陈宏芳_原子物理学_答案

h = 12.4 keV/c ∆x
L
1 fm =10-15 m L = 10-6 m
∆p ≥
h ∆x
∆p ≈ p , E =
p2 h2 ≥ 2 m 2 mL2
8
(1) (2) (3) (4)
L =0.1nm, m=511KeV, E=150eV L =10fm, m=511KeV, E=1.5×1010 =15 GeV L =10fm, m=940MeV E=8.1 MeV
E=− Rhc n2
R=
me
me e 4 2 3 8ε 0 hc
1.3.2
rn =
4πε 0 h 2 2 n me e 2
n =1 2
3 ….
6
υn =
l
e2 1 ⋅ 4πε 0 h n
n =1
2 3 ….
l = meυ n rn = nh
n = 1 2 3 … ..
2.1
632.8 nm
E=
2.2 2.2 eV
2
4
t = 2.03µm
1.10 10.19 eV 485 nm
E=
hc 1240eV ⋅ nm = 2.56eV = λ 485nm 13.6 − 10.19 − 2.56 = 0.85eV
99.2nm
1.11 108nm
121.5nm
R = 1.097 × 10 7 m −1 ~ ν
RΒιβλιοθήκη Baidu
0.9189, 0.8401, 0.7503 k = 3.5
2.7 (1) (2) 50 keV 12.4 GeV 1 GeV = 109 eV (1)
λ=
h = 5.5 × 10−3 nm 2me Ek
(2)
λ=
hc E −m c
2 2 4
= 1×10−7 nm
2.8
15 eV
0.1 nm
∆x = 0.1nm
∆p ≥
2.9 (1) (2) (3) (4) L=1Å L = 10 fm L = 10fm m = 10-6 g
1.0 mg/cm2 1.0 cm2 Au A
4 3
ρ
1.93×10 kg/m
dn dσ = Nt × × dΩ n dΩ
dσ e 2 2 zZ 2 1 =( ) ( ) 4 dΩ 4πε 0 4 E sin (θ / 2)
= 1.29 × 10 −24 m 2
dσ = dσ dΩ = 1.29 × 10 − 26 m 2 dΩ
(2)
n = 1.55 × 10 5
(3) 10 180
n = 6.75 × 10 6
2
2.16 × 10 7
0 1.6 (1) (2) (3) (1) 10 He+ Li++
1.48 × 10 7
n2 rn = a0 , Z
Z=1: r1 = 0.53 × 10 −10 m ,
vn =
αc Z n
3
R M = R∞
M me + M
~=R ( 1 ) ν M 22
~ me + 4 M P 2741940 ν MP H = ⋅ = ~ ν m + M 4 M 2743059 He e P P
M P : me = 1837.5
1.9 6.0 MeV 1.0×10-4 600
π dσ e 2 2 zZ 2 1 dσ = ) ( ) sin θdθ ⋅ dΩ = 2π ∫ ( 4 π / 3 dΩ 4πε 0 4 E sin (θ / 2)
100 90.4=9.6 MeV
v=
2E = 0.14c M
1350
2.4
0.071 nm
X
∆λ = λ '− λ =
h (1 − cos θ ) me c
θ = 135o
∆λ = 0.004nm
λ ' = 0.071 + 0.004 = 0.075nm
E=
2.5
1240eV ⋅ nm = 16.5keV 0.075
L = 10−6 m , m = 10−6 g , E 2.2×10
46
J
2.10 4 eV
h 1 = κ2 2m(V0 − E )
1 197.3MeV ⋅ fm = = 9.76 ×104 fm κ2 2 × 0.511MeV × 4eV
2.11 m u (x) = sin kx
Ek = −
E=
2.12 (1) (2) m
∆x∆p ≈ h
≈ hω
−
h 2 d 2u + Vu = Eu 2m dx 2
−( mω 2 )x 2h
−
1 h2 ∂2 u( x ) + mω 2 x 2 u( x ) = Eu( x ) 2 2 m ∂x 2
(2)
u 0 ( x) = e
(3)
E
hω 2 3hω E1 = 2 1 E = mω 2 x 2 2 E0 =
h2 ∂2 ; 2m ∂x 2
E k u( x ) = −
h2 ∂2 h2k 2 = u ( x ) u( x ) 2m ∂x 2 2m
h2k 2 2m
V(x) = 1/2 mω2x2
u 0 ( x) = e
−(
mω 2 )x 2h
mω
2
mω −( 2h ) x u1 ( x ) = 2 xe h
(3) (1)
2Mc
∆λ
2
= E 0 (1 −
E0 2Mc
2
) = E 0 (1 − 7 × 10 −9 )
6.6×10-7 nm
n=4 (1) (2) (1) n 4 (2) 97.4nm 102.7 nm 487 nm 653 nm 1911 nm
12.75 eV 121.8 nm
12.75 eV
1.18
m=n-1
θ =π /2
D=(
zZe 2 1 )( ) 4πε 0 E
1.32 ×10−13 m
b=
D ⋅ cot(θ / 2) = 0.66 ×10 −13 m 2
rm =
D 1 [1 + ] = 1.59 ×10−13 m 2 sin(θ / 2)
1
1.3 600 10 cm 197 Z 79
100 keV
R 2.5 2
~ = R( 1 − 1 ) ν k2
Z Z=2,
~ = 4 R( 1 − 1 ) ν m2 n2
m =2, n = 4,5,7
4
n=3, n=6, 1.12 45 1.13
λ 164.1nm λ = 102.5nm
n=10
n=8 9
12.2eV 13.6eV 3.4eV, 1.5eV 12.2eV n=1 n=2 n=3 12.1eV, 10.2eV, 1.9eV 102.6 121.8 653 nm
1.14 (1) (2) (3) (1) (2)
r1 = 2a 0 = 1.06 × 10 −10 m = 0.106nm
(3)
E = 13.6 / 2 = 6.8eV E = 3.4 / 2 = 1.7eV 1 R = R∞ =0.548×107 m-1, 2 1240eV ⋅ nm = 243nm λ= 5.1eV
E=
−e 2 (3R 2 − r 2 ) 8πε 0 R3 e2 3 8πε 0 R
R 0.16 nm
13.6eV
(2)
e2 r F= 4πε 0 R 3
&& F = me r
&& r e2 2π c 2 = = ω2 = ( ) 3 r 4πε 0 me R λ
R = 2.95 ×10−10 m
1.2 T = 0.87MeV
dn = dσ ⋅ N ⋅ t = 1.29 × 10 − 26 m 2 × 3.06 × 1018 / cm 2 n = 0.039%
1.4 T = 1.2 MeV
θ =π /3
π
dσ =
π dσ e 2 2 zZ 2 1 ) ( ) sin θdθ ⋅ dΩ = 2π ∫ ( 4 π / 2 4πε dΩ 4 E sin (θ / 2) 0
7
(1) 50 eV (2 50 eV (3) 50 eV 1 50eV
940 MeV
λ=
h 1240 = = 24.8nm 50 p
(2)
50eV
λ= λ=
h = 0.174nm 2me Ek h = 0.004 nm 2mn Ek
0.025 eV
(3)
50eV
2.6 1877 MeV
λ=
h = 0.13nm 2mn Ek
ν = Rc (
1 m
2
−
1 n
2
) = Rc (
1 ( n − 1)
2
−
1 n
2
) = Rc
1 n
2
(
1 (1 − 1 / n) 2
− 1)
n >>1
ν =
2 Rc n3
1.3.1 1.3.2
E=−
e2 1 2π 2 Rc 2 1 / 3 (e 4 m e ⋅ ( ) ) =− 2/3 2 ⋅ 4πε 0 r 2(4πε 0 ) n3
E = 13.6eV E = 54.4eV E = 122.4eV
v1 = 4.4 × 10 6 m ⋅ s −1 ,
Li++ Z=3: r1 = 0.18 × 10 −10 m ,
v1 = 6.6 × 10 6 m ⋅ s −1 ,
(2)
T = 13.6eV T = 54.4eV T = 122.4eV
U = 10.2eV
(3)
λ=
hc = 121.6nm ∆E
U = 40.8eV
hc = 30.4nm λ= ∆E U = 91.8eV hc = 13.5nm λ= ∆E
1.7 13.6 eV B++++ n=2 n =1
B ++++
n=2 1.8 n=1
13.6 × 5 2 = 340eV
255 eV He+ 2741940 2743059m-1
207 me
1.15 (1) (2) (1)
µ
2
Z=1
µ
H
207 * 1836 me = 186 me r1 = a0/186 2.85×10- 4 nm 207 + 1836 1 (b) E = − m( αc ) 2 E -2530 eV 2 207 * 1836 * 2 m dµ = me = 196 me 207 + 1836 * 2 E = - 2665 eV 1.16 n=4 n=1 m=
9
x ~ ∆x
p~ ∆p
∆x =
2E mω 2
E= hω 2
E=
h2 p 2 ( ∆p) 2 = = 2m 2m 2m(∆x ) 2
r2 = 2.12 × 10 −10 m
v1 = 2.2 × 10 6 m ⋅ s −1 , v 2 = 1.1 × 10 6 m ⋅ s −1
He+ Z=2:
r1 = 0.26 × 10 −10 m ,
r2 = 1.06 × 10 −10 m v 2 = 2.2 × 10 6 m ⋅ s −1 r2 = 0.71 × 10 −10 m v 2 = 3.3 × 10 6 m ⋅ s −1
dσ = 211b
1.5 1.0 MeV 10 min. 0 0 (1) 59 61 0 (2) θ >θ0 = 60 0 (3) θ <θ0 = 10 (1) 3.6×104 /
α α
1.0µm
dσ =
61o dσ e 2 2 zZ 2 1 ⋅ dΩ = 2π ∫ o ( sin θdθ ) ( ) 4 59 dΩ 4πε 0 4 E sin (θ / 2)
1240 = 1.96eV 632.8
350.0 nm
1240 = 3.54eV 350 3.54 2.2= 1.34 eV E=
2.3 (2) 1 100 MeV 938.26 MeV 900
ν '=
1+
hν (1 − cos θ ) m p c2
ν
90o
E ' = hν ' = 90.4 MeV
1.1 (1) (2) (1) V: R
13.6eV. 0.6 µm r 0
v ∞v v R v V (r ) = ∫ E ⋅ dl + ∫ E ⋅ dl
r R
V (r ) = ∫
=
R
r
∞ e⋅r ' e dr '+ ∫ dr ' 3 R 4πε 0 R 4πε 0 r '2
e(3R 2 − r 2 ) 8πε 0 R3
= 9.83 × 10 −27 m 2
dn = dσ ⋅ N .t = 5.9 × 10 − 22 m − 2 × 9.83 × 10 − 27 m − 2 = 5.8 × 10 − 4 n
10
n = 3.6 × 10 4 ⋅ 5.8 × 10 −4 × 6 × 10 2 = 1.25 × 10 4
E0=12.75 eV ER E
E0 = E
+ ER
M
5
ER =
2 P2 E2 PR = = 2M 2M 2Mc 2
E0/Mc2
E
12.75 eV/c MV
2 E0
≈ E0 −
2 E0
2 Mc 2
1.36×10-8 c 4.1 m/s
12.75 eV/c V
E
λ 1.17
≈ E0 −
97.4 nm
dn e 2 2 zZ 2 8π ) ( ) Nt =( d sin( θ / 2 ) n 4πε 0 4E sin 3 ( θ / 2 ) e 2 2 zZ 2 90 8π ∆n ) ( ) Nt ∫ =( d sin( θ / 2 ) 3 n 4πε 0 4E 30 sin ( θ / 2 )
=1.44(eV⋅nm)2(79/4⋅E) (5.9×10 )12π ⋅t =10-4