Physics 43 HW 19 Ch 43
HW 19: Ch 43: 23, 24, 26, 27, 29, 30, 32, 59, 60
43.23. Model: The number of radioactive atoms decreases exponentially with time.
Solve:
The number of remaining 109Cd nuclei at time t is
⎛1⎞
N = N0 ⎜ ⎟
⎝ 2⎠
t/t1/2
(a) After 50 days, N = (1.00 × 1012)(1/2)50 days/462 days = 9.28 × 1011.
(b) After 500 days, N = (1.00 × 1012)(1/2)500 days/462 days = 4.72 × 1011.
(c) After 5000 days, N = (1.00 × 1012)(1/2)5000 days/462 days = 5.52 × 108.
Assess: Do not think that if half the nuclei decay during one half-life, then all will decay in two half-lives.
43.24. Model: The number of radioactive atoms decreases exponentially with time.
Solve: (a) 3H is called tritium and has Z = 1 and N = 2. 3H undergoes beta-minus decay and gives rise to the
daughter nucleus 3He, that is,
3
H →3 He + β −
(b) From Equation 43.18, the decay rate is
r=
1
τ
=
0.693
0.693
1 year
=
×
= 1.78 × 10−9 s −1
t1 2
12.33 years 3.15 × 107 s
The time constant is τ = 5.62 × 108 s. The half-life in the above calculation was obtained from Appendix C.
43.26. Solve: The activity of a radioactive sample is R = rN. From Appendix C, the half-life of 60Co is t1/2 =
5.27 yr = 1.663 × 108 s. Thus the decay rate is
r=
1
τ
=
ln 2
0.693
=
= 4.17 × 10 −9 s −1
t1/ 2 1.663 × 108 s
We know that the sample’s activity is R = 1.25 × 109 Bq = 1.25 × 109 decays/s, so the number of atoms in the
sample is
N=
R 1.25 × 109 decays/s
=
= 3.00 × 1017 atoms
r
4.17 × 109 s −1
The mass of the sample is
M = mN = (60 × 1.661 × 10–27 kg)(3.00 × 1017) = 2.99 × 10–8 kg = 299 μg
43.27. Model: The activity R of a radioactive sample is the number of decays per second.
Solve:
The rate of decay is
R=
⇒ t1 2
dN
N ( 0.693) N
= rN = =
dt
t1 2
τ
15
0.693) N ( 0.693) ( 5.0 × 10 atoms )
(
=
=
= 6.93 × 106 s ×
R
5.0 × 108 Bq
1 day
= 80.2 days ≈ 80 days
86,400 s
43.29. Solve: (a) The decay is X → 224 Ra + 4 He, so X = 228 Th.
(b) The decay is X →
207
Pb + e − + v , so X =
207
Tl.
(c) The decay is Be + e → X + v, so X = Li.
−
7
(d) The decay is X →
60
7
Ni + γ , so X =
Ni.
Pu → 235 U + 4 He.
Using Appendix C to find the masses, the energy released is
43.30. Model: The decay is
Solve:
60
239
E = Δmc 2 = ⎡⎣ m ( 239 Pu ) − m ( 235 U ) − m ( 4 He ) ⎤⎦ c 2 = [ 239.052157 u − 235.043924 u − 4.002602 u ] c 2
= ( 0.00563 u ) c 2 × 931.49 MeV/c 2 u = 5.24 MeV
Assess:
Essentially all of this energy goes into the alpha particle’s kinetic energy.
43.32. Model: The decay is 3 H → 3 He + e − + v .
Solve: Beta-minus decay leaves the daughter atom as a positive ion. However, the mass of the ion plus the
mass of the escaping electron is the mass of a neutral atom, which is what is tabulated in Appendix C. Thus the
mass loss is the mass difference between the two neutral atoms. In Appendix C we find that
m ( 3 H ) = 3.016049 u, and m ( 3 He ) = 3.016029 u. The energy released in the beta-minus decay corresponds to a
mass change of 0.000020 u. The energy released is
E = ∆mc2 = (0.000020 u)(931.49 MeV/u) = 0.0186 MeV
Assess:
The energy E is shared between the electron and the antineutrino.
43.59. Model: The number of 40K atoms decays exponentially with time.
Solve: Suppose the number of 40K atoms is N0 at the time the lava solidifies. There are no 40Ar atoms at that
time. As time passes, some 40K decay into 40Ar, but the total number of atoms locked inside the lava is
unchanged. That is, NAr + NK = N0, where NK is the remaining number of 40K atoms. Dividing by NK, we have
N Ar
N
N
1
+ 1 = 0.12 + 1 = 0 ⇒ K =
= 0.893
NK
NK
N 0 1.12
That is, 89.3% of the 40K remains at a time when the Ar/K ratio is 0.12. The 40K decays as
t /t
1/ 2
t
⎛1⎞
N K = N0 ⎜ ⎟
⇒
ln(1/2 ) = ln( N K / N 0 )
t1/ 2
⎝2⎠
ln(0.893)
⇒ t = (1.28 billion years)
= 0.21 billion years = 210 million years
ln(0.50)
43.60. Model: The number of 235U atoms decays exponentially with time.
Solve:
The 235U decays as
⎛1⎞
N U = ( N U )0 ⎜ ⎟
⎝ 2⎠
t / t1/ 2
The ratio of 235U atoms when the earth formed to the number now present is
( N U )0
1
=
= (2)t / t1/ 2 = (2) (4500 my)/(700 my) = 86.1 ≈ 86
NU
(1/2 )t / t1/ 2
That is, the abundance of 235U was 86 times larger then than it is now.