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=m Cesium: 90.1=E eV ()()19106.190.1-⨯= JSo,()()()()51910341054.6106.190.110310625.6---⨯=⨯⨯⨯=λcm orμλ654.0=m_______________________________________2、6(a) 9341055010625.6--⨯⨯==λhp2710205.1-⨯=kg-m/s 331271032.11011.9102045.1⨯=⨯⨯==--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/s or 51065.1⨯=υcm/s (c) Yes_______________________________________ 2、7(a) (i) ()()()1931106.12.11011.922--⨯⨯==mE p 2510915.5-⨯=kg-m/s925341012.110915.510625.6---⨯=⨯⨯==p h λm or o A 2.11=λ(ii)()()()1931106.1121011.92--⨯⨯=p 241087.1-⨯=kg-m/s1024341054.3108704.110625.6---⨯=⨯⨯=λm or 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--⨯⨯=p 2310532.2-⨯=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--⨯⨯==λhp2610794.7-⨯= 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--⨯⨯⨯===λνhch 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 2810054.1-⨯=kg-m/s_______________________________________2、13(a) (i) =∆∆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/s so ()()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---⨯⨯⨯=∆E 2110888.8-⨯=Jor 219211055.5106.110888.8---⨯=⨯⨯=∆E eV _______________________________________2、143223410054.11010054.1---⨯=⨯=∆=∆x p kg-m/s150010054.132-⨯=∆=∆⇒=m p m p υυ 36107-⨯=∆υm/s_______________________________________2、15(a) =∆∆t E()()1619341023.8106.18.010054.1---⨯=⨯⨯=∆t s (b) 1034105.110054.1--⨯⨯=∆=∆x p 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∂⋅-,,,2112122 and()()()()t t x j t x x V x t x m ∂ψ∂=ψ+∂ψ∂⋅-,,,2222222 Adding the two equations, we obtain()()[]t x t x x m ,,221222ψ+ψ∂∂⋅- ()()()[]t x t x x V ,,21ψ+ψ+()()[]t x t x tj ,,21ψ+ψ∂∂=which 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[]21ψ⋅ψ∂∂=tjwhich can be written as⎥⎦⎤⎢⎣⎡∂ψ∂⋅∂ψ∂+∂ψ∂ψ+∂ψ∂ψ-x x x x m 2121222221222()[]⎥⎦⎤⎢⎣⎡∂ψ∂ψ+∂ψ∂ψ=ψ⋅ψ+t t j x V 122121 Dividing by 21ψ⋅ψ, we find⎥⎦⎤⎢⎣⎡∂ψ∂∂ψ∂ψψ+∂ψ∂⋅ψ+∂ψ∂⋅ψ-x x x xm21212121222222112 ()⎥⎦⎤⎢⎣⎡∂ψ∂ψ+∂ψ∂ψ=+t t j x V 112211Since 1ψ is a solution, then()tj x V x m ∂ψ∂⋅ψ⋅=+∂ψ∂⋅ψ⋅-1121212112Subtracting these last two equations, we have⎥⎦⎤⎢⎣⎡∂ψ∂∂ψ∂ψψ+∂ψ∂⋅ψ-x x x m 212122222212t j ∂ψ∂⋅ψ⋅=221 Since 2ψ is also a solution, we have()t j x V x m ∂ψ∂⋅ψ⋅=+∂ψ∂⋅ψ⋅-2222222112 Subtracting these last two equations, we obtain()02221212=-∂ψ∂⋅∂ψ∂⋅ψψ⋅-x V xx mThis 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 A 121232=⎥⎦⎤⎢⎣⎡⎪⎭⎫ ⎝⎛--Aso 212=Aor 21=A_______________________________________2、18()1cos 22/12/12=⎰+-dx x n A π()142sin 22/12/12=⎥⎦⎤⎢⎣⎡++-ππn x n x A ⎪⎭⎫⎝⎛==⎥⎦⎤⎢⎣⎡⎪⎭⎫ ⎝⎛--211414122A Aor 2=A_______________________________________2、19Note that 10*=ψ⋅ψ⎰∞dxFunction has been normalized 、 (a) Nowdx a x a P oa o o 24exp 2⎰⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛-=dx a x a oa o o⎰⎪⎪⎭⎫⎝⎛-=42exp 2402exp 22o a o o o a x a a ⎪⎪⎭⎫ ⎝⎛-⎪⎪⎭⎫⎝⎛-=or()⎪⎭⎫ ⎝⎛--=⎥⎥⎦⎤⎢⎢⎣⎡-⎪⎪⎭⎫⎝⎛--=21exp 1142exp 1o oa 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 22exp 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-⨯= so 41037.9⨯=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 p 2610555.4-⨯=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()212100698.1-⨯=n E n Jor()19212106.1100698.1--⨯⨯=n E nor ()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 ()20210018.6-⨯=n J or()()3761.0106.110018.6219202n n E n =⨯⨯=--eV Then376.01=E eV 504.12=E eV 385.33=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 π =()()()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 E13103025.3-⨯=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 waveequation()()()()02222=-+∂∂x x V E mx x ψψWe 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 ψψThe solution is of the form ()kx B kx A x sin cos +=ψ where22 mEk =Boundary conditions:()0=x ψ at 2,2a x a x +=-= First mode solution: ()x k A x 111cos =ψ where222112ma E a k ππ=⇒=Second mode solution: ()x k B x 222sin =ψ where22222242ma E a k ππ=⇒= Third mode solution: ()x k A x 333cos =ψ where22233293ma E a k ππ=⇒= Fourth mode solution: ()x k B x 444sin =ψ where222442164ma E a k ππ=⇒= _______________________________________2、30The 3-D time-independent wave equation incartesian coordinates for ()0,,=z y x V is:()()()222222,,,,,,z z y x y z y x x z y x ∂∂+∂∂+∂∂ψψψ ()0,,22=+z y x mEψUse 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 mEDividing by XYZ and letting222 mEk =, wefind(1) 01112222222=+∂∂⋅+∂∂⋅+∂∂⋅k zZ Z y Y Y x X XWe may set01222222=+∂∂⇒-=∂∂⋅X k xX k x X X x x 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 finda n k y y π=, ....3,2,1=y nan k z z π=, ...3,2,1=z nFrom Equation (1) above, we have02222=+---k k k k z y x or222222 mEk k k k z y x ==++so that()2222222z y x n n n n n n maE E z y x ++=→π _______________________________________2、31 (a)()()()0,2,,22222=⋅+∂∂+∂∂y x mEy y x x y x ψψψSolution 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=+--mE 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 an m E y x n n yx ππ (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 -=∆π For 1,212==n n Then22223ma E π =∆(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-⨯=Jor1719361046.2106.110939.3---⨯=⨯⨯=∆E eV _______________________________________2、33(a) For region II, 0>x()()()0222222=-+∂∂x V E mx x O ψψGeneral form of the solution is()()()x jk B x jk A x 22222exp exp -+=ψ where()O V E mk -=222 Term with 2B represents incident wave andterm with 2A represents reflected wave 、 Region I, 0<x()()0212212=+∂∂x mEx x ψψGeneral form of the solution is()()()x jk B x jk A x 11111exp exp -+=ψ where212 mEk =Term involving 1B represents the transmitted 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 ()222 E 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 VE VET o o22exp 116-⎪⎪⎭⎫⎝⎛-⎪⎪⎭⎫ ⎝⎛≅ where ()222 E 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 VEV E T O O 22exp 116-⎪⎪⎭⎫⎝⎛-⎪⎪⎭⎫ ⎝⎛≅ (a) For ()o m m 067.0=()222 E 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⎪⎭⎪⎬⎫⎪⎩⎪⎨⎧⨯⨯-⨯---or9210124.4⨯=k m 1- Then⎪⎭⎫⎝⎛-⎪⎭⎫ ⎝⎛=8.02.018.02.016T()()[]109101510124.42exp -⨯⨯-⨯or51027.1-⨯=T_______________________________________2、37()a k VE VET o o22exp 116-⎪⎪⎭⎫⎝⎛-⎪⎪⎭⎫ ⎝⎛≅ where ()222 E 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)where212 mEk =Region II:()()()x k B x k A x 22222exp exp -+=ψwhere ()222 E 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 +=+ ⇒=dx d dx d 21ψψ22221111B k A k B jk A jk -=-At a x =: ⇒=32ψψ()()a k B a k A 2222exp exp -+()a jk A 13exp =⇒=dxd 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, wewant to 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()222 E 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 find222212 O mV k k =+ and()⎥⎦⎤⎢⎣⎡⎥⎦⎤⎢⎣⎡-=22222122 mE E V m k k O ()()E E V m O -⎪⎭⎫ ⎝⎛=222()()E V E V m O O ⎪⎪⎭⎫ ⎝⎛-⎪⎭⎫⎝⎛=1222 Then()()⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛-⎪⎭⎫ ⎝⎛⎪⎪⎭⎫⎝⎛=E V E V m a k mV A A A A O O O12162exp 222222*33*11()a k V EV 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 ψψ()()()x jk B x jk A x 11111exp exp -+=ψincidentreflectedwhere212 mEk = Region II: 1V V = ()()()⇒=-+∂∂02221222x V E m x x ψψ ()()()x jk B x jk A x 22222exp exp -+=ψ transmitted reflectedwhere()2122V E m k -= Region III: 2V V =()()()⇒=-+∂∂02322232x V E m x x ψψ()()x jk A x 333exp =ψtransmitted where ()2232 V 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 : ⇒=21ψψ2211B A B A +=+⇒∂∂=∂∂xx 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 from the boundary 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 ψψRegion II: 0=V , so()()0222222=+∂∂x mEx x ψψRegion III: 03=⇒∞→ψV The 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()212 E V m k O -= and222 mEk =(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 =, then2212B 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 EV O 22tan 1 or⎥⎥⎦⎤⎢⎢⎣⎡⋅-=-a mE E V EO 22tan This last equation is valid only forspecific values of the total energy E 、 The energy levels are quantized 、_______________________________________2、41 ()222424ne m E o o n ∈-=π(J)()222324n e m o o ∈-=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=drr 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、43100ψ is independent of θ and φ, so the wave equation in spherical coordinates reduces to ()()021222=-+⎪⎭⎫ ⎝⎛∂∂∂∂⋅ψψr V E m r r r r o where()r a m r e r V o o o 224 -=∈-=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 r exp 1122/51002πψ We then obtain2/5100211⎪⎪⎭⎫⎝⎛⋅-=⎪⎪⎭⎫ ⎝⎛∂∂∂∂o a r r r πψ ⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛-⎪⎪⎭⎫ ⎝⎛-⎪⎪⎭⎫ ⎝⎛-⨯o o o a r a r a r r exp exp 22 Substituting 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 2220exp 112/3=⎪⎪⎭⎫ ⎝⎛-⎪⎪⎭⎫ ⎝⎛⋅⎪⎪⎭⎫ ⎝⎛⨯o o a r a π where ()222241224oo o o a m e m E E -=∈-==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 oor ⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛-⎪⎪⎭⎫ ⎝⎛⋅o o a r a exp 112/3π0211222=⎪⎭⎪⎬⎫⎪⎩⎪⎨⎧⎪⎪⎭⎫ ⎝⎛+-++-⨯r a a a r a o o o o which gives 0 = 0 and shows that 100ψ is indeed a solution to the wave equation 、 _______________________________________ 2、44All elements are from the Group I column ofthe periodic table 、 All have one valence electron in the outer shell 、 _______________________________________。