0.0323751.15 1014cm3E F E E g E c E F 1.42 0.281.14 eV375 3 / 2 1.14 p o 7 18 exp10300 0.0323754.99 103cm 3(b) E c E F 0.0259 ln 4.7 10171.15 10 140.2154 eVE F E E g E c E F 1.42 0.21541.2046 eVp o 7 10 18 exp 1.20460.02594.42 10 2cm 3_______________________________________ 4.21(a) kT 0.0259 3750.032375 eV 300375 3 / 2 0.28n o 2.8 19 exp10300 0.0323756.86 1015cm 3E F E E g E c E F 1.12 0.280.840 eV375 3 / 20.840p o 1.04 1019 exp300 0.0323757.84 107cm 3(b) E c E F kT ln N cn o0.0259 ln2.8 10196.862 10 150.2153 eVE F E 1.12 0.2153 0.9047 eVp o 1.04 10 19 exp 0.9046680.02597.04 103 cm 3_______________________________________4.22(a) p-typeE g(b) E F E1.124 0.28 eV4p o N exp E F EkT1.04 10 19 exp 0.280.02592.10 1014cm 3E c EF E g E F E1.12 0.28 0.84 eVn o N c exp E c E FkT2.8 1019exp0.840.02592.30 105cm 3_______________________________________4.23(a) n o n iE F E FiexpkT1.5 1010 exp 0.220.02597.3313cm310p oE Fi E Fn i expkT1.5 1010 exp 0.220.02593.07 106cm 3(b) n o n iE F E FiexpkT1.8 10 6 exp 0.220.02598.80 109cm 3p o n i expE Fi E FkT1.8 106 exp 0.220.02593.68 102cm 3_______________________________________4.24(a) E F ENkT lnp o0.0259 ln1.04 10 195 10 150.1979 eV(b) E c E F E g E F E1.12 0.19788 0.92212 eV(c) n o 2.8 1019 exp 0.922120.02599.66 103cm 3(d) Holesp o(e) E Fi E F kT lnn i510 150.0259 ln1.5 10100.3294 eV _______________________________________4.25kT 0.0259 4000.034533 eV 3003 / 2N 1.04 10 19400300 1.601 1019cm 33 / 2N c 2.8 1019400300 4.3109 1019cm 30.2642 eV _______________________________________4.26(a) p o 7 1018 exp 0.250.02594.50 1014cm 3E c EF 1.42 0.25 1.17 eVn o 4.7 10 17 exp 1.170.02591.13 10 2cm 3(b)kT 0.034533eV3 / 2N 7 10184003001.078 1019cm 33 / 217 400N c 4.7 103007.236 1017cm3expn i 2 4.3109 10 19 1.601 10191.12NE F E kT lnp o19 0.0345335.67022410n i 2.381 1012 cm 3(a) E F ENkT lnp o0.034533 ln 1.601 10195 1015 0.2787 eV(b) E c E F 1.12 0.27873 0.84127 eV(c) n o 4.3109 10 19 exp 0.841270.0345331.134 109cm3(d) Holes(e) E Fi E F kT ln p on i510150.034533 ln2.381 10120.034533 ln1.078104.50 10 140.3482 eVE c EF 1.42 0.3482 1.072 eVn o 7.236 1017 exp 1 .071770. 0345332.40 104cm 3_____________________________________4.27(a) p o 1.04 1019 exp 0.250.02596.68 1014cm 3E c EF 1.12 0.25 0.870 eVn o 2.8 10 19 exp 0.8700.0259n o7.2310 4 cm 3(b)kT0.034533 eV3 / 2N 1.04 10194003001.601 1019cm 33 / 2N c 2.8 1019 4003004.311 1019cm 3NE F E kT lnp o1.60110 190.034533ln6.6810140.3482 eVE c EF 1.12 0.34820.7718 eVn o 4.311 1019 exp 0.771750.0345338.49 109cm 3_______________________________________4.282(a) n o N c F1 / 2 FFor E F E c kT 2 ,E F E c kT 2 FkT 0.5kTThen F1/ 2 F 1.0n o 2 2.8 1019 1.03.16 1019cm 3(b) n o 2 N c F1 / 2 F24.7 1017 1.05.30 1017cm 3_______________________________________ 4.29p o 2 N F1/2 F5 1019 2 1.04 1019 F1/2 FSo F1/ 2 F 4.26We find F 3.0E E FkTE EF 3.0 0.0259 0.0777 eV_______________________________________4.30E F E c 4kT(a) F 4kT kTThen F1 / 2 F 6.02N c F1 / 2n o F2 2.8 1019 6.01.90 10 20 cm 3(b) n o 2 4.7 1017 6.03.18 1018cm 3_______________________________________ 4.31For the electron concentrationn E g c E f F EThe Boltzmann approximation applies, so4 * 3 / 22m nE E cn Eh3E E FexpkTor4 2m n* 3 / 2 E c E Fexpn E h3kTE E c E E ckT expkTkTDefinexEE ckTThenn E n x K x exp xTo find maximumn E n x , setdn x 0 K 1 x 1 / 2 exp xdx 2x 1 / 21 expxorKx 1 / 2 expx1 x2which yieldsx1 E E cE E c12kTkT2For the hole concentrationp Eg E 1f F EUsing the Boltzmann approximation4 2m p * 3 / 2p EEEh 3E F EexpkT or3 / 242m *p E F Ep Eh 3expkTE E E EkTexpkTkTDefinexE EkTThenp xK x exp xTo find maximum value ofp Ep x ,setdp xUsing the results from0 dxabove,we find the maximum at1E E kT2_______________________________________4.32 (a) Silicon:We haven oN c expE cE FkTWe can writeE c E FE c E d E d E FForE c E d 0.045 eV andE dE F3kT eVwe can writen o2.8 1019 exp 0.04530.02592.8 1019exp 4.737or10 17 cm3n o2.45 We also havep oN expE F EkTAgain, we can writeE FEE FE aE aEForE FE a3kTandE aE0.045eVThenp o1.04 1019 exp 3 0.0450.02591.04 1019 exp4.737orp o9.12 10 16 cm 3(b) GaAs: assume E c E d0.0058eVThenn o4.7 1017 exp0.0058 30.025917exp 3.2244.7 10orn o1.87 1016 cm3Assume E a E 0.0345 eVThenp o71018 exp0.0345 30.02597 1018 exp 4.332orp o9.20 1016 cm 3_______________________________________ 4.33Plot_______________________________________4.34 10 151015 cm 3(a)p o415 31.5 10 10 2n o7.5 10 4 cm33 10153(b) n oN d316cm1010 2p o1.5 107.5 10 3cm 33 1016 (c)n op on i 1.5 10 10cm33(d) n i 22.8 10 19 1.041019 375300 exp1.12 3000.0259 375n i7.334 1011 cm3p o N a4 10 15 cm 37.334 10 11 2n o1.34 10 8 cm34 10 153(e) n i 22.8 10 19 1.04 10 19 4503001.12 300exp0.0259 450133n i1.722 10 cm14142n o1.722 10 1310102221.029 1014 cm 31.722 1013 2p o2.88 1012 cm 31.029 1014_______________________________________(a) p oN aN d4 101510153 1015 cm 3n i 2 1.8 10 6 2n o1.08 10 3cm 3p o3 1015(b) n oN d 3 10 16 cm 3p o1.8 10 6 2 1.08 10 4 cm33 10163(c) n o p on i1.8 10 6cm375 3(d) n i 24.7 1017 7.0 10 18300 exp1.42 3000.0259 375n i 7.580 10 8 cm 3p o N a4 1015 cm 38 2n o7.580 10 1.44 10 2 cm 34 10 153 (e) 2 4.7 10 17 7.0 18450 n i 10 300 exp1.42 3000.0259 450n i 3.853 1010 cm3n oN d10 14 cm 33.853 1010 2p o1.48 10 7 cm 310 14_______________________________________4.3610 13 cm 3(a) Ge: n i2.42(i) n oN dN dn i 22 22 10152 210152.4 13 22210or2 1015 cm 3n oN d4.35n i 2 2.4 1013 2p o2 1015n o2.88 1011 cm 3(ii) p o N a N d 10167 10153 1015 cm 32n i22.4 10 13n op o310 151.92 1011cm3(b) GaAs: n i 1.8 10 6cm3(i) n o N d2 1015 cm62p o1.8 10 1.62 10 3cm32 10 15(ii) p oN aN d3 10 15 cm 362n o1.8 101.08 10 3cm 33 1015 (c) The result implies that there is only one 33minority carrier in a volume of 10 cm ._______________________________________4.37(a) For the donor leveln d 1N d1 1exp EdE F2kT11 1 exp 0.2020.0259orn d8.85 10 4N d (b) We havef F E1E E F1expkTNowE E FE E cE c E ForE EF kT 0.245Thenf F E10.2451 exp 1 0.0259orf F E 2.87 10 5_______________________________________4.38N aN d(a) p-type(b) Silicon:10131013p oN aN d 2.5 1 or1013 cm 3p o1.5Thenn i 21.5 10 10 210 7cm 3n o1.5p o 1.5 1013 Germanium:N aN d N a N d 2p o2n i 221.5131.5 10 1322.4 101310222or3.26 10 13 cm 3p oThen2n i 22.4 10 13n o1.76 10 13p o3.264 1013cm 3Gallium Arsenide:p oN a N d1.5 10 13 cm 3and2n i 21.8 10 6n o0.216 cm 3p o1.5 1013_______________________________________4.39 (a) N d N an-type(b) n oN d N a 2 10151.2 10158 1014 cm 3n i 21.5 101022.81 10 5cm 3p o8 14n o10(c)p o N aN a N d4 1015N a 1.2 10 152 1015N a 4.8 10 15 cm31.5 10 102n o5.625 10 4cm 3 4 1015_______________________________________4.40n i21.5 101021. 153n o2 10 5 125 10cmp on o p on-type_______________________________________4.413n i 21.04 10196.0 10 18 250300 exp0.660.0259250 3001.8936 102412n i 1.376cm310 n on i 2 n i 2n o 21n i 2p o4n o 4n o1n i2Son o 6.88 1011 cm 3 ,Then p o2.75 1012cm3N a N a 2p on i 222N a22.752 10122N a21.8936 10 24227.5735 10 242.752 10 12 N aN a2N a 21.8936 10 242so that N a 2.064 1012cm 3_______________________________________4.42Plot_______________________________________4.43Plot_______________________________________4.44Plot_______________________________________ 4.45N d N aN dN a 2n o2n i 2214141.1 1014 2 10 1.2 102 2 10141.2 1014 2n i 221.1 10144 10 1324 10132n i 24.9 10 271.6 10 27n i2so n i5.74 10 13 cm 3p on i 23.3 10 273 133n o 1.1 10 1410 cm_______________________________________4.46(a)N a N d p-typeMajority carriers are holesp o N a N d16163 101.5 101.5 1016 cm 3Minority carriers are electrons210 10 2n on i 1.5 1.5 10 4 cm 3p o 1.5 1016(b) Boron atoms must be addedp o N a N aN d5 1016N a 3 10161.5 1016So N a3.5 10 16 cm 31.5 10 102n o4.5 10 3cm 35 10 16_______________________________________4.47p on i (a)n-type(b) p on i 2 n on i 2n op o1.5 10 1021016 cm3n o4 1.125 2 10electrons are majoritycarriersp o2 10 4cm3holes are minority carriers(c) n oN d N a1.125 101615N d 7 10so N d1.825 1016 cm3_______________________________________4.48E Fi E FkT lnp on iFor GermaniumT (K)kT (eV)n i (cm 3)200 0.01727 2.16 1010400 0.03453 8.60 1410 6000.05183.82 1016N aN a 2p o n i 2and22N a10 15 cm 3T (K)p o (cm3)E Fi EF (eV)200 1.0 1015 0.1855 4001.49 1015 0.01898 6003.87 10160.000674_______________________________________4.49(a) E c E FkT lnN cN d0.0259 ln 2.8 1019N dFor 1014cm 3 , E cE F 0.3249eV15 cm 3 ,E cE F0.2652eV1016cm 3, E c E F 0.2056eV 101017 cm 3 , E c E F0.1459eV(b) E F E FikT lnN dn i0.0259 lnN d1.51010For 1014cm 3 , E FE Fi 0.2280 eV15cm 3, E F E Fi 0.2877 eV10 1016 cm 3 , E F E Fi 0.3473 eV 1017 cm 3 ,E F E Fi0.4070 eV_______________________________________ 4.50N d N d 2(a) n on i 222n o1.05N d1.05 10 15 cm 31.05 10150.5 10 1520.5 10152n i2son i 25.25 10 28Now3n i 22.8 1019 1.04 1019T300exp1.120.0259 T 30035.25 10 28 2.912 10 38 T300exp 12972.973TBy trial and error, T 536.5K(b) At T 300 K,E c EF kT ln N cn oE c EF 0.0259 ln 2.8 1019 1015T 536.5 K, 0.2652 eVAt536.5kT0.02590.046318 eV3003 / 2N c 2.8 1019 536.53006.696 1019cm 3E c E FN c kT lnn oE c E F6.696 10 19 0.046318 ln10151.050.5124 eVthen E c E F 0.2472 eV(c)Closer to the intrinsic energy level._______________________________________4.51p oE Fi EF kT lnn iAt T 200K, kT 0.017267 eVT 400 K, kT 0.034533 eVT 600 K, kT 0.0518 eV At T 200K,22.8 10191019 200n i 1.04300exp1.120.017267n i 7.638 10 4 cm 3At T 400 K,3n i 2 2.8 1019 1.04 10 19 4003001.12exp0.034533n i 2.381 1012 cm 3At T 600 K,322.8 1019 19 600n i 1.04 10300exp 1.120.0518n i 9.740 1014 cm 3At T 200 K and T 400 K,p o N a 3 1015 cm 3At T 600 K,N a N a2p o n i22 23 15 3 10 15 2 9.740 10 1410 22 23.288 1015cm3Then, T 200K, E Fi E F 0.4212eVT 400K,E Fi EF 0.2465 eVT600K,E Fi EF 0.0630 eV_______________________________________4.52(a)N a N aE Fi EF kT ln 0.0259 ln6n i 1.8 10For N a10 14 cm 3 ,E FiE F0.4619 eVN a 10 15 cm 3,E FiE F0.5215 eV163,N a 10 cmE FiE F0.5811 eVN a 10 17cm 3,E FiE F 0.6408 eV(b)E FEN7.0 1018kT ln0.0259 lnN aN aFor N a10 14 cm 3 ,E F E0.2889 eVN a 10 15 cm 3 ,E FE0.2293 eV163,N a 10 cmE F E0.1697 eVN a 10 17 cm3,E F E 0.1100 eV_______________________________________ 4.53(a) E Fi3 m *p E midgapkT ln4m n *3 0.0259 ln 104 orE Fi E midgap 0.0447 eV(b) Impurity atoms to be added soE midgap EF 0.45 eV(i) p-type, so add acceptor atoms(ii)E Fi EF 0.0447 0.45 0.4947 eVThenp oE FiE Fn i expkT10 5exp 0.49470.0259 or10 13 cm3p o N a1.97_______________________________________4.54n oN d N aN c expE c E FkTsoN d 5 10 15 2.8 10 19 exp0.2150.025951015 6.95 1015orcm 3N d 1.2 1016_______________________________________4.55(a) Silicon(i) E cE F N ckT lnN d0.0259 ln 2.8 10 190.2188 eV6 1015(ii) E cE F0.2188 0.0259 0.1929 eVN dN c expE c E FkT2.8 10 19 exp0.19290.0259N d1.631 1016 cm3N d 6 1015N d1.031 10 16 cm 3Additional donor atoms(b) GaAs(i) E c E F0.0259 ln4.7101710150.15936eV(ii) E cE F0.15936 0.0259 0.13346 eVN d4.7 1017 exp0.133460.02592.718 1015 cm 3N d 1015N d1.718 10 15 cm3Additionaldonor atoms_______________________________________ 4.56(a) E Fi E FN kT lnN a0.0259 ln 1.04 10190.1620 eV2 1016(b) E F E Fi kT ln N c N d0.0259 ln 2.8 1019 0.1876 eV2 10 16(c) For part (a);p o 2 1016 cm 3n i2 1.5 1010 2n op o 2 10161.125 104cm3For part (b):3n o 2 1016 cmn i 2 1.5 1010 2p on o 2 10 161.125 104cm3_______________________________________ 4.57n oE F E Fin i expkT1.8 10 6 exp 0.550.02593.0 1015cm 3Add additional acceptor impuritiesn o N d N a3 10 15 7 10 15 N aN a 4 10 15 cm 3_______________________________________(a) E Fi E F kT lnpon i0.02593 10 150.3161 eVln10 101.5(b) E F E Fin okT lnn i0.02593 10160.3758 eVln10 101.5(c) E F E Fi(d) E Fi E Fp okT lnn i0.0259 375 ln 4 1015300 7.334 10 110.2786 eV(e) E F E Fi kT lnnon i140.0259 450 ln 1.029 10300 1.722 10 130.06945eV_______________________________________4.59(a) E F ENkT lnp o0.0259 ln7.0 10180.2009 eV3 1015(b) E F E 0.0259 l n7.0 10 181.08 10 41.360 eV(c) E F E 0.0259 l n 7.0 10181.8 10 60.7508 eV4.58(d) E F E 0.0259 375300ln 7.0 10 18 375 300 3 / 24 10 150.2526 eV(e) E F E 0.0259 450 300ln 7.0 10 18 450 300 3/ 21.48 10 71.068 eV_______________________________________4.60n-typeE F E Fi kT ln n o n i0.02591.125 10 16ln100.3504 eV1.5 10______________________________________ 4.61N a N a 2 p o 22 2 n i5.08 1015 5 101525 10 15 2n i225.08 10 15 2.5 10 15 22.5 1015 2n i26.6564 10 30 6.25 10 30 n i2n i 2 4.064 10 29n i2 N c N expE gkTkT 0.02593500.030217 eV3003502N c 1.2 10 19 1.633 1019 cm 33003502N 1.8 1019 2.45 10 19 cm 3300Now4.064 10 29 1.633 1019 2.45 1019E gexp0.030217SoE g 0.030217 ln 1.633 10 19 2.45 10 194.064 10 29E g 0.6257 eV_______________________________________4.62(a) Replace Ga atoms Silicon acts as adonorN d0.05 7 1015 3.5 10 14 cm 3Replace As atoms Silicon acts asanacceptorN a 0.95 7 1015 6.65 10 15 cm 3(b) N a N d p-type(c) p o N a N d 6.65 1015 3.5 10146.3 1015cm 3n i 2 1. 810 6 2n o 5.14 10 4 cm 3 p o 6 .3 1015(d) E Fi E F kT ln p o n i0.0259 ln 6.3 10 150.5692 eV1.8 10 6_______________________________________。
半导体物理与器件第四版课后习题答案1
______________________________________________________________________________________Chapter 1Problem Solutions1.1 (a)fcc: 8 corner atoms 18/1atom6 face atoms32/1atomsTotal of 4 atoms per unit cell (b)bcc: 8 corner atoms 18/1atom1 enclosed atom=1 atom Total of 2 atoms per unit cell(c)Diamond: 8 corner atoms 18/1atom6 faceatoms 32/1atoms4 enclosedatoms= 4 atomsTotal of 8 atoms per unit cell_______________________________________ 1.2 (a)Simple cubic lattice: r a 2Unit cell vol33382rra1 atom per cell, so atom vol 3413r ThenRatio%4.52%10083433rr(b)Face-centered cubic latticerd aa rd22224Unit cell vol 33321622rr a4 atoms per cell, so atom vol3443r ThenRatio%74%10021634433rr (c)Body-centered cubic latticeraa rd3434Unit cell vol 3334ra2 atoms per cell, so atom vol 3423r ThenRatio%68%1003434233r r (d)Diamond lattice Body diagonal raa rd3838Unit cell vol3338r a8 atoms per cell, so atom vol 3483r ThenRatio%34%1003834833rr _______________________________________1.3(a)oA a43.5; From Problem 1.2d,ra38Then oAa r176.18343.583Center of one silicon atom to center ofnearest neighboroAr 35.22______________________________________________________________________________________ (b)Number density22381051043.58cm 3(c)Mass density23221002.609.28105..AN W t At N 33.2grams/cm3_______________________________________1.4(a)4 Ga atoms per unit cell Number density381065.54Density of Ga atoms 221022.2cm34 As atoms per unit cell Density of As atoms 221022.2cm3(b)8 Ge atoms per unit cell Number density381065.58Density of Ge atoms221044.4cm3_______________________________________ 1.5From Figure 1.15 (a)aa d4330.0232oAd 447.265.54330.0(b)aa d7071.022oAd 995.365.57071.0_______________________________________1.674.5423232222sin a a 5.109_______________________________________ 1.7(a) Simple cubic: oAr a 9.32(b)fcc:oAr a515.524(c) bcc:oA r a 503.434(d) diamond:oAra007.9342_______________________________________ 1.8 (a)Br 2035.122035.12oBAr 4287.0(b)oAa 07.2035.12(c)A-atoms: # of atoms1818Density381007.21231013.1cm3B-atoms: # of atoms3216Density381007.23231038.3cm3_______________________________________ 1.9(a)oAr a 5.42# of atoms1818Number density38105.412210097.1cm3______________________________________________________________________________________Mass density AN W t At N ..23221002.65.12100974.1228.0gm/cm3(b)oAr a196.534# of atoms 21818Number density3810196.5222104257.1cm3Mass density23221002.65.12104257.1296.0gm/cm3_______________________________________ 1.10From Problem 1.2, percent volume of fcc atoms is 74%; Therefore after coffee is ground,Volume = 0.74 cm3_______________________________________1.11(b)oAa 8.20.18.1(c)Na: Density38108.22/1221028.2cm3Cl: Density221028.2cm3(d)Na: At. Wt. = 22.99 Cl: At. Wt. = 35.45 So, mass per unit cell23231085.41002.645.352199.2221Then mass density21.2108.21085.43823grams/cm3_______________________________________ 1.12(a)oAa 88.122.223Then oA a 62.4Density of A:22381001.11062.41cm3Density of B:22381001.11062.41cm3(b)Same as (a) (c)Same material_______________________________________ 1.13oAa619.438.122.22(a) For 1.12(a), A-atomsSurface density28210619.411a1410687.4cm2For 1.12(b), B-atoms: oAa 619.4Surface density14210687.41acm2For 1.12(a) and (b), Same material(b) For 1.12(a), A-atoms;oAa 619.4Surface density212a1410315.3cm2B-atoms;Surface density______________________________________________________________________________________14210315.321a cm 2For 1.12(b), A-atoms;oAa 619.4Surface density212a1410315.3cm2B-atoms;Surface density14210315.321acm2For 1.12(a) and (b), Same material_______________________________________ 1.14 (a)Vol. Density31oaSurface Density212oa(b)Same as (a)_______________________________________ 1.15 (i)(110) plane(see Figure 1.10(b))(ii) (111) plane(see Figure 1.10(c))(iii) (220) plane,1,1,21,21Same as (110) plane and [110]direction(iv) (321) plane6,3,211,21,31Intercepts of plane at6,3,2sq p [321] direction is perpendicular to(321) plane_______________________________________1.16(a)31311,31,11(b)12141,21,41_______________________________________ 1.17Intercepts: 2, 4, 331,41,21(634) plane_______________________________________ 1.18(a)oAa d 28.5(b)oAa d734.322(c)oAa d048.333_______________________________________ 1.19(a) Simple cubic(i) (100) plane:Surface density2821073.411a141047.4cm 2(ii) (110) plane:Surface density212a141016.3cm 2(iii) (111) plane: Area of planebh21where oAa b 689.62Now2222243222a a a hSooAh793.573.426______________________________________________________________________________________Area of plane881079304.51068923.62116103755.19cm 2Surface density16103755.19613141058.2cm2(b) bcc(i) (100) plane:Surface density 1421047.41acm2(ii) (110) plane: Surface density222a141032.6cm 2(iii) (111) plane:Surface density16103755.19613141058.2cm2(c) fcc(i) (100) plane:Surface density 1421094.82acm2(ii) (110) plane: Surface density222a141032.6cm 2(iii) (111) plane:Surface density16103755.19213613151003.1cm2_______________________________________ 1.20 (a)(100) plane: - similar to a fcc:Surface density281043.52141078.6cm 2(b)(110) plane:Surface density281043.524141059.9cm2(c)(111) plane: Surface density281043.5232141083.7cm2_______________________________________1.21oAr a703.6237.2424(a)#/cm338310703.64216818a2210328.1cm3(b)#/cm222124142a210703.62281410148.3cm2(c)oA a d74.422703.622(d)# of atoms2213613Area of plane: (see Problem 1.19)oAa b4786.92oAa h2099.826Area88102099.8104786.92121bh______________________________________________________________________________________15108909.3cm2#/cm215108909.32=141014.5cm2oAa d87.333703.633_______________________________________ 1.22Density of silicon atoms 22105cm3and4 valence electrons per atom, soDensity of valence electrons 23102cm3_______________________________________ 1.23Density of GaAs atoms22381044.41065.58cm3An average of 4 valence electrons peratom,SoDensity of valence electrons231077.1cm3_______________________________________ 1.24 (a)%10%10010510532217(b)%104%10010510262215_______________________________________ 1.25 (a)Fraction by weight7221610542.106.2810582.10102(b)Fraction by weight5221810208.206.2810598.3010_______________________________________ 1.26Volume density 1631021dcm3So610684.3dcmoAd 4.368We haveoo Aa 43.5Then85.6743.54.368oa d _______________________________________ 1.27Volume density 1531041dcm 3So61030.6dcmoAd630We have oo Aa 43.5Then11643.5630oa d _______________________________________。
半导体物理与器件第四版课后习题答案(完整教资)
Chapter 1Problem Solutions1.1 (a) fcc: 8 corner atoms 18/1=⨯atom 6 face atoms 32/1=⨯atomsTotal of 4 atoms per unit cell (b) bcc: 8 corner atoms 18/1=⨯atom1 enclosed atom =1 atomTotal of 2 atoms per unit cell (c) Diamond: 8 corner atoms 18/1=⨯atom 6 face atoms 32/1=⨯atoms4 enclosed atoms = 4 atomsTotal of 8 atoms per unit cell_______________________________________ 1.2 (a) Simple cubic lattice: r a 2=Unit cell vol ()33382r r a === 1 atom per cell, so atom vol ()⎪⎪⎭⎫⎝⎛=3413r π ThenRatio %4.52%10083433=⨯⎪⎪⎭⎫ ⎝⎛=rr π (b) Face-centered cubic latticer da a r d ⋅==⇒==22224Unit cell vol ()33321622r r a ⋅=⋅==4 atoms per cell, so atom vol()⎪⎪⎭⎫⎝⎛=3443r π ThenRatio ()%74%10021634433=⨯⋅⎪⎪⎭⎫⎝⎛=r r π (c) Body-centered cubic latticer a a r d ⋅=⇒==3434 Unit cell vol 3334⎪⎪⎭⎫⎝⎛⋅==r a 2 atoms per cell, so atom vol()⎪⎪⎭⎫⎝⎛=3423r πThenRatio ()%68%1003434233=⨯⎪⎪⎭⎫ ⎝⎛⎪⎪⎭⎫⎝⎛=r r π (d) Diamond lattice Body diagonalr a a r d ⋅=⇒===3838Unit cell vol 3338⎪⎪⎭⎫⎝⎛==r a 8 atoms per cell, so atom vol ()⎪⎪⎭⎫⎝⎛=3483r π ThenRatio ()%34%1003834833=⨯⎪⎪⎭⎫ ⎝⎛⎪⎪⎭⎫⎝⎛=r r π _______________________________________ 1.3(a)oA a 43.5=; From Problem 1.2d,r a ⋅=38Then ()o A a r 176.18343.583===Center of one silicon atom to center ofnearest neighbor oA r 35.22== (b) Number density()22381051043.58⨯=⨯=-cm 3-(c)Mass density()()()23221002.609.28105..⨯⨯===A N W t At N ρ 33.2=⇒ρ grams/cm 3_______________________________________ 1.4 (a) 4 Ga atoms per unit cellNumber density ()381065.54-⨯=⇒Density of Ga atoms221022.2⨯=cm 3-4 As atoms per unit cell ⇒Density of As atoms 221022.2⨯=cm 3- (b) 8 Ge atoms per unit cellNumber density ()381065.58-⨯=⇒Density of Ge atoms 221044.4⨯=cm 3-_______________________________________ 1.5From Figure 1.15(a)()a a d 4330.0232=⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛= =()()oA d 447.265.54330.0=⇒ (b)()a a d 7071.022=⎪⎭⎫⎝⎛=()()oA d 995.365.57071.0=⇒= _______________________________________ 1.6︒=⇒==⎪⎭⎫ ⎝⎛74.5423232222sin θθa a︒=⇒5.109θ_______________________________________ 1.7(a) Simple cubic: oA r a 9.32== (b) fcc: oA r a 515.524==(c) bcc: oA ra 503.434==(d) diamond: ()oA r a 007.9342==_______________________________________ 1.8(a)()()B r 2035.122035.12+= oB A r 4287.0= (b) ()oA a 07.2035.12==(c)A-atoms: # of atoms 1818=⨯= Density ()381007.21-⨯=231013.1⨯=cm 3-B-atoms: # of atoms 3216=⨯=Density ()381007.23-⨯=231038.3⨯= cm 3- _______________________________________ 1.9 (a)oA r a 5.42==# of atoms 1818=⨯= Number density ()38105.41-⨯=2210097.1⨯=cm 3-Mass density ()AN W t At N ..==ρ ()()23221002.65.12100974.1⨯⨯==228.0gm/cm 3(b)o A ra 196.534==# of atoms 21818=+⨯Number density ()3810196.52-⨯=22104257.1⨯=cm 3-Mass density ()()23221002.65.12104257.1⨯⨯==ρ296.0=gm/cm 3_______________________________________ 1.10From Problem 1.2, percent volume of fcc atoms is 74%; Therefore after coffee is ground,Volume = 0.74 cm 3_______________________________________ 1.11(b)oA a 8.20.18.1=+= (c)Na: Density ()()38108.22/1-⨯=221028.2⨯=cm 3-Cl: Density 221028.2⨯=cm 3- (d) Na: At. Wt. = 22.99Cl: At. Wt. = 35.45So, mass per unit cell ()()23231085.41002.645.352199.2221-⨯=⨯⎪⎭⎫⎝⎛+⎪⎭⎫ ⎝⎛= Then mass density()21.2108.21085.43823=⨯⨯=--ρ grams/cm 3_______________________________________ 1.12 (a)()()oA a 88.122.223=+=Then oA a 62.4= Density of A:()22381001.11062.41⨯=⨯=-cm 3-Density of B: ()22381001.11062.41⨯=⨯=-cm 3-(b) Same as (a) (c) Same material_______________________________________ 1.13()()o A a 619.438.122.22=+=(a) For 1.12(a), A-atoms Surface density ()28210619.411-⨯==a 1410687.4⨯=cm 2-For 1.12(b), B-atoms: oA a 619.4= Surface density 14210687.41⨯==acm 2- For 1.12(a) and (b), Same material(b) For 1.12(a), A-atoms; o A a 619.4= Surface density212a =1410315.3⨯=cm 2-B-atoms;Surface density14210315.321⨯==a cm 2- For 1.12(b), A-atoms; o A a 619.4= Surface density212a =1410315.3⨯=cm 2-B-atoms;Surface density14210315.321⨯==a cm 2- For 1.12(a) and (b), Same material_______________________________________ 1.14(a) Vol. Density 31oa =Surface Density 212oa=(b) Same as (a)_______________________________________ 1.15 (i) (110) plane(see Figure 1.10(b))(ii) (111) plane(see Figure 1.10(c))(iii) (220) plane ⇒()0,1,1,21,21⇒⎪⎭⎫⎝⎛∞Same as (110) plane and [110] direction(iv) (321) plane ()6,3,211,21,31⇒⎪⎭⎫⎝⎛⇒Intercepts of plane at 6,3,2===s q p[321] direction is perpendicular to (321) plane_______________________________________ 1.16(a)()31311,31,11⇒⎪⎭⎫⎝⎛(b)()12141,21,41⇒⎪⎭⎫⎝⎛_______________________________________ 1.17Intercepts: 2, 4, 3 ⇒⎪⎭⎫⎝⎛⇒31,41,21(634) plane_______________________________________ 1.18(a) oA a d 28.5==(b) o A a d 734.322==(c) o A a d 048.333==_______________________________________ 1.19 (a) Simple cubic(i) (100) plane:Surface density ()2821073.411-⨯==a141047.4⨯=cm 2- (ii) (110) plane:Surface density 212a =141016.3⨯=cm 2- (iii) (111) plane:Area of plane bh 21=where oA a b 689.62== Now ()()2222243222a a a h =⎪⎪⎭⎫⎝⎛-= So ()o A h 793.573.426==Area of plane ()()881079304.51068923.621--⨯⨯= 16103755.19-⨯=cm 2Surface density 16103755.19613-⨯⨯=141058.2⨯=cm 2- (b) bcc(i) (100) plane:Surface density 1421047.41⨯==a cm 2- (ii) (110) plane:Surface density 222a =141032.6⨯=cm 2- (iii) (111) plane:Surface density 16103755.19613-⨯⨯=141058.2⨯=cm 2- (c) fcc(i) (100) plane:Surface density 1421094.82⨯==acm 2-(ii) (110) plane:Surface density 222a =141032.6⨯=cm 2- (iii) (111) plane:Surface density 16103755.19213613-⨯⨯+⨯=151003.1⨯=cm 2-_______________________________________ 1.20 (a) (100) plane: - similar to a fcc:Surface density ()281043.52-⨯=141078.6⨯=cm 2- (b) (110) plane:Surface density ()281043.524-⨯=141059.9⨯=cm 2- (c) (111) plane:Surface density ()()281043.5232-⨯= 141083.7⨯=cm 2-_______________________________________ 1.21()o A r a 703.6237.2424===(a) #/cm 3()38310703.64216818-⨯=⨯+⨯=a 2210328.1⨯=cm 3-(b) #/cm 222124142a ⨯+⨯= ()210703.6228-⨯=1410148.3⨯=cm 2- (c) ()o A a d 74.422703.622===(d)# of atoms 2213613=⨯+⨯=Area of plane: (see Problem 1.19)oA a b 4786.92==o A ah 2099.826==Area ()()88102099.8104786.92121--⨯⨯==bh 15108909.3-⨯=cm 2#/cm 215108909.32-⨯= =141014.5⨯ cm 2-()o A a d 87.333703.633===_______________________________________ 1.22Density of silicon atoms 22105⨯=cm 3- and4 valence electrons per atom, so Density of valence electrons 23102⨯=cm 3-_______________________________________ 1.23Density of GaAs atoms()22381044.41065.58⨯=⨯=-cm 3- An average of 4 valence electrons per atom,SoDensity of valence electrons231077.1⨯=cm 3-_______________________________________ 1.24(a) %10%10010510532217-=⨯⨯⨯ (b) %104%10010510262215-⨯=⨯⨯⨯ _______________________________________ 1.25 (a) Fraction by weight()()()()7221610542.106.2810582.10102-⨯=⨯⨯≅ (b) Fraction by weight()()()()5221810208.206.2810598.3010-⨯=⨯≅ _______________________________________ 1.26Volume density 1631021⨯==dcm 3-So 610684.3-⨯=d cm oA d 4.368=⇒ We have oo A a 43.5=Then85.6743.54.368==o a d _______________________________________ 1.27Volume density 1531041⨯==dcm 3-So 61030.6-⨯=d cm oA d 630=⇒ We have oo A a 43.5= Then11643.5630==o a d _______________________________________。
半导体物理与器件第四版课后习题答案(供参考)
Chapter 44.1 where cO N and O N υ are the values at 300 K.4.2Plot_______________________________________4.3By trial and error, 5.367≅T K(b)By trial and error, 5.417≅T K_______________________________________4.4At 200=T K, ()⎪⎭⎫⎝⎛=3002000259.0kT017267.0=eVAt 400=T K, ()⎪⎭⎫ ⎝⎛=3004000259.0kT 034533.0=eVoror 318.1=g E eV Now so 371041.9⨯=o co N N υcm 6-_______________________________________4.5 For 200=T K, 017267.0=kT eV For 300=T K, 0259.0=kT eVFor 400=T K, 034533.0=kT eV(a) For 200=T K,(b) For 300=T K,(c) For 400=T K,_______________________________________4.6Let x E E c =-Then ⎪⎭⎫ ⎝⎛-∝kT x x f g F c exp To find the maximum value: which yields The maximum value occurs at (b)Let x E E =-υ Then ()⎪⎭⎫⎝⎛-∝-kT x x f g F exp 1υ To find the maximum value Same as part (a). Maximum occurs at or _______________________________________4.7 wherekT E E c 41+= and 22kT E E c += Then or _______________________________________ 4.8 Plot _______________________________________ 4.9 Plot _______________________________________4.10 Silicon: o p m m 56.0*=, o n m m 08.1*=0128.0-=-midgap Fi E E eV Germanium: o p m m 37.0*=,o n m m 55.0*=0077.0-=-midgap Fi EE eVGallium Arsenide: o p m m 48.0*=, 0382.0+=-midgapFi E E eV _______________________________________4.12 63.10-⇒meV 47.43+⇒meV _______________________________________ 4.13 Let ()==K E g c constant Then Let kT E E c-=η so that ηd kT dE ⋅= We can writeso that The integral can then be written as which becomes_______________________________________ 4.14Let ()()c c E E C E g -=1 for c E E ≥ Then LetkTE E c-=η so that ηd kT dE ⋅=We can write Then orWe find that So_______________________________________ 4.15We have ⎪⎪⎭⎫⎝⎛=∈*1m m a r o r o For germanium, 16=∈r , o m m 55.0*= Then orThe ionization energy can be written as ()6.132*⎪⎪⎭⎫⎝⎛∈∈⎪⎪⎭⎫ ⎝⎛=s o o m m E eV ()()029.06.131655.02=⇒=E eV_______________________________________ 4.16We have ⎪⎪⎭⎫⎝⎛=∈*1m m a r o r o For gallium arsenide, 1.13=∈r , ThenThe ionization energy is or0053.0=E eV_______________________________________ 4.172148.0=eV90518.02148.012.1=-=eV 31090.6⨯=cm 3- (a) Holes338.0=eV_______________________________________ 4.18162.0=eV 958.0162.012.1=-=eV 31041.2⨯=cm 3-365.0=eV_______________________________________ 4.198436.0=eV2764.0=-υE E F eV1410414.2⨯=cm 3- (a) p-type_______________________________________ 4.20(a) ()032375.03003750259.0=⎪⎭⎫⎝⎛=kT eV141015.1⨯=cm 3-14.1=eV31099.4⨯=cm 3-2154.0=eV 2046.1=eV 21042.4-⨯=cm 3-_______________________________________ 4.21(a) ()032375.03003750259.0=⎪⎭⎫⎝⎛=kT eV151086.6⨯= cm 3-840.0=eV 71084.7⨯=cm 3-2153.0=eV9047.02153.012.1=-=-υE E F eV 31004.7⨯=cm 3-_______________________________________ 4.22(a) p-type(b) 28.0412.14===-g F E E E υeV141010.2⨯=cm 3- 84.028.012.1=-=eV 51030.2⨯=cm 3-_______________________________________ 4.23131033.7⨯=cm 3- 61007.3⨯=cm 3- 91080.8⨯=cm 3-21068.3⨯=cm 3-_______________________________________ 4.241979.0=eV 92212.019788.012.1=-=eV 31066.9⨯=cm 3- (a) Holes3294.0=eV _______________________________________4.25()034533.03004000259.0=⎪⎭⎫⎝⎛=kT eV1910601.1⨯=cm 3-19103109.4⨯=cm 3-1210381.2⨯=⇒i n cm 3- 2787.0=eV(a) 84127.027873.012.1=-=-F c E E eV 910134.1⨯=cm 3- (b) Holes2642.0=eV _______________________________________ 4.26141050.4⨯=cm 3-17.125.042.1=-=-F c E E eV 21013.1-⨯=cm 3- (a) 034533.0=kT eV1910078.1⨯=cm 3- 1710236.7⨯=cm 3-3482.0=eV072.13482.042.1=-=-F c E E eV41040.2⨯=cm 3-_____________________________________ 4.27141068.6⨯=cm 3-870.025.012.1=-=-F c E E eV 41023.7⨯=o n cm 3- (a) 034533.0=kT eV1910601.1⨯=cm 3- 1910311.4⨯=cm 3- 3482.0=eV7718.03482.012.1=-=-F c E E eV 91049.8⨯=cm 3-_______________________________________ 4.28(a) ()F c o F N n ηπ2/12=For 2kT E E c F +=, Then ()0.12/1≅F F η 191016.3⨯=cm 3-(b) ()F c o F N n ηπ2/12=171030.5⨯=cm 3-_______________________________________ 4.29So ()26.42/1='FF η We find kTE E FF-=≅'υη0.3()()0777.00259.00.3==-F E E υeV_______________________________________ 4.30(a) 44==-=kTkTkT E E c F F ηThen ()0.62/1≅F F η 201090.1⨯=cm 3-(b) ()()0.6107.4217⨯=πo n181018.3⨯=cm 3-_______________________________________ 4.31For the electron concentrationThe Boltzmann approximation applies, so or Define ThenTo find maximum ()()x n E n →, set orwhich yieldsFor the hole concentrationUsing the Boltzmann approximation or Define ThenTo find maximum value of ()()x p E p '→, set()0=''x d x dp Using the results from above,we find the maximum at_______________________________________ 4.32(a) Silicon: We have We can write For045.0=-d c E E eV andkT E E F d 3=-eV we can write or171045.2⨯=o n cm 3-We also haveAgain, we can write ForkT E E a F 3=- and045.0=-υE E a eVThen or161012.9⨯=o p cm 3-(b) GaAs: assume 0058.0=-d c E E eV Then or161087.1⨯=o n cm 3-Assume 0345.0=-υE E a eV Then or161020.9⨯=o p cm 3-_______________________________________ 4.33Plot_______________________________________ 4.34(a) 151510310154⨯=-⨯=o p cm 3-()415210105.7103105.1⨯=⨯⨯=on cm 3-(b) 16103⨯==d o N n cm 3- ()316210105.7103105.1⨯=⨯⨯=op cm 3-(c) 10105.1⨯===i o o n p n cm 3- 1110334.7⨯=⇒i n cm 3-15104⨯==a o N p cm 3-()8152111034.110410334.7⨯=⨯⨯=o n cm 3-1310722.1⨯=⇒i n cm 3- 1410029.1⨯=cm 3-()12142131088.210029.110722.1⨯=⨯⨯=op cm 3-_______________________________________ 4.3515103⨯=cm 3- ()3152621008.1103108.1-⨯=⨯⨯==o i o p n n cm 3-(a) 16103⨯==d o N n cm 3-()416261008.1103108.1-⨯=⨯⨯=op cm 3-(b) 6108.1⨯===i o o n p n cm 3-810580.7⨯=⇒i n cm 3- 15104⨯==a o N p cm 3-()215281044.110410580.7⨯=⨯⨯=on cm 3-1010853.3⨯=⇒i n cm 3- 1410==d o N n cm 3-()7142101048.11010853.3⨯=⨯=op cm 3-_______________________________________ 4.36(a) Ge: 13104.2⨯=i n cm 3-(i)2222i d d o n N N n +⎪⎪⎭⎫⎝⎛+= or15102⨯=≅d o N n cm 3-111088.2⨯= cm 3- (ii)151610710⨯-=-≅d a o N N p 15103⨯=cm 3- 111092.1⨯=cm 3- (b) GaAs: 6108.1⨯=i n cm 3- (i)15102⨯=≅d o N n cm()315261062.1102108.1-⨯=⨯⨯=op cm 3-(ii)15103⨯=-≅d a o N N p cm 3-()315261008.1103108.1-⨯=⨯⨯=on cm 3-(c) The result implies that there is only one minority carrier in a volume of 310cm 3. _______________________________________ 4.37(a) For the donor level or(b) We have Now or Then or_______________________________________ 4.38(a) ⇒>d a N N p-type (b) Silicon: or13105.1⨯=o p cm 3- Then()7132102105.1105.1105.1⨯=⨯⨯==o i o p n n cm 3- Germanium: or131026.3⨯=o p cm 3- Then()131321321076.110264.3104.2⨯=⨯⨯==o i o p n n cm 3-Gallium Arsenide:13105.1⨯=-=d a o N N p cm 3- and()216.0105.1108.113262=⨯⨯==o i o p n n cm 3- _______________________________________ 4.39(a) ⇒>a d N N n-type14108⨯=cm 3-()51421021081.2108105.1⨯=⨯⨯==o i o n n p cm 3-15108.4⨯='⇒aN cm 3-()41521010625.5104105.1⨯=⨯⨯=o n cm 3-_______________________________________ 4.40()155210210125.1102105.1⨯=⨯⨯==o i o p n n cm 3- ⇒>o o p n n-type_______________________________________ 4.411210376.1⨯=⇒i n cm 3-So 111088.6⨯=o n cm 3-, Then 121075.2⨯=o p cm 3-so that 1210064.2⨯=a N cm 3-_______________________________________ 4.42Plot_______________________________________ 4.43Plot_______________________________________ 4.44Plot_______________________________________ 4.45so 131074.5⨯=i n cm 3-1314272103101.1103.3⨯=⨯⨯==o i o n n p cm 3- _______________________________________ 4.46(a) ⇒>d a N N p-typeMajority carriers are holes 16105.1⨯=cm 3-Minority carriers are electrons ()4162102105.1105.1105.1⨯=⨯⨯==o i o p n n cm 3- (b) Boron atoms must be addedSo 16105.3⨯='aN cm 3-()316210105.4105105.1⨯=⨯⨯=on cm 3-_______________________________________ 4.47(a) ⇒<<i o n p n-type on ()16421010125.1102105.1⨯=⨯⨯=cm 3-⇒electrons are majority carriers4102⨯=o p cm 3-⇒holes are minority carriers so 1610825.1⨯=d N cm 3-_______________________________________ 4.482222i a a o n N N p +⎪⎪⎭⎫⎝⎛+= and1510=N cm 3-4.49For 1410cm 3-, 3249.0=-F c E E eV 1510cm 3-, 2652.0=-F c E E eV 1610cm 3-, 2056.0=-F c E E eV 1710cm 3-, 1459.0=-F c E E eV For 1410cm 3-, 2280.0=-Fi F E E eV 1510cm 3-, 2877.0=-Fi F E E eV 1610cm 3-, 3473.0=-Fi F E E eV 1710cm 3-, 4070.0=-Fi F E E eV _______________________________________ 4.50151005.105.1⨯==d o N n cm 3- so 2821025.5⨯=i n NowBy trial and error, 5.536=T K (a) At 300=T K,2652.0=eV At 5.536=T K,()046318.03005.5360259.0=⎪⎭⎫⎝⎛=kT eV1910696.6⨯=cm 3-5124.0=eV then ()2472.0=-∆F c E E eV (b) Closer to the intrinsic energy level._______________________________________ 4.51At 200=T K, 017267.0=kT eV 400=T K, 034533.0=kT eV 600=T K, 0518.0=kT eV At 200=T K,410638.7⨯=⇒i n cm 3- At 400=T K,1210381.2⨯=⇒i n cm 3- At 600=T K,1410740.9⨯=⇒i n cm 3- At 200=T K and 400=T K, 15103⨯==a o N p cm 3- At 600=T K,1510288.3⨯=cm 3-Then, 200=T K, 4212.0=-F Fi E E eV400=T K, 2465.0=-F Fi E E eV600=T K, 0630.0=-F Fi E E eV_______________________________________ 4.52(a)For 1410=a N cm 3-,4619.0=-F Fi E E eV1510=a N cm 3-,5215.0=-F Fi E E eV1610=a N cm 3-,5811.0=-F Fi E E eV1710=a N cm 3-,6408.0=-F Fi E E eV (b)For 1410=a N cm 3-, 2889.0=-υE E F eV1510=a N cm 3-,2293.0=-υE E F eV1610=a N cm 3-,1697.0=-υE E F eV1710=a N cm 3-,1100.0=-υE E F eV_______________________________________ 4.53 or0447.0+=-midgap Fi E E eV(a) Impurity atoms to be added so 45.0=-F midgap E E eV(i) p-type, so add acceptor atoms (ii)4947.045.00447.0=+=-F Fi E E eV Then or131097.1⨯==a o N p cm 3-_______________________________________ 4.54 so or16102.1⨯=d N cm 3-_______________________________________ 4.55(a) Silicon(i)⎪⎪⎭⎫⎝⎛=-d c F c N N kT E E ln()2188.0106108.2ln 0259.01519=⎪⎪⎭⎫⎝⎛⨯⨯=eV(ii)1929.00259.02188.0=-=-F c E E eV1610631.1⨯=d N cm 3-15106⨯+'=dN 1610031.1⨯='⇒dN cm 3- Additionaldonor atoms(b) GaAs(i)()⎪⎪⎭⎫ ⎝⎛⨯=-151710107.4ln 0259.0F c E E 15936.0=eV(ii)13346.00259.015936.0=-=-F c E E eV1510718.2⨯=cm 3-1510+'=dN 1510718.1⨯='⇒dN cm 3- Additionaldonor atoms _______________________________________4.56()⎪⎪⎭⎫ ⎝⎛⨯⨯=16191021004.1ln 0259.01620.0=eV ()1876.0102108.2ln 0259.01619=⎪⎪⎭⎫ ⎝⎛⨯⨯=eV(a) For part (a);16102⨯=o p cm 3- 410125.1⨯=cm 3- For part (b):16102⨯=o n cm 3- 410125.1⨯=cm 3-_______________________________________ 4.5715100.3⨯=cm 3- Add additional acceptor impurities 15104⨯=⇒a N cm 3-_______________________________________ 4.58()3161.0105.1103ln 0259.01015=⎪⎪⎭⎫⎝⎛⨯⨯=eV()3758.0105.1103ln 0259.01016=⎪⎪⎭⎫⎝⎛⨯⨯=eV2786.0=eV 06945.0=eV _______________________________________ 4.59 ()2009.0103100.7ln 0259.01518=⎪⎪⎭⎫ ⎝⎛⨯⨯=eV 360.1=eV7508.0=eV 2526.0=eV 068.1=eV_______________________________________ 4.60n-type ()3504.0105.110125.1ln 0259.01016=⎪⎪⎭⎫⎝⎛⨯⨯=eV ______________________________________ 4.61 ()030217.03003500259.0=⎪⎭⎫ ⎝⎛=kT eV ()1921910633.1300350102.1⨯=⎪⎭⎫ ⎝⎛⨯=c N cm 3-()192191045.2300350108.1⨯=⎪⎭⎫ ⎝⎛⨯=υN cm 3- Now So6257.0=⇒g E eV_______________________________________ 4.62(a) Replace Ga atoms ⇒Silicon acts as adonor()()1415105.310705.0⨯=⨯=d N cm 3-Replace As atoms ⇒Silicon acts as anacceptor()()15151065.610795.0⨯=⨯=a N cm 3-(b) ⇒>d a N N p-type(c) 1415105.31065.6⨯-⨯=-=d a o N N p 15103.6⨯=cm 3-()4152621014.5103.6108.1-⨯=⨯⨯==o i o p n n cm 3-()5692.0108.1103.6ln 0259.0615=⎪⎪⎭⎫⎝⎛⨯⨯=eV_______________________________________。
半导体物理与器件第四版第一章课后答案
Number density
5.196 10
2
8 3
1.4257 10 22 cm 3
Mass density
1.4257 10 12.5
22
6.02 10 23
0.296 gm/cm 3 _______________________________________
1.10 From Problem 1.2, percent volume of fcc atoms is 74%; Therefore after coffee is ground, Volume = 0.74 cm 3 _______________________________________ 1.11 (b) a 1.8 1.0 2.8 A (c) Na: Density
o
Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 1 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ (b) Number density a 8 2 5 10 22 cm 3 3 2 5.43 10 8 sin 2 54.74 3 2 2 a 3 (c) Mass density 2 22 N At.Wt. 5 10 28.09 109.5 NA 6.02 10 23 _______________________________________ 2.33 grams/cm 3 1.7 _______________________________________ o (a) Simple cubic: a 2r 3.9 A
半导体物理与器件第四版课后习题答案3
Chapter 33.1If o a were to increase, the bandgap energy would decrease and the material would begin to behave less like a semiconductor and more like a metal. If o a were to decrease, the bandgap energy would increase and thematerial would begin to behave more like an insulator._______________________________________ 3.2Schrodinger's wave equation is:()()()t x x V xt x m ,,2222ψ⋅+∂ψ∂- ()tt x j ∂ψ∂=, Assume the solution is of the form:()()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-=ψt E kx j x u t x exp , Region I: ()0=x V . Substituting theassumed solution into the wave equation, we obtain:()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-⎩⎨⎧∂∂-t E kx j x jku x m exp 22 ()⎪⎭⎪⎬⎫⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-∂∂+t E kx j x x u exp ()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-⋅⎪⎭⎫ ⎝⎛-=t E kx j x u jE j exp which becomes()()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-⎩⎨⎧-t E kx j x u jk m exp 222 ()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-∂∂+t E kx j x x u jkexp 2 ()⎪⎭⎪⎬⎫⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-∂∂+t E kx j x x u exp 22 ()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-+=t E kx j x Eu exp This equation may be written as()()()()0222222=+∂∂+∂∂+-x u mE x x u x x u jk x u kSetting ()()x u x u 1= for region I, the equation becomes:()()()()021221212=--+x u k dx x du jk dxx u d α where222mE=α Q.E.D.In Region II, ()O V x V =. Assume the same form of the solution:()()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-=ψt E kx j x u t x exp , Substituting into Schrodinger's wave equation, we find:()()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-⎩⎨⎧-t E kx j x u jk m exp 222 ()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-∂∂+t E kx j x x u jkexp 2 ()⎪⎭⎪⎬⎫⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-∂∂+t E kx j x x u exp 22 ()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-+t E kx j x u V O exp ()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-=t E kx j x Eu exp This equation can be written as:()()()2222x x u x x u jk x u k ∂∂+∂∂+- ()()02222=+-x u mEx u mV OSetting ()()x u x u 2= for region II, this equation becomes()()dx x du jk dxx u d 22222+ ()022222=⎪⎪⎭⎫ ⎝⎛+--x u mV k O α where again222mE=α Q.E.D._______________________________________3.3We have()()()()021221212=--+x u k dx x du jk dxx u d α Assume the solution is of the form: ()()[]x k j A x u -=αexp 1()[]x k j B +-+αexp The first derivative is()()()[]x k j A k j dxx du --=ααexp 1 ()()[]x k j B k j +-+-ααexp and the second derivative becomes()()[]()[]x k j A k j dxx u d --=ααexp 2212 ()[]()[]x k j B k j +-++ααexp 2Substituting these equations into the differential equation, we find()()[]x k j A k ---ααexp 2()()[]x k j B k +-+-ααexp 2(){()[]x k j A k j jk --+ααexp 2()()[]}x k j B k j +-+-ααexp ()()[]{x k j A k ---ααexp 22 ()[]}0exp =+-+x k j B α Combining terms, we obtain()()()[]222222αααα----+--k k k k k ()[]x k j A -⨯αexp()()()[]222222αααα--++++-+k k k k k ()[]0exp =+-⨯x k j B α We find that00= Q.E.D. For the differential equation in ()x u 2 and the proposed solution, the procedure is exactly the same as above._______________________________________ 3.4We have the solutions ()()[]x k j A x u -=αexp 1()[]x k j B +-+αexp for a x <<0 and()()[]x k j C x u -=βexp 2()[]x k j D +-+βexp for 0<<-x b .The first boundary condition is ()()0021u u =which yields0=--+D C B AThe second boundary condition is201===x x dx dudx du which yields()()()C k B k A k --+--βαα()0=++D k β The third boundary condition is ()()b u a u -=21 which yields()[]()[]a k j B a k j A +-+-ααexp exp ()()[]b k j C --=βexp()()[]b k j D -+-+βexp and can be written as()[]()[]a k j B a k j A +-+-ααexp exp ()[]b k j C ---βexp()[]0exp =+-b k j D β The fourth boundary condition isbx a x dx dudx du -===21 which yields()()[]a k j A k j --ααexp()()[]a k j B k j +-+-ααexp ()()()[]b k j C k j ---=ββexp()()()[]b k j D k j -+-+-ββexp and can be written as ()()[]a k j A k --ααexp()()[]a k j B k +-+-ααexp()()[]b k j C k ----ββexp()()[]0exp =+++b k j D k ββ_______________________________________ 3.5(b) (i) First point: πα=aSecond point: By trial and error, πα729.1=a (ii) First point: πα2=aSecond point: By trial and error, πα617.2=a_______________________________________3.6(b) (i) First point: πα=aSecond point: By trial and error, πα515.1=a (ii) First point: πα2=aSecond point: By trial and error, πα375.2=a_______________________________________ 3.7ka a aaP cos cos sin =+'αααLet y ka =, x a =α Theny x x xP cos cos sin =+'Consider dy dof this function.()[]{}y x x x P dy d sin cos sin 1-=+⋅'- We find()()()⎭⎬⎫⎩⎨⎧⋅+⋅-'--dy dx x x dy dx x x P cos sin 112y dydxx sin sin -=- Theny x x x x x P dy dx sin sin cos sin 12-=⎭⎬⎫⎩⎨⎧-⎥⎦⎤⎢⎣⎡+-'For πn ka y ==, ...,2,1,0=n 0sin =⇒y So that, in general,()()dk d ka d a d dy dxαα===0 And 22 mE=α Sodk dEm mE dk d ⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛=-22/122221 α This implies thatdk dE dk d ==0α for an k π= _______________________________________ 3.8(a) πα=a 1π=⋅a E m o 212()()()()2103123422221102.41011.9210054.12---⨯⨯⨯==ππa m E o19104114.3-⨯=J From Problem 3.5 πα729.12=aπ729.1222=⋅a E m o()()()()2103123422102.41011.9210054.1729.1---⨯⨯⨯=πE18100198.1-⨯=J 12E E E -=∆1918104114.3100198.1--⨯-⨯= 19107868.6-⨯=Jor 24.4106.1107868.61919=⨯⨯=∆--E eV(b) πα23=aπ2223=⋅a E m o()()()()2103123423102.41011.9210054.12---⨯⨯⨯=πE18103646.1-⨯=J From Problem 3.5, πα617.24=aπ617.2224=⋅a E m o()()()()2103123424102.41011.9210054.1617.2---⨯⨯⨯=πE18103364.2-⨯=J 34E E E -=∆1818103646.1103364.2--⨯-⨯= 1910718.9-⨯=Jor 07.6106.110718.91919=⨯⨯=∆--E eV_______________________________________3.9(a) At π=ka , πα=a 1π=⋅a E m o 212()()()()2103123421102.41011.9210054.1---⨯⨯⨯=πE19104114.3-⨯=JAt 0=ka , By trial and error, πα859.0=a o ()()()()210312342102.41011.9210054.1859.0---⨯⨯⨯=πoE19105172.2-⨯=J o E E E -=∆11919105172.2104114.3--⨯-⨯= 2010942.8-⨯=Jor 559.0106.110942.81920=⨯⨯=∆--E eV (b) At π2=ka , πα23=aπ2223=⋅a E m o()()()()2103123423102.41011.9210054.12---⨯⨯⨯=πE18103646.1-⨯=JAt π=ka . From Problem 3.5, πα729.12=aπ729.1222=⋅a E m o()()()()2103123422102.41011.9210054.1729.1---⨯⨯⨯=πE18100198.1-⨯=J23E E E -=∆1818100198.1103646.1--⨯-⨯= 19104474.3-⨯=Jor 15.2106.1104474.31919=⨯⨯=∆--E eV_______________________________________3.10(a) πα=a 1π=⋅a E m o 212()()()()2103123421102.41011.9210054.1---⨯⨯⨯=πE19104114.3-⨯=JFrom Problem 3.6, πα515.12=aπ515.1222=⋅a E m o()()()()2103123422102.41011.9210054.1515.1---⨯⨯⨯=πE1910830.7-⨯=J 12E E E -=∆1919104114.310830.7--⨯-⨯= 19104186.4-⨯=Jor 76.2106.1104186.41919=⨯⨯=∆--E eV (b) πα23=aπ2223=⋅a E m o()()()()2103123423102.41011.9210054.12---⨯⨯⨯=πE18103646.1-⨯=JFrom Problem 3.6, πα375.24=aπ375.2224=⋅a E m o()()()()2103123424102.41011.9210054.1375.2---⨯⨯⨯=πE18109242.1-⨯=J 34E E E -=∆1818103646.1109242.1--⨯-⨯= 1910597.5-⨯=Jor 50.3106.110597.51919=⨯⨯=∆--E eV_____________________________________3.11(a) At π=ka , πα=a 1π=⋅a E m o 212()()()()2103123421102.41011.9210054.1---⨯⨯⨯=πE19104114.3-⨯=JAt 0=ka , By trial and error, πα727.0=a oπ727.022=⋅a E m o o()()()()210312342102.41011.9210054.1727.0---⨯⨯⨯=πo E19108030.1-⨯=Jo E E E -=∆11919108030.1104114.3--⨯-⨯= 19106084.1-⨯=Jor 005.1106.1106084.11919=⨯⨯=∆--E eV (b) At π2=ka , πα23=aπ2223=⋅a E m o()()()()2103123423102.41011.9210054.12---⨯⨯⨯=πE18103646.1-⨯=JAt π=ka , From Problem 3.6,πα515.12=aπ515.1222=⋅a E m o()()()()2103423422102.41011.9210054.1515.1---⨯⨯⨯=πE1910830.7-⨯=J23E E E -=∆191810830.7103646.1--⨯-⨯= 1910816.5-⨯=Jor 635.3106.110816.51919=⨯⨯=∆--E eV_______________________________________3.12For 100=T K, ()()⇒+⨯-=-1006361001073.4170.124gE164.1=g E eV200=T K, 147.1=g E eV 300=T K, 125.1=g E eV 400=T K, 097.1=g E eV 500=T K, 066.1=g E eV 600=T K, 032.1=g E eV_______________________________________3.13The effective mass is given by1222*1-⎪⎪⎭⎫⎝⎛⋅=dk E d mWe have()()B curve dkE d A curve dk E d 2222> so that ()()B curve m A curve m **<_______________________________________ 3.14The effective mass for a hole is given by1222*1-⎪⎪⎭⎫ ⎝⎛⋅=dk E d m p We have that()()B curve dkEd A curve dk E d 2222> so that ()()B curve m A curve m p p **<_______________________________________ 3.15Points A,B: ⇒<0dk dEvelocity in -x directionPoints C,D: ⇒>0dk dEvelocity in +x directionPoints A,D: ⇒<022dk Ednegative effective massPoints B,C: ⇒>022dkEd positive effective mass _______________________________________3.16For A: 2k C E i =At 101008.0+⨯=k m 1-, 05.0=E eV Or ()()2119108106.105.0--⨯=⨯=E J So ()2101211008.0108⨯=⨯-C3811025.1-⨯=⇒CNow ()()38234121025.1210054.12--*⨯⨯==C m 311044.4-⨯=kgor o m m ⋅⨯⨯=--*31311011.9104437.4o m m 488.0=* For B: 2k C E i =At 101008.0+⨯=k m 1-, 5.0=E eV Or ()()2019108106.15.0--⨯=⨯=E JSo ()2101201008.0108⨯=⨯-C 3711025.1-⨯=⇒CNow ()()37234121025.1210054.12--*⨯⨯==C m 321044.4-⨯=kg or o m m ⋅⨯⨯=--*31321011.9104437.4o m m 0488.0=*_______________________________________ 3.17For A: 22k C E E -=-υ()()()2102191008.0106.1025.0⨯-=⨯--C 3921025.6-⨯=⇒C()()39234221025.6210054.12--*⨯⨯-=-=C m31108873.8-⨯-=kgor o m m ⋅⨯⨯-=--*31311011.9108873.8o m m 976.0--=* For B: 22k C E E -=-υ()()()2102191008.0106.13.0⨯-=⨯--C 382105.7-⨯=⇒C()()3823422105.7210054.12--*⨯⨯-=-=C m3210406.7-⨯-=kgor o m m ⋅⨯⨯-=--*31321011.910406.7o m m 0813.0-=*_______________________________________ 3.18(a) (i) νh E =or ()()341910625.6106.142.1--⨯⨯==h E ν1410429.3⨯=Hz(ii) 141010429.3103⨯⨯===νλc E hc 51075.8-⨯=cm 875=nm(b) (i) ()()341910625.6106.112.1--⨯⨯==h E ν1410705.2⨯=Hz(ii) 141010705.2103⨯⨯==νλc410109.1-⨯=cm 1109=nm_______________________________________ 3.19(c) Curve A: Effective mass is a constantCurve B: Effective mass is positive around 0=k , and is negativearound 2π±=k . _______________________________________ 3.20()[]O O k k E E E --=αcos 1 Then()()()[]O k k E dkdE ---=ααsin 1()[]O k k E -+=ααsin 1 and()[]O k k E dk E d -=ααcos 2122Then221222*11 αE dk Ed m o k k =⋅== or212*αE m =_______________________________________ 3.21(a) ()[]3/123/24lt dn m m m =*()()[]3/123/264.1082.04oom m =o dn m m 56.0=*(b)o o l t cnm m m m m 64.11082.02123+=+=*oo m m 6098.039.24+=o cn m m 12.0=*_______________________________________ 3.22(a) ()()[]3/22/32/3lh hh dp m m m +=*()()[]3/22/32/3082.045.0o om m +=[]o m ⋅+=3/202348.030187.0o dp m m 473.0=*(b) ()()()()2/12/12/32/3lh hh lh hh cpm m m m m ++=*()()()()om ⋅++=2/12/12/32/3082.045.0082.045.0 o cp m m 34.0=*_______________________________________ 3.23For the 3-dimensional infinite potential well, ()0=x V when a x <<0, a y <<0, and a z <<0. In this region, the wave equation is:()()()222222,,,,,,z z y x y z y x x z y x ∂∂+∂∂+∂∂ψψψ()0,,22=+z y x mEψ Use separation of variables technique, so let ()()()()z Z y Y x X z y x =,,ψSubstituting into the wave equation, we have222222zZXY y Y XZ x X YZ ∂∂+∂∂+∂∂ 022=⋅+XYZ mEDividing by XYZ , we obtain021*********=+∂∂⋅+∂∂⋅+∂∂⋅ mEz Z Z y Y Y x X XLet01222222=+∂∂⇒-=∂∂⋅X k x X k x X X xx The solution is of the form: ()x k B x k A x X x x cos sin +=Since ()0,,=z y x ψ at 0=x , then ()00=X so that 0=B .Also, ()0,,=z y x ψ at a x =, so that ()0=a X . Then πx x n a k = where ...,3,2,1=x n Similarly, we have2221y k y Y Y -=∂∂⋅ and 2221z k zZ Z -=∂∂⋅From the boundary conditions, we find πy y n a k = and πz z n a k = where...,3,2,1=y n and ...,3,2,1=z n From the wave equation, we can write022222=+---mE k k k z y xThe energy can be written as()222222⎪⎭⎫ ⎝⎛++==a n n n m E E z y x n n n z y x π _______________________________________ 3.24The total number of quantum states in the 3-dimensional potential well is given (in k-space) by()332a dk k dk k g T ⋅=ππ where222 mEk =We can then writemEk 2=Taking the differential, we obtaindE Em dE E m dk ⋅⋅=⋅⋅⋅⋅=2112121 Substituting these expressions into the density of states function, we have()dE E mmE a dE E g T ⋅⋅⋅⎪⎭⎫ ⎝⎛=212233 ππ Noting thatπ2h=this density of states function can be simplified and written as()()dE E m h a dE E g T ⋅⋅=2/33324π Dividing by 3a will yield the density of states so that()()E h m E g ⋅=32/324π _______________________________________ 3.25For a one-dimensional infinite potential well,222222k a n E m n ==*π Distance between quantum states()()aa n a n k k n n πππ=⎪⎭⎫ ⎝⎛=⎪⎭⎫ ⎝⎛+=-+11Now()⎪⎭⎫ ⎝⎛⋅=a dkdk k g T π2NowE m k n *⋅=21dE Em dk n⋅⋅⋅=*2211 Then()dE Em a dE E g n T ⋅⋅⋅=*2212 π Divide by the "volume" a , so ()Em E g n *⋅=21πSo()()()()()EE g 31341011.9067.0210054.11--⨯⋅⨯=π ()EE g 1810055.1⨯=m 3-J 1-_______________________________________ 3.26(a) Silicon, o n m m 08.1=*()()c nc E E h m E g -=*32/324π()dE E E h m g kTE E c nc c c⋅-=⎰+*232/324π()()kT E E c nc cE E h m 22/332/33224+*-⋅⋅=π()()2/332/323224kT hm n⋅⋅=*π ()()[]()()2/33342/33123210625.61011.908.124kT ⋅⋅⨯⨯=--π ()()2/355210953.7kT ⨯=(i) At 300=T K, 0259.0=kT eV()()19106.10259.0-⨯= 2110144.4-⨯=J Then ()()[]2/3215510144.4210953.7-⨯⨯=c g25100.6⨯=m 3- or 19100.6⨯=c g cm 3-(ii) At 400=T K, ()⎪⎭⎫⎝⎛=3004000259.0kT034533.0=eV()()19106.1034533.0-⨯= 21105253.5-⨯=J Then()()[]2/32155105253.5210953.7-⨯⨯=c g2510239.9⨯=m 3- or 191024.9⨯=c g cm 3-(b) GaAs, o nm m 067.0=*()()[]()()2/33342/33123210625.61011.9067.024kT g c ⋅⋅⨯⨯=--π ()()2/3542102288.1kT ⨯=(i) At 300=T K, 2110144.4-⨯=kT J ()()[]2/3215410144.42102288.1-⨯⨯=c g2310272.9⨯=m 3- or 171027.9⨯=c g cm 3-(ii) At 400=T K, 21105253.5-⨯=kT J ()()[]2/32154105253.52102288.1-⨯⨯=c g2410427.1⨯=m 3-181043.1⨯=c g cm 3-_______________________________________ 3.27(a) Silicon, o p m m 56.0=* ()()E E h mE g p-=*υυπ32/324()dE E E h mg E kTE p⋅-=⎰-*υυυυπ332/324()()υυυπE kTE pE E hm 32/332/33224-*-⎪⎭⎫ ⎝⎛-=()()[]2/332/333224kT hmp-⎪⎭⎫ ⎝⎛-=*π ()()[]()()2/33342/33133210625.61011.956.024kT ⎪⎭⎫ ⎝⎛⨯⨯=--π ()()2/355310969.2kT ⨯=(i)At 300=T K, 2110144.4-⨯=kT J ()()[]2/3215510144.4310969.2-⨯⨯=υg2510116.4⨯=m3-or 191012.4⨯=υg cm 3- (ii)At 400=T K, 21105253.5-⨯=kT J()()[]2/32155105253.5310969.2-⨯⨯=υg2510337.6⨯=m3-or 191034.6⨯=υg cm 3- (b) GaAs, o p m m 48.0=*()()[]()()2/33342/33133210625.61011.948.024kT g ⎪⎭⎫ ⎝⎛⨯⨯=--πυ ()()2/3553103564.2kT ⨯=(i)At 300=T K, 2110144.4-⨯=kT J()()[]2/3215510144.43103564.2-⨯⨯=υg2510266.3⨯=m 3- or 191027.3⨯=υg cm 3-(ii)At 400=T K, 21105253.5-⨯=kT J()()[]2/32155105253.53103564.2-⨯⨯=υg2510029.5⨯=m 3-or 191003.5⨯=υg cm 3-_______________________________________ 3.28(a) ()()c nc E E h m E g -=*32/324π()()[]()c E E -⨯⨯=--3342/33110625.61011.908.124πc E E -⨯=56101929.1 For c E E =; 0=c g1.0+=c E E eV; 4610509.1⨯=c g m 3-J 1-2.0+=c E E eV; 4610134.2⨯=m 3-J 1-3.0+=c E E eV; 4610614.2⨯=m 3-J 1- 4.0+=c E E eV; 4610018.3⨯=m 3-J 1- (b) ()E E h m g p-=*υυπ32/324()()[]()E E -⨯⨯=--υπ3342/33110625.61011.956.024E E -⨯=υ55104541.4 For υE E =; 0=υg1.0-=υE E eV; 4510634.5⨯=υg m 3-J 1-2.0-=υE E eV; 4510968.7⨯=m 3-J 1-3.0-=υE E eV; 4510758.9⨯=m 3-J 1-4.0-=υE E eV; 4610127.1⨯=m 3-J 1-_______________________________________ 3.29(a) ()()68.256.008.12/32/32/3=⎪⎭⎫ ⎝⎛==**pnc m m g g υ(b) ()()0521.048.0067.02/32/32/3=⎪⎭⎫ ⎝⎛==**pncmm g g υ_______________________________________3.30 Plot_______________________________________ 3.31(a) ()()()!710!7!10!!!-=-=i i i i i N g N g W()()()()()()()()()()()()1201238910!3!7!78910===(b) (i) ()()()()()()()()12!10!101112!1012!10!12=-=i W 66=(ii) ()()()()()()()()()()()()1234!8!89101112!812!8!12=-=i W 495=_______________________________________ 3.32()⎪⎪⎭⎫ ⎝⎛-+=kT E E E f F exp 11(a) kT E E F =-, ()()⇒+=1exp 11E f()269.0=E f (b) kT E E F 5=-, ()()⇒+=5exp 11E f()31069.6-⨯=E f(c) kT E E F 10=-, ()()⇒+=10exp 11E f ()51054.4-⨯=E f_______________________________________ 3.33()⎪⎪⎭⎫ ⎝⎛-+-=-kT E E E f F exp 1111or()⎪⎪⎭⎫ ⎝⎛-+=-kT E E E f F exp 111(a) kT E E F =-, ()269.01=-E f (b) kT E E F 5=-, ()31069.61-⨯=-E f(c) kT E E F 10=-, ()51054.41-⨯=-E f_______________________________________3.34(a) ()⎥⎦⎤⎢⎣⎡--≅kT E E f F F exp c E E =; 61032.90259.030.0exp -⨯=⎥⎦⎤⎢⎣⎡-=F f 2kT E c +; ()⎥⎦⎤⎢⎣⎡+-=0259.020259.030.0exp F f 61066.5-⨯=kT E c +; ()⎥⎦⎤⎢⎣⎡+-=0259.00259.030.0exp F f 61043.3-⨯=23kT E c +; ()()⎥⎦⎤⎢⎣⎡+-=0259.020259.0330.0exp F f 61008.2-⨯=kT E c 2+; ()()⎥⎦⎤⎢⎣⎡+-=0259.00259.0230.0exp F f 61026.1-⨯=(b) ⎥⎦⎤⎢⎣⎡-+-=-kT E E f F F exp 1111()⎥⎦⎤⎢⎣⎡--≅kT E E F exp υE E =; ⎥⎦⎤⎢⎣⎡-=-0259.025.0exp 1F f 51043.6-⨯= 2kT E -υ; ()⎥⎦⎤⎢⎣⎡+-=-0259.020259.025.0exp 1F f 51090.3-⨯=kT E -υ; ()⎥⎦⎤⎢⎣⎡+-=-0259.00259.025.0exp 1F f 51036.2-⨯=23kTE -υ; ()()⎥⎦⎤⎢⎣⎡+-=-0259.020259.0325.0exp 1F f 51043.1-⨯= kT E 2-υ;()()⎥⎦⎤⎢⎣⎡+-=-0259.00259.0225.0exp 1F f 61070.8-⨯=_______________________________________3.35()()⎥⎦⎤⎢⎣⎡-+-=⎥⎦⎤⎢⎣⎡--=kT E kT E kT E E f F c F F exp exp and()⎥⎦⎤⎢⎣⎡--=-kT E E f F F exp 1 ()()⎥⎦⎤⎢⎣⎡---=kT kT E E F υexp So ()⎥⎦⎤⎢⎣⎡-+-kT E kT E F c exp ()⎥⎦⎤⎢⎣⎡+--=kT kT E E F υexp Then kT E E E kT E F F c +-=-+υOr midgap c F E E E E =+=2υ_______________________________________ 3.3622222ma n E n π =For 6=n , Filled state()()()()()2103122234610121011.92610054.1---⨯⨯⨯=πE18105044.1-⨯=Jor 40.9106.1105044.119186=⨯⨯=--E eV For 7=n , Empty state()()()()()2103122234710121011.92710054.1---⨯⨯⨯=πE1810048.2-⨯=Jor 8.12106.110048.219187=⨯⨯=--E eV Therefore 8.1240.9<<F E eV_______________________________________ 3.37(a) For a 3-D infinite potential well()222222⎪⎭⎫ ⎝⎛++=a n n n mE z y x π For 5 electrons, the 5th electron occupies the quantum state 1,2,2===z y x n n n ; so()2222252⎪⎭⎫ ⎝⎛++=a n n n m E z y x π()()()()()21031222223410121011.9212210054.1---⨯⨯++⨯=π1910761.3-⨯=Jor 35.2106.110761.319195=⨯⨯=--E eV For the next quantum state, which is empty, the quantum state is 2,2,1===z y x n n n . This quantum state is at the same energy, so 35.2=F E eV(b) For 13 electrons, the 13th electronoccupies the quantum state 3,2,3===z y x n n n ; so ()()()()()2103122222341310121011.9232310054.1---⨯⨯++⨯=πE 1910194.9-⨯=Jor 746.5106.110194.9191913=⨯⨯=--E eVThe 14th electron would occupy the quantum state 3,3,2===z y x n n n . This state is at the same energy, so 746.5=F E eV_______________________________________ 3.38The probability of a state at E E E F ∆+=1 being occupied is()⎪⎭⎫ ⎝⎛∆+=⎪⎪⎭⎫ ⎝⎛-+=kT E kT E E E f F exp 11exp 11111 The probability of a state at E E E F ∆-=2being empty is()⎪⎪⎭⎫ ⎝⎛-+-=-kT E E E f F 222exp 1111⎪⎭⎫ ⎝⎛∆-+⎪⎭⎫ ⎝⎛∆-=⎪⎭⎫ ⎝⎛∆-+-=kT E kT E kT E exp 1exp exp 111or()⎪⎭⎫ ⎝⎛∆+=-kT E E f exp 11122so ()()22111E f E f -= Q.E.D. _______________________________________3.39(a) At energy 1E , we want01.0exp 11exp 11exp 1111=⎪⎪⎭⎫ ⎝⎛-+⎪⎪⎭⎫ ⎝⎛-+-⎪⎪⎭⎫ ⎝⎛-kT E E kT E E kT E E F F FThis expression can be written as01.01exp exp 111=-⎪⎪⎭⎫ ⎝⎛-⎪⎪⎭⎫ ⎝⎛-+kT E E kT E E F F or()⎪⎪⎭⎫⎝⎛-=kT E E F 1exp 01.01Then()100ln 1kT E E F += orkT E E F 6.41+= (b)At kT E E F 6.4+=, ()()6.4exp 11exp 1111+=⎪⎪⎭⎫ ⎝⎛-+=kT E E E f F which yields()01.000990.01≅=E f_______________________________________ 3.40 (a)()()⎥⎦⎤⎢⎣⎡--=⎥⎦⎤⎢⎣⎡--=0259.050.580.5exp exp kT E E f F F 61032.9-⨯=(b) ()060433.03007000259.0=⎪⎭⎫⎝⎛=kT eV31098.6060433.030.0exp -⨯=⎥⎦⎤⎢⎣⎡-=F f (c) ()⎥⎦⎤⎢⎣⎡--≅-kT E E f F F exp 1 ⎥⎦⎤⎢⎣⎡-=kT 25.0exp 02.0or 5002.0125.0exp ==⎥⎦⎤⎢⎣⎡+kT ()50ln 25.0=kTor()()⎪⎭⎫⎝⎛===3000259.0063906.050ln 25.0T kT which yields 740=T K_______________________________________ 3.41 (a)()00304.00259.00.715.7exp 11=⎪⎭⎫ ⎝⎛-+=E for 0.304%(b) At 1000=T K, 08633.0=kT eV Then()1496.008633.00.715.7exp 11=⎪⎭⎫ ⎝⎛-+=E for 14.96%(c) ()997.00259.00.785.6exp 11=⎪⎭⎫ ⎝⎛-+=E for 99.7% (d)At F E E =, ()21=E f for all temperatures_______________________________________ 3.42(a) For 1E E =()()⎥⎦⎤⎢⎣⎡--≅⎪⎪⎭⎫ ⎝⎛-+=kT E E kTE E E fF F11exp exp 11Then()611032.90259.030.0exp -⨯=⎪⎭⎫ ⎝⎛-=E fFor 2E E =, 82.030.012.12=-=-E E F eV Then()⎪⎭⎫ ⎝⎛-+-=-0259.082.0exp 1111E for()⎥⎦⎤⎢⎣⎡⎪⎭⎫ ⎝⎛---≅-0259.082.0exp 111E f141078.10259.082.0exp -⨯=⎪⎭⎫ ⎝⎛-=(b) For 4.02=-E E F eV,72.01=-F E E eVAt 1E E =,()()⎪⎭⎫⎝⎛-=⎥⎦⎤⎢⎣⎡--=0259.072.0exp exp 1kT E E E f F or()131045.8-⨯=E f At 2E E =,()()⎥⎦⎤⎢⎣⎡--=-kT E E E f F 2exp 1 ⎪⎭⎫ ⎝⎛-=0259.04.0expor()71096.11-⨯=-E f_______________________________________ 3.43(a) At 1E E =()()⎪⎭⎫⎝⎛-=⎥⎦⎤⎢⎣⎡--=0259.030.0exp exp 1kT E E E f F or()61032.9-⨯=E fAt 2E E =, 12.13.042.12=-=-E E F eV So()()⎥⎦⎤⎢⎣⎡--=-kT E E E f F 2exp 1 ⎪⎭⎫ ⎝⎛-=0259.012.1expor()191066.11-⨯=-E f (b) For 4.02=-E E F ,02.11=-F E E eV At 1E E =,()()⎪⎭⎫⎝⎛-=⎥⎦⎤⎢⎣⎡--=0259.002.1exp exp 1kT E E E f F or()181088.7-⨯=E f At 2E E =,()()⎥⎦⎤⎢⎣⎡--=-kT E E E f F 2exp 1 ⎪⎭⎫ ⎝⎛-=0259.04.0expor ()71096.11-⨯=-E f_______________________________________ 3.44()1exp 1-⎥⎦⎤⎢⎣⎡⎪⎪⎭⎫ ⎝⎛-+=kTE E E f Fso()()2exp 11-⎥⎦⎤⎢⎣⎡⎪⎪⎭⎫ ⎝⎛-+-=kT E E dE E df F⎪⎪⎭⎫ ⎝⎛-⎪⎭⎫⎝⎛⨯kT E E kT F exp 1or()2exp 1exp 1⎥⎦⎤⎢⎣⎡⎪⎪⎭⎫ ⎝⎛-+⎪⎪⎭⎫ ⎝⎛-⎪⎭⎫⎝⎛-=kT E E kT E E kT dE E df F F (a) At 0=T K, For()00exp =⇒=∞-⇒<dE dfE E F()0exp =⇒+∞=∞+⇒>dEdfE E FAt -∞=⇒=dEdfE E F(b) At 300=T K, 0259.0=kT eVFor F E E <<, 0=dE dfFor F E E >>, 0=dEdfAt F E E =,()()65.91110259.012-=+⎪⎭⎫ ⎝⎛-=dE df (eV)1-(c) At 500=T K, 04317.0=kT eVFor F E E <<, 0=dE dfFor F E E >>, 0=dEdfAt F E E =,()()79.511104317.012-=+⎪⎭⎫ ⎝⎛-=dE df (eV)1- _______________________________________ 3.45(a) At midgap E E =,()⎪⎪⎭⎫⎝⎛+=⎪⎪⎭⎫ ⎝⎛-+=kT E kTE E E f g F2exp 11exp 11Si: 12.1=g E eV, ()()⎥⎦⎤⎢⎣⎡+=0259.0212.1exp 11E for()101007.4-⨯=E fGe: 66.0=g E eV ()()⎥⎦⎤⎢⎣⎡+=0259.0266.0exp 11E for()61093.2-⨯=E f GaAs: 42.1=g E eV ()()⎥⎦⎤⎢⎣⎡+=0259.0242.1exp 11E for()121024.1-⨯=E f(b) Using the results of Problem 3.38, the answers to part (b) are exactly the same as those given in part (a)._______________________________________3.46(a) ()⎥⎦⎤⎢⎣⎡--=kT E E f F F exp ⎥⎦⎤⎢⎣⎡-=-kT 60.0exp 108or ()810ln 60.0+=kT()032572.010ln 60.08==kT eV ()⎪⎭⎫⎝⎛=3000259.0032572.0Tso 377=T K(b) ⎥⎦⎤⎢⎣⎡-=-kT 60.0exp 106()610ln 60.0+=kT()043429.010ln 60.06==kT ()⎪⎭⎫⎝⎛=3000259.0043429.0Tor 503=T K_______________________________________ 3.47(a) At 200=T K,()017267.03002000259.0=⎪⎭⎫⎝⎛=kT eV⎪⎪⎭⎫ ⎝⎛-+==kT E E f F F exp 1105.019105.01exp =-=⎪⎪⎭⎫ ⎝⎛-kT E E F()()()19ln 017267.019ln ==-kT E E F 05084.0=eV By symmetry, for 95.0=F f , 05084.0-=-F E E eVThen ()1017.005084.02==∆E eV (b) 400=T K, 034533.0=kT eV For 05.0=F f , from part (a),()()()19ln 034533.019ln ==-kT E E F 10168.0=eVThen ()2034.010168.02==∆E eV _______________________________________。
半导体物理与器件第四版课后习题答案
半导体物理与器件第四版课后习题答案第一章半导体材料基础知识1.1 小题一根据题目描述，当n=5时，半导体材料的载流子浓度为’n=2.5×1015cm(-3)’，求势垒能为多少？解答：根据势垒能公式E_g = E_c - E_v其中E_g为势垒能，E_c为导带底，E_v为价带顶。
根据载流子浓度和温度的关系n = 2 * （2 * pi * m_e * k * T / h^2）^(3/2) * e^(-E_g / (2 * k * T))其中m_e为载流子质量，k为玻尔兹曼常数，T为绝对温度。
可以得到E_g = -2 * k * T * ln(n / (2 * （2 * pi * m_e * k * T / h^2）^(3/2)))代入已知条件，计算得到势垒能为E_g = -2 * 1.38 * 10^(-23) * 300 * ln(2.5 * 10^15 / (2 * （2 * pi * 9.1 * 10^(-31) * 1.38 * 10^ (-23) * 300 / (6.63 * 10^(-34))^2）^(3/2)))1.1 小题二根据题目描述，当势垒能E_g=1.21eV时，求温度为多少时，载流子浓度为’n=5.0×1015cm(-3)’？解答：按照1.1 小题一的公式，可以求出温度TT = E_g / (2 * k * ln(n / (2 * （2 * pi * m_e * k * T / h^2）^(3/2))))将已知数据代入公式，计算得到温度T = 1.21 / (2 * 1.38 * 10^(-23) * ln(5 * 10^15/ (2 * （2 * pi * 9.1 * 10^(-31) * 1.38 * 10^(-2 3) * T / (6.63 * 10^(-34))^2）^(3/2))))第二章半导体材料与器件基本特性2.1 小题一根据题目描述，当Si掺杂浓度[N_b]为5×10^15 cm(-3)和[P_e]为2×1017 cm^(-3)，求Si中的载流子浓度和导电类型。
半导体物理与器件第四版课后习题答案8
Chapter 8In forward biasThenor(a)For , thenormVmV(b)For , thenormVmV_______________________________________ 8.2cmcm(a) V,cmorcm(b) V,cmcm(c) V_______________________________________ 8.3cmcm(a) V,cmcm(b) Vcmcm(c)_______________________________________8.4(a)cmcm(i)orV(ii) n-region - lower doped side(b)cmcm(i)V(ii) p-region - lower doped side_______________________________________8.5(a)A/cmAor mA(b)A/cmAor mA(c) mA_______________________________________ 8.6For an silicon diodeorA(a) For V,A(b) For V,orA_______________________________________8.7A/cm(a)Aor mA(b)A_______________________________________8.8(a)A/cmA(i)A(ii)A(iii)A_______________________________________ 8.9We haveor we can write this asso thatIn reverse bias, is negative, so at, we haveormV_______________________________________ 8.10Case 1:AmAmA/cmCase 2:or mAmA/cmCase 3:SoVThenmACase 4:mAcm_______________________________________8.11(a)or(b) From part (a),or_______________________________________ 8.12The cross-sectional area iscmWe havewhich yieldsA/cmWe can writeWe wantor=which yieldsNowWe findcmandcm_______________________________________ 8.13Plot_______________________________________8.14(a)We haveandsoor(b) Using Einstein's relation, we can writeWe haveandAlsoThen_______________________________________8.15(a) p-side;oreVAlso on the n-side;oreV(b) We can findcm/scm/sNoworA/cmThenorAWe findorAA(c) The hole current isor(A)Then_______________________________________8.16(a)A(b)A(c)VVcm(d)A(e)AANowAThenA_______________________________________8.17(a) The excess hole concentration is given by We findcmandcm mThenorcm(b) We haveAt cm,orA/cm(c) We haveWe can determine thatcm and mThenorA/cmWe can also findA/cmThen at m,orA/cm_______________________________________ 8.188. (a) Problem 8.7orV(b) Problem 8.8orV_______________________________________ 8.19The excess electron concentration is given by The total number of excess electrons isWe may note thatThenWe find thatcm/s and mAlsocmThenorThen, we find the total number of excesselectrons in the p-region to be:(a)V,(b)V,(c)V,Similarly, the total number of excess holes in the n-region is found to beWe find thatcm/s and mAlsocmThenSo(a)V,(b)V,(c)V,_______________________________________ 8.20ThensoorWe then haveorThenoreV_______________________________________ 8.21(a) We havewhich can be written in the formor(b) Taking the ratioFor K, ,For K, ,(i) Germanium: eVor(ii) Silicon: eVor_______________________________________ 8.22Plot_______________________________________ 8.23First case:or VNowKSecond case:orNowBy trial and error,KThe reverse-bias current is limiting factor._______________________________________ 8.24cmor m;(a)(i)orV(ii)AAAor mA(b) (i)orV(ii)AAAor mA_______________________________________8.25(a) We can write for the n-regionThe general solution is of the formThe boundary condition at givesand the boundary condition atgivesFrom this equation, we haveThen, from the first boundary condition, weobtainWe then obtainwhich can be written asWe can also findThe solution can now be written asor finally(b)=Then_______________________________________ 8.26For the temperature range K,neglect the change in and .ThenTaking the ratio of currents, but maintaining a constant, we haveWe then haveWe haveK , V andeV, VK ,eV, VK ,eV, VFor K ,which yieldsVFor K ,which yieldsV_______________________________________8.27(a) We can writewhere C is a constant, independent oftemperature.As a first approximation, neglect thevariation of and with temperatureover the range of interest. We can then write where is another constant, independent oftemperature. We findor_______________________________________8.28(a)A(b)We findVandcmThenA(c)_______________________________________8.29(a) Set ,socmThenBy trial and error,KWe haveThenAor A(b) From Problem 8.28AASoV_______________________________________ 8.30cm/scm/s(a)(i)A(ii)A(iii)A(iv)A(b)Vcm(i)ThenA(ii)A(iii)A(iv)A_______________________________________ 8.31Using results from Problem 8.30, we findV, A,A, AV, AA, AV, AA, AV, AA, AV. AA, A_______________________________________ 8.32Plot_______________________________________ 8.33Plot_______________________________________ 8.34We have thatLet andWe can writeandWe also haveso thatThenDefineandThen the recombination rate can be written as orTo find the maximum recombination rate, set orwhich simplifies toThe denominator is not zero, so we haveorThen the maximum recombination ratebecomesorwhich can be written asIf , then we can neglect the (-1)term in the numerator and the (+1) term in the denominator, so we finally haveQ.E.D. _______________________________________ 8.35We haveIn this case, cms and isa constant through the space charge region. ThenorVAlsoorcmThenorA/cm_______________________________________8.36orA/cmNoworwhich can be written asWe findorV_______________________________________8.37(a)For nF(b)For nF_______________________________________8.38(a) , For mAC(b) For mAC_______________________________________8.39For a diode,NowSandFWe havewhereWe obtainkHz ,kHz ,MHz ,MHz ,_______________________________________ 8.40Reverse biasVF(V) (pF)10 1.5555 2.1233 2.6241 3.8180 5.7476.6508.179Forward biasForThenA(V) (F) + (F) = (F)0.20.40.60_______________________________________ 8.41For a diode, , thenNowF/AThenorsAt 1 mA,orF_______________________________________8.42(a)(i)orAor mA(ii)V(iii)(b)(i)Aor mA(ii)V(iii)_______________________________________8.43(a) p-region:soorn-region:soorThe total resistance isor(b)which yieldsmA_______________________________________ 8.44orWe can write(a) (i) For mA,or V(ii) For mA,or V(b) Set(i) For mA,or V(ii) For mA,or V_______________________________________8.45(a)or AV(b) AV_______________________________________8.46(a)orwhich yields(b)which yields_______________________________________8.47(a) IfThen we haveorWe find(b) If , thenwhich yields_______________________________________8.48(a) erferf= erferfThen(b) erfBy trial and error,_______________________________________ 8.49pF atpF at VWe haves , mAandmASoorsAlsopFThe time constant issNow, the turn-off time iss_______________________________________ 8.50VWe findwhich yieldscm_______________________________________ 8.51Sketch_______________________________________ 8.53From Figure 7.15, cmLet cmcmThen VAor cm_______________________________________。
半导体物理及器件第四版课后习题答案解析
Chapter 44.1⎪⎪⎭⎫ ⎝⎛-=kTE N N n gc i exp 2υ ⎪⎪⎭⎫ ⎝⎛-⎪⎭⎫ ⎝⎛=kT E T N N g O cO exp 3003υwhere cO N and O N υ are the values at300 K.(a) SiliconT (K)kT (eV) i n (cm 3-) 200 400 60001727.0 03453.0 0518.041068.7⨯ 121038.2⨯ 141074.9⨯(b) Germanium (c) GaAs T (K)i n (cm 3-) i n (cm 3-) 200 400 600101016.2⨯ 141060.8⨯ 161082.3⨯38.1 91028.3⨯ 121072.5⨯_______________________________________ 4.2Plot_______________________________________ 4.3(a) ⎪⎪⎭⎫⎝⎛-=kT E N N n g c i exp 2υ ()()()319192113001004.1108.2105⎪⎭⎫⎝⎛⨯⨯=⨯T()()⎥⎦⎤⎢⎣⎡-⨯3000259.012.1exp T()3382330010912.2105.2⎪⎭⎫ ⎝⎛⨯=⨯T()()()()⎥⎦⎤⎢⎣⎡-⨯T 0259.030012.1expBy trial and error, 5.367≅T K(b)()252122105.2105⨯=⨯=i n()()()()()⎥⎦⎤⎢⎣⎡-⎪⎭⎫⎝⎛⨯=T T 0259.030012.1exp 30010912.2338By trial and error, 5.417≅T K_______________________________________ 4.4At 200=T K, ()⎪⎭⎫⎝⎛=3002000259.0kT017267.0=eVAt 400=T K, ()⎪⎭⎫⎝⎛=3004000259.0kT034533.0=eV ()()()()17222102210025.31040.11070.7200400⨯=⨯⨯=ii nn⎥⎦⎤⎢⎣⎡-⎥⎦⎤⎢⎣⎡-⨯⎪⎭⎫⎝⎛⎪⎭⎫ ⎝⎛=017267.0exp 034533.0exp 30020030040033g g E E⎥⎦⎤⎢⎣⎡-=034533.0017267.0exp 8g g E E()[]9578.289139.57exp 810025.317-=⨯g Eor()1714.38810025.3ln 9561.2817=⎪⎪⎭⎫⎝⎛⨯=g E or 318.1=g E eVNow()32103004001070.7⎪⎭⎫⎝⎛=⨯o co N N υ⎪⎭⎫⎝⎛-⨯034533.0318.1exp()()172110658.2370.210929.5-⨯=⨯o co N N υso 371041.9⨯=o co N N υcm 6- _______________________________________4.5()()⎪⎭⎫ ⎝⎛-=⎪⎭⎫ ⎝⎛-⎪⎭⎫ ⎝⎛-=kT kT kT A n B n i i 20.0exp 90.0exp 10.1exp For 200=T K, 017267.0=kT eVFor 300=T K, 0259.0=kT eVFor 400=T K, 034533.0=kT eV(a) For 200=T K,()()610325.9017267.020.0exp -⨯=⎪⎭⎫ ⎝⎛-=A n B n i i (b) For 300=T K, ()()41043.40259.020.0exp -⨯=⎪⎭⎫ ⎝⎛-=A n B n i i (c) For 400=T K, ()()31005.3034533.020.0exp -⨯=⎪⎭⎫ ⎝⎛-=A n B n i i _______________________________________4.6 (a)()⎥⎦⎤⎢⎣⎡---∝kT E E E E f g F c F c exp()⎥⎦⎤⎢⎣⎡---∝kT E E E E c c exp ()⎥⎦⎤⎢⎣⎡--⨯kT E E F c exp Let x E E c =- Then ⎪⎭⎫ ⎝⎛-∝kT x x f g F c exp To find the maximum value: ()⎪⎭⎫ ⎝⎛-∝-kT x x dx f g d F c exp 212/1 0exp 12/1=⎪⎭⎫ ⎝⎛-⋅-kT x x kT which yields 2212/12/1kTx kT x x=⇒= The maximum value occurs at2kTE E c += (b) ()()⎥⎦⎤⎢⎣⎡---∝-kT E E E E f g F F exp 1υυ ()⎥⎦⎤⎢⎣⎡---∝kT E E E E υυexp()⎥⎦⎤⎢⎣⎡--⨯kT E E F υexp Let x E E =-υThen ()⎪⎭⎫ ⎝⎛-∝-kT x x f g F exp 1υTo find the maximum value()[]0exp 1=⎥⎦⎤⎢⎣⎡⎪⎭⎫⎝⎛-∝-kT x x dx d dx f g d F υ Same as part (a). Maximum occurs at2kTx =or2kTE E -=υ_______________________________________ 4.7()()()()⎥⎦⎤⎢⎣⎡---⎥⎦⎤⎢⎣⎡---=kT E E E E kT E E E E E n E n c c c c 221121exp exp wherekT E E c 41+= and 22kTE E c += Then()()()⎥⎦⎤⎢⎣⎡--=kT E E kT kTE n E n 2121exp 24()5.3exp 22214exp 22-=⎥⎦⎤⎢⎣⎡⎪⎭⎫ ⎝⎛--=or()()0854.021=E n E n_______________________________________ 4.8Plot_______________________________________ 4.9Plot_______________________________________4.10⎪⎪⎭⎫ ⎝⎛=-**ln 43n pmidgap Fi m m kT E ESilicon: o p m m 56.0*=, o n m m 08.1*=0128.0-=-midgap Fi E E eV Germanium: o p m m 37.0*=,o n m m 55.0*=0077.0-=-midgap Fi E E eV Gallium Arsenide: o p m m 48.0*=,o n m m 067.0*= 0382.0+=-midgap Fi E E eV_______________________________________4.11()⎪⎪⎭⎫ ⎝⎛=-c midgap Fi N N kT E E υln 21()()kT kT 4952.0108.21004.1ln 211919-=⎪⎪⎭⎫ ⎝⎛⨯⨯=T (K) kT (eV)(midgap Fi E E -)(eV )200 400 600 01727.0 03453.0 0518.0 0086.0- 0171.0- 0257.0-_______________________________________4.12(a) ⎪⎪⎭⎫ ⎝⎛=-**ln 43n pmidgap Fi m m kT E E()⎪⎭⎫⎝⎛=21.170.0ln 0259.04363.10-⇒meV (b) ()⎪⎭⎫⎝⎛=-080.075.0ln 0259.043midgap Fi E E47.43+⇒meV_______________________________________4.13Let ()==K E g c constant Then()()dE E fE g n FE co c⎰∞=dE kT E E Kc E F⎰∞⎪⎪⎭⎫⎝⎛-+=exp 11()dE kT E E K cE F ⎰∞⎥⎦⎤⎢⎣⎡--≅exp LetkTE E c-=η so that ηd kT dE ⋅= We can write()()c F c F E E E E E E -+-=- so that()()()η-⋅⎥⎦⎤⎢⎣⎡--=⎥⎦⎤⎢⎣⎡--exp exp exp kT E E kT E E F c F The integral can then be written as()()ηηd kT E E kT K n F c o ⎰∞-⎥⎦⎤⎢⎣⎡--⋅⋅=0exp exp which becomes()⎥⎦⎤⎢⎣⎡--⋅⋅=kT E E kT K n F c o exp _______________________________________4.14Let ()()c c E E C E g -=1 for c E E ≥ Then()()dE E fE g n FE co c⎰∞=()dE kT E E E E C c E Fc ⎰∞⎪⎪⎭⎫⎝⎛-+-=exp 11()()dE kT E E E E C F E C c⎥⎦⎤⎢⎣⎡---≅⎰∞exp 1Let kTE E c-=η so that ηd kT dE ⋅= We can write()()F c c F E E E E E E -+-=-Then()⎥⎦⎤⎢⎣⎡--=kT E E C n F c o exp 1()()dE kT E E E E c E c c⎥⎦⎤⎢⎣⎡---⨯⎰∞exp or()⎥⎦⎤⎢⎣⎡--=kT E E C n F c o exp 1 ()()()[]()ηηηd kT kT -⨯⎰∞exp 0We find that()()()11exp exp 0+=---=-∞∞⎰ηηηηηdSo()()⎥⎦⎤⎢⎣⎡--=kT E E kT C n F c o exp 21 _______________________________________4.15We have ⎪⎪⎭⎫⎝⎛=∈*1m m a r o r o For germanium, 16=∈r , o m m 55.0*= Then()()()53.02955.01161=⎪⎭⎫⎝⎛=o a roroA r 4.151=The ionization energy can be written as()6.132*⎪⎪⎭⎫⎝⎛∈∈⎪⎪⎭⎫ ⎝⎛=s o o m m E eV ()()029.06.131655.02=⇒=E eV_______________________________________4.16We have ⎪⎪⎭⎫ ⎝⎛=∈*1m m a r o r o For gallium arsenide, 1.13=∈r , o m m 067.0*= Then ()()oA r 10453.0067.011.131=⎪⎭⎫ ⎝⎛=The ionization energy is()()()6.131.13067.06.1322*=⎪⎪⎭⎫ ⎝⎛∈∈⎪⎪⎭⎫ ⎝⎛=s o o m m E or0053.0=E eV_______________________________________4.17(a) ⎪⎪⎭⎫⎝⎛=-o c F c n N kT E E ln()⎪⎪⎭⎫⎝⎛⨯⨯=1519107108.2ln 0259.02148.0=eV (b) ()F c g F E E E E E --=-υ90518.02148.012.1=-=eV(c) ()⎥⎦⎤⎢⎣⎡--=kT E E N p F o υυexp()⎥⎦⎤⎢⎣⎡-⨯=0259.090518.0exp 1004.11931090.6⨯=cm 3- (d) Holes(e) ⎪⎪⎭⎫⎝⎛=-i o Fi F n n kT E E ln()⎪⎪⎭⎫⎝⎛⨯⨯=1015105.1107ln 0259.0338.0=eV_______________________________________4.18(a)⎪⎪⎭⎫⎝⎛=-o F p N kT E E υυln()⎪⎪⎭⎫⎝⎛⨯⨯=16191021004.1ln 0259.0162.0=eV (b) ()υE E E E E F g F c --=-958.0162.012.1=-=eV(c)()⎪⎭⎫⎝⎛-⨯=0259.0958.0exp 108.219o n31041.2⨯=cm 3-(d) ⎪⎪⎭⎫⎝⎛=-i o F Fi n p kT E E ln()⎪⎪⎭⎫⎝⎛⨯⨯=1016105.1102ln 0259.0365.0=eV_______________________________________4.19(a) ⎪⎪⎭⎫⎝⎛=-o c F c n N kT E E ln()⎪⎪⎭⎫⎝⎛⨯⨯=519102108.2ln 0259.08436.0=eV ()F c g F E E E E E --=-υ 8436.012.1-= 2764.0=-υE E F eV (b)()⎪⎭⎫⎝⎛-⨯=0259.027637.0exp 1004.119o p1410414.2⨯=cm 3- (c) p-type_______________________________________4.20(a)()032375.03003750259.0=⎪⎭⎫⎝⎛=kT eV()⎥⎦⎤⎢⎣⎡-⎪⎭⎫ ⎝⎛⨯=032375.028.0exp 300375107.42/317o n 141015.1⨯=cm 3-()28.042.1-=--=-F c g F E E E E E υ 14.1=eV()⎥⎦⎤⎢⎣⎡-⎪⎭⎫⎝⎛⨯=032375.014.1exp 3003751072/318o p 31099.4⨯=cm 3- (b) ()⎪⎪⎭⎫⎝⎛⨯⨯=-14171015.1107.4ln 0259.0F c E E2154.0=eV()2154.042.1-=--=-F c g F E E E E E υ 2046.1=eV()⎥⎦⎤⎢⎣⎡-⨯=0259.02046.1exp 10718o p 21042.4-⨯=cm 3-_______________________________________4.21(a) ()032375.03003750259.0=⎪⎭⎫⎝⎛=kT eV()⎥⎦⎤⎢⎣⎡-⎪⎭⎫⎝⎛⨯=032375.028.0exp 300375108.22/319o n 151086.6⨯= cm 3-()28.012.1-=--=-F c g F E E E E E υ 840.0=eV()⎥⎦⎤⎢⎣⎡-⎪⎭⎫⎝⎛⨯=032375.0840.0exp 3003751004.12/319o p 71084.7⨯=cm 3-(b) ⎪⎪⎭⎫⎝⎛=-o c F c n N kT E E ln()⎪⎪⎭⎫⎝⎛⨯⨯=151910862.6108.2ln 0259.02153.0=eV9047.02153.012.1=-=-υE E F eV()⎥⎦⎤⎢⎣⎡-⨯=0259.0904668.0exp 1004.119o p 31004.7⨯=cm 3-_______________________________________4.22(a) p-type(b) 28.0412.14===-g F E E E υeV()⎥⎦⎤⎢⎣⎡--=kT E E N p F o υυexp ()⎥⎦⎤⎢⎣⎡-⨯=0259.028.0exp 1004.119141010.2⨯=cm 3- ()υE E E E E F g F c --=- 84.028.012.1=-=eV()⎥⎦⎤⎢⎣⎡--=kT E E N n F c c o exp ()⎥⎦⎤⎢⎣⎡-⨯=0259.084.0exp 108.21951030.2⨯=cm 3-_______________________________________4.23(a) ⎥⎦⎤⎢⎣⎡-=kT E E n n Fi Fi o exp()⎥⎦⎤⎢⎣⎡⨯=0259.022.0exp 105.110 131033.7⨯=cm 3-⎥⎦⎤⎢⎣⎡-=kT E E n p F Fii o exp()⎥⎦⎤⎢⎣⎡-⨯=0259.022.0exp 105.11061007.3⨯=cm 3-(b) ⎥⎦⎤⎢⎣⎡-=kT E E n n Fi Fi o exp ()⎥⎦⎤⎢⎣⎡⨯=0259.022.0exp 108.16 91080.8⨯=cm 3-⎥⎦⎤⎢⎣⎡-=kT E E n p F Fii o exp()⎥⎦⎤⎢⎣⎡-⨯=0259.022.0exp 108.16 21068.3⨯=cm 3-_______________________________________4.24(a) ⎪⎪⎭⎫⎝⎛=-o F p N kT E E υυln()⎪⎪⎭⎫⎝⎛⨯⨯=151********.1ln 0259.01979.0=eV (b) ()υE E E E E F g F c --=- 92212.019788.012.1=-=eV(c) ()⎥⎦⎤⎢⎣⎡-⨯=0259.092212.0exp 108.219o n31066.9⨯=cm 3- (d) Holes(e) ⎪⎪⎭⎫⎝⎛=-i o F Fi n p kT E E ln()⎪⎪⎭⎫⎝⎛⨯⨯=1015105.1105ln 0259.03294.0=eV _______________________________________4.25()034533.03004000259.0=⎪⎭⎫⎝⎛=kT eV()2/3193004001004.1⎪⎭⎫⎝⎛⨯=υN1910601.1⨯=cm 3-()2/319300400108.2⎪⎭⎫⎝⎛⨯=c N19103109.4⨯=cm 3-()()1919210601.1103109.4⨯⨯=i n⎥⎦⎤⎢⎣⎡-⨯034533.012.1exp24106702.5⨯= 1210381.2⨯=⇒i n cm 3- (a)⎪⎪⎭⎫ ⎝⎛=-oF pN kT E E υυln ()⎪⎪⎭⎫⎝⎛⨯⨯=151910510601.1ln 034533.02787.0=eV(b) 84127.027873.012.1=-=-F c E E eV(c)()⎥⎦⎤⎢⎣⎡-⨯=034533.084127.0exp 103109.419o n910134.1⨯=cm 3- (d) Holes(e) ⎪⎪⎭⎫⎝⎛=-i o F Fi n p kT E E ln()⎪⎪⎭⎫⎝⎛⨯⨯=121510381.2105ln 034533.02642.0=eV _______________________________________4.26(a) ()⎥⎦⎤⎢⎣⎡-⨯=0259.025.0exp 10718o p141050.4⨯=cm 3-17.125.042.1=-=-F c E E eV()⎥⎦⎤⎢⎣⎡-⨯=0259.017.1exp 107.417o n21013.1-⨯=cm 3-(b) 034533.0=kT eV()2/318300400107⎪⎭⎫⎝⎛⨯=υN1910078.1⨯=cm 3- ()2/317300400107.4⎪⎭⎫ ⎝⎛⨯=c N1710236.7⨯=cm 3-⎪⎪⎭⎫⎝⎛=-o F p N kT E E υυln()⎪⎪⎭⎫⎝⎛⨯⨯=14191050.410078.1ln 034533.03482.0=eV072.13482.042.1=-=-F c E E eV()⎥⎦⎤⎢⎣⎡-⨯=034533.007177.1exp 10236.717o n 41040.2⨯=cm 3-_____________________________________4.27(a) ()⎥⎦⎤⎢⎣⎡-⨯=0259.025.0exp 1004.119o p141068.6⨯=cm 3-870.025.012.1=-=-F c E E eV()⎥⎦⎤⎢⎣⎡-⨯=0259.0870.0exp 108.219o n41023.7⨯=o n cm 3- (b)034533.0=kT eV()2/3193004001004.1⎪⎭⎫⎝⎛⨯=υN1910601.1⨯=cm 3- ()2/319300400108.2⎪⎭⎫⎝⎛⨯=c N1910311.4⨯=cm 3-⎪⎪⎭⎫⎝⎛=-o F p N kT E E υυln()⎪⎪⎭⎫⎝⎛⨯⨯=14191068.610601.1ln 034533.03482.0=eV7718.03482.012.1=-=-F c E E eV()⎥⎦⎤⎢⎣⎡-⨯=034533.077175.0exp 10311.419o n 91049.8⨯=cm 3-_______________________________________4.28(a) ()F c o F N n ηπ2/12=For 2kT E E c F +=,5.02==-=kTkT kT E E c F F η Then ()0.12/1≅F F η()()0.1108.2219⨯=πo n191016.3⨯=cm 3-(b) ()F c o F N n ηπ2/12=()()0.1107.4217⨯=π171030.5⨯=cm 3-_______________________________________4.29()Fo F N p ηπυ'=2/12()()FF ηπ'⨯=⨯2/119191004.12105So ()26.42/1='FF η We find kTE E FF-=≅'υη0.3 ()()0777.00259.00.3==-F E E υeV _______________________________________4.30(a) 44==-=kTkTkT E E c F F ηThen ()0.62/1≅F F η()F c o F N n ηπ2/12=()()0.6108.2219⨯=π201090.1⨯=cm 3-(b) ()()0.6107.4217⨯=πo n181018.3⨯=cm 3-_______________________________________4.31For the electron concentration ()()()E f E g E n F c = The Boltzmann approximation applies, so ()()c n E E hmE n -=32/3*24π()⎥⎦⎤⎢⎣⎡--⨯kT E E F exp or ()()()⎥⎦⎤⎢⎣⎡--=kT E E hm E n F c nexp 2432/3*π ()⎥⎦⎤⎢⎣⎡---⨯kT E E kT E E kTc c exp DefinekTE E x c-=Then ()()()x x K x n E n -=→exp To find maximum ()()x n E n →, set()()x x K dx x dn -⎢⎣⎡==-exp 2102/1+()()⎥⎦⎤--x x exp 12/1or()⎥⎦⎤⎢⎣⎡--=-x x Kx 21exp 02/1which yieldskT E E kT E E x c c 2121+=⇒-==For the hole concentration ()()()[]E f E g E p F -=1υUsing the Boltzmann approximation ()()E E hm E p p-=υπ32/3*24()⎥⎦⎤⎢⎣⎡--⨯kT E E F exp or()()()⎥⎦⎤⎢⎣⎡--=kT E E h m E p F pυπexp 2432/3* ()⎥⎦⎤⎢⎣⎡---⨯kT E E kT E E kTυυexp DefinekTEE x -='υThen ()()x x K x p '-''='exp To find maximum value of()()x p E p '→, set()0=''x d x dp Using the resultsfrom above,we find the maximum atkT E E 21-=υ_______________________________________4.32(a) Silicon: We have()⎥⎦⎤⎢⎣⎡--=kT E E N n F c c o exp We can write()()F d d c F c E E E E E E -+-=- For045.0=-d c E E eV and kT E E F d 3=-eVwe can write()⎥⎦⎤⎢⎣⎡--⨯=30259.0045.0exp 108.219o n()()737.4exp 108.219-⨯=or171045.2⨯=o n cm 3- We also have()⎥⎦⎤⎢⎣⎡--=kT E E N p F o υυexp Again, we can write()()υυE E E E E E a a F F -+-=- ForkT E E a F 3=- and045.0=-υE E a eVThen()⎥⎦⎤⎢⎣⎡--⨯=0259.0045.03exp 1004.119o p ()()737.4exp 1004.119-⨯=or161012.9⨯=o p cm 3- (b)GaAs: assume 0058.0=-d c E E eVThen()⎥⎦⎤⎢⎣⎡--⨯=30259.00058.0exp 107.417o n ()()224.3exp 107.417-⨯= or161087.1⨯=o n cm 3- Assume 0345.0=-υE E a eV Then()⎥⎦⎤⎢⎣⎡--⨯=30259.00345.0exp 10718o p ()()332.4exp 10718-⨯= or161020.9⨯=o p cm 3-_______________________________________4.33Plot_______________________________________4.34(a)151510310154⨯=-⨯=o p cm 3- ()415210105.7103105.1⨯=⨯⨯=o n cm 3-(b)16103⨯==d o N n cm 3-()316210105.7103105.1⨯=⨯⨯=o p cm 3-(c)10105.1⨯===i o o n p n cm 3-(d) ()()3191923003751004.1108.2⎪⎭⎫⎝⎛⨯⨯=in()()()()⎥⎦⎤⎢⎣⎡-⨯3750259.030012.1exp1110334.7⨯=⇒i n cm 3- 15104⨯==a o N p cm 3-()8152111034.110410334.7⨯=⨯⨯=on cm 3-(e) ()()3191923004501004.1108.2⎪⎭⎫⎝⎛⨯⨯=in()()()()⎥⎦⎤⎢⎣⎡-⨯4500259.030012.1exp1310722.1⨯=⇒i n cm 3-()2132141410722.1210210⨯+⎪⎪⎭⎫ ⎝⎛+=o n1410029.1⨯=cm 3-()12142131088.210029.110722.1⨯=⨯⨯=op cm 3-_______________________________________4.35(a)151510104-⨯=-=d a o N N p 15103⨯=cm 3-()3152621008.1103108.1-⨯=⨯⨯==o i o p n n cm 3-(b)16103⨯==d o N n cm 3-()416261008.1103108.1-⨯=⨯⨯=o p cm 3-(c)6108.1⨯===i o o n p n cm 3-(d) ()()318172300375100.7107.4⎪⎭⎫⎝⎛⨯⨯=in()()()()⎥⎦⎤⎢⎣⎡-⨯3750259.030042.1exp810580.7⨯=⇒i n cm 3- 15104⨯==a o N p cm 3-()215281044.110410580.7⨯=⨯⨯=o n cm 3-(e) ()()318172300450100.7107.4⎪⎭⎫⎝⎛⨯⨯=in()()()()⎥⎦⎤⎢⎣⎡-⨯4500259.030042.1exp1010853.3⨯=⇒i n cm 3- 1410==d o N n cm 3-()7142101048.11010853.3⨯=⨯=op cm 3-_______________________________________4.36(a) Ge: 13104.2⨯=i n cm 3-(i)2222i d d o n N N n +⎪⎪⎭⎫⎝⎛+=()21321515104.221022102⨯+⎪⎪⎭⎫⎝⎛⨯+⨯=or15102⨯=≅d o N n cm 3- ()152132102104.2⨯⨯==o i o n n p111088.2⨯= cm 3-(ii)151610710⨯-=-≅d a o N N p 15103⨯=cm 3-()152132103104.2⨯⨯==o i o p n n111092.1⨯=cm 3- (b)GaAs: 6108.1⨯=i n cm 3- (i)15102⨯=≅d o N n cm()315261062.1102108.1-⨯=⨯⨯=op cm 3-(ii)15103⨯=-≅d a o N N p cm 3-()315261008.1103108.1-⨯=⨯⨯=on cm 3-(c) The result implies that there is only one minority carrier in a volume of 310cm 3._______________________________________4.37(a) For the donor level⎪⎪⎭⎫ ⎝⎛-+=kT E E N n F d d d exp 2111⎪⎭⎫ ⎝⎛+=0259.020.0exp 2111or41085.8-⨯=d d N n(b) We have ()⎪⎪⎭⎫ ⎝⎛-+=kTE E E f FF exp 11Now()()F c c F E E E E E E -+-=- or245.0+=-kT E E FThen ()⎪⎭⎫ ⎝⎛++=0259.0245.01exp 11E f For()51087.2-⨯=E f F_______________________________________4.38(a) ⇒>d a N N p-type(b) Silicon:1313101105.2⨯-⨯=-=d a o N N p or13105.1⨯=o p cm 3- Then()7132102105.1105.1105.1⨯=⨯⨯==o i o p n n cm 3- Germanium:2222i da d a o n N N N N p +⎪⎪⎭⎫ ⎝⎛-+-=()21321313104.22105.12105.1⨯+⎪⎪⎭⎫⎝⎛⨯+⎪⎪⎭⎫ ⎝⎛⨯=or131026.3⨯=o p cm 3-Then()131321321076.110264.3104.2⨯=⨯⨯==o i o p n n cm 3-Gallium Arsenide:13105.1⨯=-=d a o N N p cm 3- and()216.0105.1108.113262=⨯⨯==o io p n n cm 3- _______________________________________4.39(a) ⇒>a d N N n-type(b)1515102.1102⨯-⨯=-≅a d o N N n14108⨯=cm 3-()51421021081.2108105.1⨯=⨯⨯==o i o n n p cm 3-(c)()d a ao N N N p -+'≅ 151515102102.1104⨯-⨯+'=⨯aN 15108.4⨯='⇒aN cm 3- ()41521010625.5104105.1⨯=⨯⨯=on cm 3-_______________________________________4.40()155210210125.1102105.1⨯=⨯⨯==o i o p n n cm 3- ⇒>o o p n n-type_______________________________________4.41()()318192300250100.61004.1⎪⎭⎫⎝⎛⨯⨯=in()()⎥⎦⎤⎢⎣⎡-⨯3002500259.066.0exp24108936.1⨯=1210376.1⨯=⇒i n cm 3- 2222414i o o i o i o n n n n p n n =⇒==i o n n 21=⇒ So 111088.6⨯=o n cm 3-,Then 121075.2⨯=o p cm 3-2222i aa o n N N p +⎪⎪⎭⎫ ⎝⎛+= 212210752.2⎪⎪⎭⎫ ⎝⎛-⨯a N 242108936.12⨯+⎪⎪⎭⎫ ⎝⎛=a N()21224210752.2105735.7⎪⎪⎭⎫⎝⎛+⨯-⨯a a N N242108936.12⨯+⎪⎪⎭⎫ ⎝⎛=a Nso that 1210064.2⨯=a N cm 3- _______________________________________4.42Plot_______________________________________4.43Plot_______________________________________4.44Plot_______________________________________4.452222i ad a d o n N N N N n +⎪⎪⎭⎫ ⎝⎛-+-= 2102.1102101.1141414⨯-⨯=⨯2214142102.1102i n +⎪⎪⎭⎫⎝⎛⨯-⨯+()()221321314104104101.1i n +⨯=⨯-⨯22727106.1109.4i n +⨯=⨯ so 131074.5⨯=i n cm 3-1314272103101.1103.3⨯=⨯⨯==o i o n n p cm 3- _______________________________________4.46(a) ⇒>d a N N p-typeMajority carriers are holes1616105.1103⨯-⨯=-=d a o N N p 16105.1⨯=cm 3-Minority carriers are electrons()4162102105.1105.1105.1⨯=⨯⨯==o i o p n n cm 3- (b) Boron atoms must be addedd a ao N N N p -+'=161616105.1103105⨯-⨯+'=⨯aN So 16105.3⨯='aN cm 3-()316210105.4105105.1⨯=⨯⨯=on cm 3-_______________________________________4.47 (a) ⇒<<i o n p n-type (b)oi o o i o p n n n n p 22=⇒=on ()16421010125.1102105.1⨯=⨯⨯=cm 3-⇒electrons are majority carriers4102⨯=o p cm 3-⇒holes are minority carriers(c) a d o N N n -=151610710125.1⨯-=⨯d N so 1610825.1⨯=d N cm 3-_______________________________________4.48⎪⎪⎭⎫⎝⎛=-i o F Fi n p kT E E ln For Germanium T (K)kT (eV) i n (cm 3-) 200 400 60001727.003453.0 0518.0101016.2⨯ 141060.8⨯ 161082.3⨯2222i a a o n N N p +⎪⎪⎭⎫⎝⎛+= and1510=a N cm 3- T (K) o p (cm 3-) ()F Fi E E -(eV) 200 400 60015100.1⨯ 151049.1⨯ 161087.3⨯1855.0 01898.0 000674.0_______________________________________4.49(a) ⎪⎪⎭⎫⎝⎛=-d c F c N N kT E E ln()⎪⎪⎭⎫⎝⎛⨯=d N 19108.2ln 0259.0 For 1410cm 3-, 3249.0=-F c E E eV 1510cm 3-, 2652.0=-F c E E eV 1610cm 3-, 2056.0=-F c E E eV 1710cm 3-, 1459.0=-F c E E eV (b)⎪⎪⎭⎫⎝⎛=-i d Fi F n N kT E E ln()⎪⎪⎭⎫⎝⎛⨯=10105.1ln 0259.0d N For 1410cm 3-, 2280.0=-Fi F E E eV 1510cm 3-, 2877.0=-Fi F E E eV 1610cm 3-, 3473.0=-Fi F E E eV 1710cm 3-, 4070.0=-Fi F E E eV _______________________________________4.50 (a)2222i d d o n N N n +⎪⎪⎭⎫ ⎝⎛+= 151005.105.1⨯==d o N n cm 3- ()21515105.01005.1⨯-⨯()2215105.0i n +⨯=so 2821025.5⨯=i nNow()()3191923001004.1108.2⎪⎭⎫ ⎝⎛⨯⨯=T n i()()⎥⎦⎤⎢⎣⎡-⨯3000259.012.1exp T()3382830010912.21025.5⎪⎭⎫ ⎝⎛⨯=⨯T⎥⎦⎤⎢⎣⎡-⨯T 973.12972exp By trial and error, 5.536=T K(b) At 300=T K,⎪⎪⎭⎫⎝⎛=-o c F c n N kT E E ln()⎪⎪⎭⎫⎝⎛⨯=-151910108.2ln 0259.0F c E E2652.0=eV At 5.536=T K,()046318.03005.5360259.0=⎪⎭⎫⎝⎛=kT eV()2/3193005.536108.2⎪⎭⎫ ⎝⎛⨯=c N1910696.6⨯=cm 3-⎪⎪⎭⎫⎝⎛=-o c F c n N kT E E ln()⎪⎪⎭⎫⎝⎛⨯⨯=-15191005.110696.6ln 046318.0F c E E5124.0=eV then ()2472.0=-∆F c E E eV(c) Closer to the intrinsic energylevel._______________________________________4.51⎪⎪⎭⎫⎝⎛=-i o F Fi n p kT E E ln At 200=T K, 017267.0=kT eV 400=T K, 034533.0=kT eV 600=T K, 0518.0=kT eVAt 200=T K,()()3191923002001004.1108.2⎪⎭⎫⎝⎛⨯⨯=in⎥⎦⎤⎢⎣⎡-⨯017267.012.1exp 410638.7⨯=⇒i n cm 3- At 400=T K,()()3191923004001004.1108.2⎪⎭⎫⎝⎛⨯⨯=in⎥⎦⎤⎢⎣⎡-⨯034533.012.1exp 1210381.2⨯=⇒i n cm 3- At 600=T K,()()3191923006001004.1108.2⎪⎭⎫⎝⎛⨯⨯=i n⎥⎦⎤⎢⎣⎡-⨯0518.012.1exp 1410740.9⨯=⇒in cm 3-At 200=T K and 400=T K, 15103⨯==a o N p cm 3- At 600=T K,2222i aa o n N N p +⎪⎪⎭⎫ ⎝⎛+=()2142151510740.921032103⨯+⎪⎪⎭⎫ ⎝⎛⨯+⨯=1510288.3⨯=cm 3-Then, 200=T K, 4212.0=-F Fi E E eV 400=T K, 2465.0=-F Fi E E eV 600=T K, 0630.0=-F Fi E E eV_______________________________________4.52(a)()⎪⎪⎭⎫⎝⎛⨯=⎪⎪⎭⎫ ⎝⎛=-6108.1ln 0259.0ln a i a F Fi N n N kT E E For 1410=a N cm 3-,4619.0=-F Fi E E eV1510=a N cm 3-,5215.0=-F Fi E E eV1610=a N cm 3-,5811.0=-F Fi E E eV1710=a N cm 3-,6408.0=-F Fi E E eV(b) ()⎪⎪⎭⎫ ⎝⎛⨯=⎪⎪⎭⎫ ⎝⎛=-a aF N NN kT E E 18100.7ln 0259.0ln υυ For 1410=a N cm 3-,2889.0=-υE E F eV1510=a N cm 3-,2293.0=-υE E F eV 1610=a N cm 3-,1697.0=-υE E F eV1710=a N cm 3-, 1100.0=-υE E F eV _______________________________________4.53(a) ⎪⎪⎭⎫ ⎝⎛=-**ln 43n pmidgap Fi m m kT E E()()10ln 0259.043= or0447.0+=-midgap Fi E E eV (b) Impurity atoms to be added so 45.0=-F midgap E E eV (i) p-type, so add acceptor atoms(ii)4947.045.00447.0=+=-F Fi E E eV Then⎪⎪⎭⎫⎝⎛-=kT E E n p F Fi i o exp()⎪⎭⎫⎝⎛=0259.04947.0exp 105or131097.1⨯==a o N p cm 3-_______________________________________4.54()⎥⎦⎤⎢⎣⎡--=-=kT E E N N N n F c c a d o exp so()⎪⎭⎫⎝⎛-⨯+⨯=0259.0215.0exp 108.21051915d N15151095.6105⨯+⨯=or16102.1⨯=d N cm 3-_______________________________________4.55(a) Silicon(i)⎪⎪⎭⎫⎝⎛=-d c F c N N kT E E ln()2188.0106108.2ln 0259.01519=⎪⎪⎭⎫⎝⎛⨯⨯=eV(ii)1929.00259.02188.0=-=-F c E E eV()⎥⎦⎤⎢⎣⎡--=kT E E N N F c c d exp()⎥⎦⎤⎢⎣⎡-⨯=0259.01929.0exp 108.2191610631.1⨯=d N cm 3-15106⨯+'=dN 1610031.1⨯='⇒d N cm 3- Additionaldonor atoms(b) GaAs(i)()⎪⎪⎭⎫⎝⎛⨯=-151710107.4ln 0259.0F c E E15936.0=eV(ii)13346.00259.015936.0=-=-F c E E eV()⎥⎦⎤⎢⎣⎡-⨯=0259.013346.0exp 107.417d N1510718.2⨯=cm 3-1510+'=dN 1510718.1⨯='⇒dN cm 3- Additionaldonor atoms_______________________________________4.56(a) ⎪⎪⎭⎫⎝⎛=-a F Fi N N kT E E υln()⎪⎪⎭⎫⎝⎛⨯⨯=16191021004.1ln 0259.01620.0=eV(b) ⎪⎪⎭⎫⎝⎛=-d c Fi F N N kT E E ln()1876.0102108.2ln 0259.01619=⎪⎪⎭⎫⎝⎛⨯⨯=eV(c) For part (a);16102⨯=o p cm 3- ()162102102105.1⨯⨯==o i o p n n410125.1⨯=cm 3-For part (b):16102⨯=o n cm 3-()162102102105.1⨯⨯==o i o n n p 410125.1⨯=cm 3- _______________________________________4.57⎥⎦⎤⎢⎣⎡-=kTE E n n FiFi o exp ()⎥⎦⎤⎢⎣⎡⨯=0259.055.0exp 108.1615100.3⨯=cm 3-Add additional acceptor impurities a d o N N n -=a N -⨯=⨯1515107103 15104⨯=⇒a N cm 3-_______________________________________4.58(a) ⎪⎪⎭⎫⎝⎛=-i o F Fi n p kT E E ln()3161.0105.1103ln 0259.01015=⎪⎪⎭⎫⎝⎛⨯⨯=eV(b)⎪⎪⎭⎫⎝⎛=-i o Fi F n n kT E E ln()3758.0105.1103ln 0259.01016=⎪⎪⎭⎫⎝⎛⨯⨯=eV(c) Fi F E E =(d)⎪⎪⎭⎫⎝⎛=-i o F Fi n p kT E E ln()⎪⎪⎭⎫ ⎝⎛⨯⨯⎪⎭⎫ ⎝⎛=111510334.7104ln 3003750259.0 2786.0=eV(e) ⎪⎪⎭⎫⎝⎛=-i o Fi F n n kT E E ln()⎪⎪⎭⎫ ⎝⎛⨯⨯⎪⎭⎫ ⎝⎛=131410722.110029.1ln 3004500259.0 06945.0=eV_______________________________________4.59 (a)⎪⎪⎭⎫⎝⎛=-o F p N kT E E υυln()2009.0103100.7ln 0259.01518=⎪⎪⎭⎫⎝⎛⨯⨯=eV(b) ()⎪⎪⎭⎫⎝⎛⨯⨯=--4181008.1100.7ln 0259.0υE E F360.1=eV(c) ()⎪⎪⎭⎫⎝⎛⨯⨯=-618108.1100.7ln 0259.0υE E F7508.0=eV(d) ()⎪⎭⎫⎝⎛=-3003750259.0υE E F()()⎥⎥⎦⎤⎢⎢⎣⎡⨯⨯⨯152/318104300375100.7ln 2526.0=eV(e) ()⎪⎭⎫⎝⎛=-3004500259.0υE E F()()⎥⎥⎦⎤⎢⎢⎣⎡⨯⨯⨯72/3181048.1300450100.7ln 068.1=eV_______________________________________4.60n-type⎪⎪⎭⎫⎝⎛=-i o Fi F n n kT E E ln()3504.0105.110125.1ln 0259.01016=⎪⎪⎭⎫⎝⎛⨯⨯=eV______________________________________4.612222i aa o n N N p +⎪⎪⎭⎫ ⎝⎛+= 21051008.51515⨯=⨯22152105i n +⎪⎪⎭⎫ ⎝⎛⨯+()21515105.21008.5⨯-⨯()2215105.2i n +⨯=230301025.6106564.6i n +⨯=⨯ 29210064.4⨯=⇒in⎥⎦⎤⎢⎣⎡-=kT E N N n g c i exp 2υ()030217.03003500259.0=⎪⎭⎫⎝⎛=kT eV()1921910633.1300350102.1⨯=⎪⎭⎫ ⎝⎛⨯=c N cm 3-()192191045.2300350108.1⨯=⎪⎭⎫ ⎝⎛⨯=υN cm 3- Now()()1919291045.210633.110064.4⨯⨯=⨯⎥⎦⎤⎢⎣⎡-⨯030217.0exp g ESo()()()⎥⎦⎤⎢⎣⎡⨯⨯⨯=29191910064.41045.210633.1ln 030217.0g E6257.0=⇒g E eV_______________________________________4.62(a) Replace Ga atoms ⇒Silicon acts as adonor()()1415105.310705.0⨯=⨯=d N cm 3- Replace As atoms ⇒Silicon acts as anacceptor()()15151065.610795.0⨯=⨯=a N cm 3-(b) ⇒>d a N N p-type(c) 1415105.31065.6⨯-⨯=-=d a o N N p 15103.6⨯=cm 3-()4152621014.5103.6108.1-⨯=⨯⨯==o i o p n n cm 3- (d)⎪⎪⎭⎫⎝⎛=-i o F Fi n p kT E E ln()5692.0108.1103.6ln 0259.0615=⎪⎪⎭⎫⎝⎛⨯⨯=eV_______________________________________。
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Chapter 33、1If were to increase, the bandgap energywould decrease and the material would begin to behave less like a semiconductor and more like a metal、 If were to decrease, thebandgap energy would increase and thematerial would begin to behave more like an insulator、_______________________________________ 3、2Schrodinger's wave equation is:Assume the solution is of the form:Region I: 、 Substituting theassumed solution into the wave equation, we obtain:which beesThis equation may be written asSetting for region I, the equationbees:whereQ、E、D、In Region II, 、 Assume the sameform of the solution:Substituting into Schrodinger's waveequation, we find:This equation can be written as:Setting for region II, thisequation beeswhere againQ、E、D、_______________________________________ 3、3We haveAssume the solution is of the form:The first derivative isand the second derivative beesSubstituting these equations into thedifferential equation, we findbining terms, we obtainWe find thatQ、E、D、 For the differential equation in and theproposed solution, the procedure is exactly the same as above、_______________________________________ 3、4We have the solutionsfor andfor 、The first boundary condition iswhich yieldsThe second boundary condition iswhich yieldsThe third boundary condition iswhich yieldsand can be written asThe fourth boundary condition iswhich yieldsand can be written as_______________________________________3、5(b) (i) First point:Second point: By trial and error,(ii) First point:Second point: By trial and error,_______________________________________3、6(b) (i) First point:Second point: By trial and error,(ii) First point:Second point: By trial and error,_______________________________________ 3、7Let ,ThenConsider of this function、We findThenFor ,So that, in general,AndSoThis implies thatfor_______________________________________ 3、8(a)JFrom Problem 3、5JJor eV(b)JFrom Problem 3、5,JJor eV_______________________________________3、9(a)At ,JAt , By trial and error,JJor eV(b)At ,JAt 、 From Problem 3、5,JJor eV_______________________________________3、10(a)JFrom Problem 3、6,JJor eV(b)JFrom Problem 3、6,JJor eV_____________________________________3、11(a)At ,JAt , By trial and error,JJor eV(b)At ,JAt , From Problem 3、6,JJor eV_______________________________________ 3、12 For K,eVK, eVK, eVK, eVK, eVK, eV_______________________________________ 3、13The effective mass is given byWe haveso that_______________________________________ 3、14The effective mass for a hole is given byWe have thatso that_______________________________________ 3、15Points A,B: velocity in -x directionPoints C,D: velocity in +x directionPoints A,D:negative effective mass Points B,C:positive effective mass _______________________________________ 3、16For A:At m, eVOr JSoNowkgorFor B:At m, eVOr JSoNowkgor_______________________________________ 3、17For A:kgorFor B:kgor_______________________________________3、18(a)(i)orHz(ii)cmnm(b)(i)Hz(ii)cmnm_______________________________________3、19(c)Curve A: Effective mass is a constantCurve B: Effective mass is positivearound , and is negativearound 、_______________________________________ 3、20ThenandThenor_______________________________________3、21(a)(b)_______________________________________3、22(a)(b)_______________________________________ 3、23For the 3-dimensional infinite potential well, when , , and、 In this region, the wave equationis:Use separation of variables technique, so let Substituting into the wave equation, we have Dividing by , we obtainLetThe solution is of the form:Since at , thenso that 、Also, at , so that、 Then whereSimilarly, we haveandFrom the boundary conditions, we findandwhereandFrom the wave equation, we can writeThe energy can be written as_______________________________________ 3、24The total number of quantum states in the3-dimensional potential well is given(in k-space) bywhereWe can then writeTaking the differential, we obtainSubstituting these expressions into the density of states function, we haveNoting thatthis density of states function can besimplified and written asDividing by will yield the density ofstates so that_______________________________________ 3、25For a one-dimensional infinite potential well, Distance between quantum statesNowNowThenDivide by the "volume" a, soSomJ_______________________________________3、26(a) Silicon,(i) At K, eVJThenmor cm(ii) At K,eVJThenmor cm(b) GaAs,(i) At K, Jmor cm(ii) At K, Jmcm_______________________________________ 3、27(a)Silicon,(i)At K, Jmor cm(ii)At K, Jmor cm(b)GaAs,(i)At K, Jmor cm(ii)At K, Jmor cm_______________________________________3、28(a)For ;eV; mJeV; mJeV; mJeV; mJ(b)For ;eV; mJeV; mJeV; mJeV; mJ_______________________________________3、29(a)(b)_______________________________________ 3、30Plot_______________________________________3、31(a)(b)(i)(ii)_______________________________________3、32(a),(b),(c),_______________________________________ 3、33or(a),(b),(c),_______________________________________3、34(a);;;;;(b);;;;;_______________________________________ 3、35andSo ThenOr_______________________________________ 3、36For , Filled stateJor eVFor , Empty stateJor eVTherefore eV_______________________________________3、37(a)For a 3-D infinite potential wellFor 5 electrons, the 5th electron occupiesthe quantum state ; soJor eVFor the next quantum state, which is empty, the quantum state is 、 This quantum state is at the same energy, soeV(b)For 13 electrons, the 13th electronoccupies the quantum state; soJor eVThe 14th electron would occupy the quantum state 、 This state is at the same energy, soeV_______________________________________ 3、38The probability of a state atbeing occupied isThe probability of a state atbeing empty isorso Q、E、D、_______________________________________3、39(a)At energy , we wantThis expression can be written asorThenor(b)At ,which yields_______________________________________3、40(a)(b) eV(c)ororwhich yields K_______________________________________3、41(a)or 0、304%(b)At K, eVThenor 14、96%(c)or 99、7%(d)At , for all temperatures_______________________________________3、42(a)ForThenFor , eVThenor(b)For eV,eVAt ,orAt ,or_______________________________________3、43(a)AtorAt , eVSoor(b)For ,eVAt ,orAt ,or_______________________________________ 3、44soor(a)At K, ForAt(b)At K, eVFor ,For ,At ,(eV)(c)At K, eVFor ,For ,At ,(eV)_______________________________________3、45(a)At ,Si: eV,orGe: eVorGaAs: eVor(b) Using the results of Problem 3、38, theanswers to part (b) are exactly the same asthose given in part (a)、_______________________________________3、46(a)oreVso K(b)or K_______________________________________3、47(a)At K,eVeVBy symmetry, for ,eVThen eV(b)K, eVFor , from part (a),eVThen eV_______________________________________。