原子物理学-习题解答(第五章)
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3 3 1 H → 2 He
3 M (1 H ) = 3.016050u
r =7.1fm r =4.4fm
5.12
β
M ( 3 He) = 3.016029u
∆M = 0.000021u = 0.0196MeV
β
5.13 β
0.0196MeV
12.33
(a),
1 ml
β
-
A = Nλ
N = 2×(0.001/22.4)6.02×1023 =5.38×1019
σ = 21000 b (1b =10-24cm2)
0.01%
N Aρ A
Cd
8.7 g/cm3
Φ = Φ0e
0.01%,
−σ ⋅t ⋅
94.8µ m
5.20 0.0156%.
20500 km3 0.0156%, 0.0312%.
2.05 × 1019 g
2.05 × 1019 ÷ 9 × 3.12 ×10−4 g
28
B / A = 7.07 MeV
5.4
34 16 S
M = 33.967865 u
∆m( 34 S ) = 16 M ( 1H ) + 18mn − M ( 34 S )
= 0.313305u
B / A = 931.494 × 0.313305 / 34 = 8.58( MeV )
5.5 0.5 MeV, 2.0 MeV, 10 MeV 100MeV
3
H , 3 He, 4 He
3
H
∆m = M ( 1 H ) + 2 M n − M ( 3 H ) = 0.009105u
B / A = 2.83MeV
3
He
∆m = 2M ( 1 H ) + M n − M ( 3 He) = 0.008286u
B / A = 2.57 MeV
4
He
∆m = 2M ( 1H ) + 2 M n − M ( 4 He) = 0.030377u
K
(3) (2) K L 3 K
16
31 keV
X
661 37 624 KeV 661 6 655 KeV L K
Ba
31keV
X
5.16
O + d →17 O + p
M(16O)=15.994915 u
M(17O)=16.999133 u M(2H)=2.014102 u
2 17 ∆E = M ( 16 8O ) + M ( 1H ) − M ( 8O ) − m p − me
140 140 140 Xe → 140 55 Cs → 56 Ba → 57 La → 58 Ce
5.7
63 29 Cu
63 29 Cu
64 29 Cu
2p3/2 j = 3/2
64 29 Cu
l 1 2p3/2
(-1)l = -1
3/21f5/2
I = jn − j p
1
1
( −1 ) n
l +l p
T1/2 = 4.5×109
5.24
0.150MeV
3
H (d , n) 4 He
90o
0o
32
Q= (Mx+MX)c2-( My+MY)c2 Q 17.59 MeV 0 90 En(1+1/4)=17.59+0.15(1-2/4) En = 14.1 MeV 0 0 5.5.5 En = 14.9 MeV
2.15 ×1024 MeV 1.5 ×1039 MeV =2.4×1026 J
5.21
241 95 Am 241 95 Am
2I 1
130 52Te
I =5/2
5.22
β-
130 54 Xe
2.5 MeV 5.23
238 238
U
α
1 mg 238U
740
α
U
A = λN = 740 / min
ln 2 1 × 10 −3 ⋅ 6.02 × 10 23 ⋅ = 740 T1 / 2 238
=(-1)1+3=+1
29
5.8
7
7
Li
Li
3
4
1 p3/ 2
7
Li
3/2
5.9
9 4
9 4 Be
14 7N
37 17 Cl
Be N Cl
210
1 p3/ 2
1 p1/ 2 1 d3/ 2 Po α
210 84
3/2
1 1 3/2 +1 1
14 7
37 17
5.10
5.3MeV
Po →
206 82
Pb + α
Ed = Tα (1 +
mα ) = 5.4( MeV ) M Pb
5.11
27 13 Al
α
5.3MeV
8.6MeV
α
R = r0 A1/ 3 = 4.35 fm
Z1 Z 2 e 2 = 5.67 MeV Vc = 1/ 3 1/ 3 ) r0 ( A1 + A2
5.3MeV 8.6MeV
1/ 2 0.4279MeV = 1/ 2 0.0304MeV
En = 0.183 MeV
0.00092 MeV
33
r0= 1.45fm
R = r0 A1/ 3
4 2
r0 = 1.45 fm
H Ag Au
A=4
R = 2.30 fm
107 47 238 92
R = 6.88 fm R = 8.99 fm
3 5.3 H 3He 4He B/A M(3H) =3.016050 u M(3He) =3.016029 u M(4He) =4.002603 u M(1H) =1.007825 u Mn =1.008665 u
= 0.002059u = 1.92 MeV
5.17 U 4.5×109a
238
235
U
99.27% 235U 7.05×108a
0.72%
α
238
U
T
9 8
T
exp(−T ln 2 / T1/ 2 )
e −T ln 2 / 4.5×10 / e −T ln 2 / 7.05×10 = 0.9927 / 0.0072
3
5.25
H ( p, n) 3He
1 (1) (2)
Q 1.120 MeV
-0.764 MeV 300
Eth =
(2)
− (m B + mb )Q 3+1 = 0.764 = 1.019 MeV m B + mb − m a 3 +1−1
/2 E1 n
− (1.12)1 / 2 (cos 30) ± 1.12(cos 2 30) + 4[3 × (−0.764) + 2 × 1.12] = 4
E = me c 2 + E k
pc = ( E 2 − m 2 c 4 )
λ =h/ p λ = h / p =1.63×10-3 nm
Ek = 0.5MeV E=1.01MeV p=0.87MeV/c Ek=2.0MeV: Ek =10MeV Ek =100MeV β β
λ = h / p = 5×10-4 nm
λ=
ln 2 =1.78×10-9 /s T1 / 2
30
A = 5.38(1.78)×1010= 9.6×1010 /s
210 88 Bi
5.14
RaE 0.33 MeV
4.0mg W
5.01
(d)
β
N=
4mg × 6.02 ×1023 / mol = 1.15 ×1019 −1 210 g ⋅ mol
T = 5.94×109 a 5.18 (16.1±0.3) (5.40±0.14)%
31
8g (5.0±0.1) λN0 = 16.1 λN=λN0exp(-0.693T/T1/2)
14
(9.5±0.1) C 5730 a λN = (9.5-5)/8
T=
5.19
113
5730 4.5 ) = 3600 × ln( 0.693 16.1 × 8 × 0.054
5.1 1000Gs (3) 18.2 cm
1000V 1
(2)
mv = qBR
1 2 mv = eU 2
v=
2eU = 1.1×105 m / s qBR
m = qBR / v = 2.65 ×10−26 kg
2.65 ×10 −26 A= = 16 1.66 ×10−27
5.2
4 2 He 107 47 Ag 238 92 U
ln 2 = 1.6 ×10 −6 / s T1/ 2
λ=
A = N λ = 1.84 ×1013 / s
W = ε ⋅ A = 0.97 J / s
5.15 (1) 0.515MeV (2)
137 36 Ba 137 55 Cs
β γ
-
1.76 MeV 0.661 MeV EK = 37 keV L EL = 6 keV
p=10.5 MeV/c
λ = h / p =1.18×10-4
λ=பைடு நூலகம்
hc = 1.24 ×10−5 nm E
57
5.6
Ni
140
Xe
β
β
Z=
A=57 A=140 Z=26 Z=58
58 28
A 1.98 + 0.0155 A2 / 3
Fe Ce
58 Ni → 27 Co →
58 26
Fe
140 54
3 M (1 H ) = 3.016050u
r =7.1fm r =4.4fm
5.12
β
M ( 3 He) = 3.016029u
∆M = 0.000021u = 0.0196MeV
β
5.13 β
0.0196MeV
12.33
(a),
1 ml
β
-
A = Nλ
N = 2×(0.001/22.4)6.02×1023 =5.38×1019
σ = 21000 b (1b =10-24cm2)
0.01%
N Aρ A
Cd
8.7 g/cm3
Φ = Φ0e
0.01%,
−σ ⋅t ⋅
94.8µ m
5.20 0.0156%.
20500 km3 0.0156%, 0.0312%.
2.05 × 1019 g
2.05 × 1019 ÷ 9 × 3.12 ×10−4 g
28
B / A = 7.07 MeV
5.4
34 16 S
M = 33.967865 u
∆m( 34 S ) = 16 M ( 1H ) + 18mn − M ( 34 S )
= 0.313305u
B / A = 931.494 × 0.313305 / 34 = 8.58( MeV )
5.5 0.5 MeV, 2.0 MeV, 10 MeV 100MeV
3
H , 3 He, 4 He
3
H
∆m = M ( 1 H ) + 2 M n − M ( 3 H ) = 0.009105u
B / A = 2.83MeV
3
He
∆m = 2M ( 1 H ) + M n − M ( 3 He) = 0.008286u
B / A = 2.57 MeV
4
He
∆m = 2M ( 1H ) + 2 M n − M ( 4 He) = 0.030377u
K
(3) (2) K L 3 K
16
31 keV
X
661 37 624 KeV 661 6 655 KeV L K
Ba
31keV
X
5.16
O + d →17 O + p
M(16O)=15.994915 u
M(17O)=16.999133 u M(2H)=2.014102 u
2 17 ∆E = M ( 16 8O ) + M ( 1H ) − M ( 8O ) − m p − me
140 140 140 Xe → 140 55 Cs → 56 Ba → 57 La → 58 Ce
5.7
63 29 Cu
63 29 Cu
64 29 Cu
2p3/2 j = 3/2
64 29 Cu
l 1 2p3/2
(-1)l = -1
3/21f5/2
I = jn − j p
1
1
( −1 ) n
l +l p
T1/2 = 4.5×109
5.24
0.150MeV
3
H (d , n) 4 He
90o
0o
32
Q= (Mx+MX)c2-( My+MY)c2 Q 17.59 MeV 0 90 En(1+1/4)=17.59+0.15(1-2/4) En = 14.1 MeV 0 0 5.5.5 En = 14.9 MeV
2.15 ×1024 MeV 1.5 ×1039 MeV =2.4×1026 J
5.21
241 95 Am 241 95 Am
2I 1
130 52Te
I =5/2
5.22
β-
130 54 Xe
2.5 MeV 5.23
238 238
U
α
1 mg 238U
740
α
U
A = λN = 740 / min
ln 2 1 × 10 −3 ⋅ 6.02 × 10 23 ⋅ = 740 T1 / 2 238
=(-1)1+3=+1
29
5.8
7
7
Li
Li
3
4
1 p3/ 2
7
Li
3/2
5.9
9 4
9 4 Be
14 7N
37 17 Cl
Be N Cl
210
1 p3/ 2
1 p1/ 2 1 d3/ 2 Po α
210 84
3/2
1 1 3/2 +1 1
14 7
37 17
5.10
5.3MeV
Po →
206 82
Pb + α
Ed = Tα (1 +
mα ) = 5.4( MeV ) M Pb
5.11
27 13 Al
α
5.3MeV
8.6MeV
α
R = r0 A1/ 3 = 4.35 fm
Z1 Z 2 e 2 = 5.67 MeV Vc = 1/ 3 1/ 3 ) r0 ( A1 + A2
5.3MeV 8.6MeV
1/ 2 0.4279MeV = 1/ 2 0.0304MeV
En = 0.183 MeV
0.00092 MeV
33
r0= 1.45fm
R = r0 A1/ 3
4 2
r0 = 1.45 fm
H Ag Au
A=4
R = 2.30 fm
107 47 238 92
R = 6.88 fm R = 8.99 fm
3 5.3 H 3He 4He B/A M(3H) =3.016050 u M(3He) =3.016029 u M(4He) =4.002603 u M(1H) =1.007825 u Mn =1.008665 u
= 0.002059u = 1.92 MeV
5.17 U 4.5×109a
238
235
U
99.27% 235U 7.05×108a
0.72%
α
238
U
T
9 8
T
exp(−T ln 2 / T1/ 2 )
e −T ln 2 / 4.5×10 / e −T ln 2 / 7.05×10 = 0.9927 / 0.0072
3
5.25
H ( p, n) 3He
1 (1) (2)
Q 1.120 MeV
-0.764 MeV 300
Eth =
(2)
− (m B + mb )Q 3+1 = 0.764 = 1.019 MeV m B + mb − m a 3 +1−1
/2 E1 n
− (1.12)1 / 2 (cos 30) ± 1.12(cos 2 30) + 4[3 × (−0.764) + 2 × 1.12] = 4
E = me c 2 + E k
pc = ( E 2 − m 2 c 4 )
λ =h/ p λ = h / p =1.63×10-3 nm
Ek = 0.5MeV E=1.01MeV p=0.87MeV/c Ek=2.0MeV: Ek =10MeV Ek =100MeV β β
λ = h / p = 5×10-4 nm
λ=
ln 2 =1.78×10-9 /s T1 / 2
30
A = 5.38(1.78)×1010= 9.6×1010 /s
210 88 Bi
5.14
RaE 0.33 MeV
4.0mg W
5.01
(d)
β
N=
4mg × 6.02 ×1023 / mol = 1.15 ×1019 −1 210 g ⋅ mol
T = 5.94×109 a 5.18 (16.1±0.3) (5.40±0.14)%
31
8g (5.0±0.1) λN0 = 16.1 λN=λN0exp(-0.693T/T1/2)
14
(9.5±0.1) C 5730 a λN = (9.5-5)/8
T=
5.19
113
5730 4.5 ) = 3600 × ln( 0.693 16.1 × 8 × 0.054
5.1 1000Gs (3) 18.2 cm
1000V 1
(2)
mv = qBR
1 2 mv = eU 2
v=
2eU = 1.1×105 m / s qBR
m = qBR / v = 2.65 ×10−26 kg
2.65 ×10 −26 A= = 16 1.66 ×10−27
5.2
4 2 He 107 47 Ag 238 92 U
ln 2 = 1.6 ×10 −6 / s T1/ 2
λ=
A = N λ = 1.84 ×1013 / s
W = ε ⋅ A = 0.97 J / s
5.15 (1) 0.515MeV (2)
137 36 Ba 137 55 Cs
β γ
-
1.76 MeV 0.661 MeV EK = 37 keV L EL = 6 keV
p=10.5 MeV/c
λ = h / p =1.18×10-4
λ=பைடு நூலகம்
hc = 1.24 ×10−5 nm E
57
5.6
Ni
140
Xe
β
β
Z=
A=57 A=140 Z=26 Z=58
58 28
A 1.98 + 0.0155 A2 / 3
Fe Ce
58 Ni → 27 Co →
58 26
Fe
140 54