Chapter 29
Nuclear Physics
Answers to Even Numbered Conceptual Questions
4.
An alpha particle is a doubly positive charged helium nucleus, is very massive and does
not penetrate very well. A beta particle is a singly negative charged electron and is very
light and only slightly more difficult to shield from. A gamma ray is a high energy photon,
or high frequency electromagnetic wave, and has high penetrating ability.
6.
Beta particles have greater penetrating ability than do alpha particles.
8.
The much larger mass of the alpha particle as compared to that of the beta particle ensures
that it will not deflect as much as does the beta, which has a mass about 7000 times
smaller.
Problem Solutions
29.8
13
Using r = rA
, with r0 = 1.2× 10−15 m = 1.2 fm , gives:
0
(a) For 24 H e,A = 4 , and
(b) For
29.12
238
92
r = ( 1.2 fm
H e,A = 238 , and r = ( 1.2 fm
13
)( 4) = 1.9 fm = 1.9× 10−15 m
13
)( 238) = 7.4 fm = 7.4× 10−15 m
∆m = Z m H + ( A − Z ) m n − m and Eb A = ∆m ( 931.5 M eV u ) A
55
25
∆m
( in u )
( in M eV )
(A – Z)
25
30
54.938 048
0.517 527
8.765
Fe
26
30
55.934 940
0.528 460
8.786
Co
27
32
58.933 198
0.555 357
8.768
Mn
56
26
59
27
( in u )
Z
Nucleus
m
Eb A
Therefore, 56
26 Fe has a greater binding energy per nucleon than its neighbors. This gives
us finer detail than is shown in Figure 29.4.
457
458
CHAPTER 29
29.15
The decay constant is λ =
ln 2
, so the activity is
T1 2
3.0 × 1016 ) ln 2
(
N ln 2
R = λN =
=
= 1.7 × 1010 decays s
4
T1 2
( 14 d ) ( 8.64× 10 s d )
⎛
⎞
1 Ci
R = ( 1.7 × 1010 decays s) ⎜
⎟ = 0.46 C i
10
⎝ 3.7 × 10 decays s⎠
or
29.20
Recall that the activity of a radioactive sample is directly proportional to the number of
radioactive nuclei present and hence, to the mass of the radioactive material present.
R
N
m
=
=
R0 N 0 m 0
and
R = R 0e− λt becom es
The decay constant is
λ=
ln 2
ln 2
=
= 0.181 d −1
T1 2 3.83 d
Thus,
m = m 0e− λt
If m 0 = 3.00 g and the elapsed time is t= 1.50 d , the mass of radioactive material
remaining is
(
)
− 0.181 d −1 ( 1.50 d )
m = m 0e− λt = ( 3.00 g ) e
29.22
Using R = R 0 e− λt , with R R 0 = 0.125 , gives λt= − ln ( R R 0 )
t= −
or
29.25
= 2.29 g
212
83
95
36
Bi →
Kr →
144
60
ln ( R R 0 )
λ
⎡ ln ( R R 0 ) ⎤
⎡ ln ( 0.125) ⎤
4
= − T1 2 ⎢
⎥ = − ( 5730 yr) ⎢
⎥ = 1.72× 10 yr
l
n
2
l
n
2
⎣
⎦
⎣⎢
⎦⎥
208
81
Tl + 24H e
95
37
Rb + −01e
N d → 24H e + 140
58 Ce
Nuclear Physics
29.30
The energy released in the decay
238
92
U → 24H e +
Th is
234
90
Q = ( ∆m ) c2 = ⎡⎣ m 238 U − ( m 4 H e + m 234 Th ) ⎤⎦ c2
= ⎡⎣ 238.050784 u − ( 4.002602 u + 234.043 583 u ) ⎤⎦ ( 931.5 M eV u )
= 4.28 M eV
29.53
From R = R 0e− λt , the elapsed time is
t= −
ln ( R R 0 )
λ
= − T1 2
ln ( R R 0 )
ln 2
= − ( 14.0 d )
ln ( 20.0 m C i200 m C i)
ln 2
= 46.5 d
459