7.3 Solutions
421
7.95 Let the proportion of
235 U and
238 U at t = 0 be x : (1 − x). Present day
radioactive atoms for the two components after time t will be
N 235 = x N 0 exp(−λ 235 t)
( 1 )
N 238 = (1 − x)N 0 exp(−λ 238 t)
( 2 )
Dividing (1) by (2)
N 235 /N 238 = 1/140 =
xe
−(λ 235 −λ 238 )t
1 − x
(3)
λ 235 = 0.693/T 1/2 , T 1/2 = 8.8 × 10
8 years
λ 238 = 0.693/T 1/2 , T 1/2 = 4.5 × 10
9 years
t = 3 × 10
9 years
Substituting these values in (3) and solving for x we get x = 1/22
Therefore, the ratio of
235 U and
238 U atoms 3 × 10
9 years ago was
1/22:21/22 or 1:21
7.96 |dN /dt| = N λ
N = 1.0 × 10
−6
× 6.02 × 10
23
/242 = 2.487 × 10
15
|dN /dt| = N λ
80 + 3/3,600 = 2.487 × 10
15
× 0.693/T 1/2
Solving for T 1/2 , we find T 1/2 = 2.15 × 10
13 s or 6.8 × 10
5 years
For α-decay: 80 = N λ α = 2.487 × 10
5
λ α
λ α = 1.01 × 10
−6 year
−1
For fission 3/3,600 = 2.487 × 10
15
λ f
λ f = 1.05 × 10
−11 year
−1
7.97 N sr = N sr
0 exp(−λ sr t)
( 1 )
N y =
λ sr N
0
sr
λ y − λ sr
[exp(λ sr t) − exp(−λ y t)]
(2)
Dividing the two equations
N y
N sr
=
λ sr
λ y − λ sr
[1 − exp(λ sr − λ y )t]
( 3 )
λ sr = 0.693/(28 × 365 × 24) = 2.825 × 10
−6 h
−1
λ y = 0.693/64 = 0.0108 h
−1
(a) For t = 1 h and using the values for the decay constants N sr /N y = 3.56×
10
5
(b) For t = 10 years, N sr /N y = 3,823
7.98 If N is the number of atoms of each component at t = 0, then at time t the
number of atoms of the two components will be
N 235 = N 0 exp(−λ 235 t)
( 1 )
N 238 = N 0 exp(−λ 238 t)
( 2 )
Dividing the two equations
N 235 /N 238 = (0.7/100) = exp −(λ 235 − λ 238 ) t
(3)
421
7.95 Let the proportion of
235 U and
238 U at t = 0 be x : (1 − x). Present day
radioactive atoms for the two components after time t will be
N 235 = x N 0 exp(−λ 235 t)
( 1 )
N 238 = (1 − x)N 0 exp(−λ 238 t)
( 2 )
Dividing (1) by (2)
N 235 /N 238 = 1/140 =
xe
−(λ 235 −λ 238 )t
1 − x
(3)
λ 235 = 0.693/T 1/2 , T 1/2 = 8.8 × 10
8 years
λ 238 = 0.693/T 1/2 , T 1/2 = 4.5 × 10
9 years
t = 3 × 10
9 years
Substituting these values in (3) and solving for x we get x = 1/22
Therefore, the ratio of
235 U and
238 U atoms 3 × 10
9 years ago was
1/22:21/22 or 1:21
7.96 |dN /dt| = N λ
N = 1.0 × 10
−6
× 6.02 × 10
23
/242 = 2.487 × 10
15
|dN /dt| = N λ
80 + 3/3,600 = 2.487 × 10
15
× 0.693/T 1/2
Solving for T 1/2 , we find T 1/2 = 2.15 × 10
13 s or 6.8 × 10
5 years
For α-decay: 80 = N λ α = 2.487 × 10
5
λ α
λ α = 1.01 × 10
−6 year
−1
For fission 3/3,600 = 2.487 × 10
15
λ f
λ f = 1.05 × 10
−11 year
−1
7.97 N sr = N sr
0 exp(−λ sr t)
( 1 )
N y =
λ sr N
0
sr
λ y − λ sr
[exp(λ sr t) − exp(−λ y t)]
(2)
Dividing the two equations
N y
N sr
=
λ sr
λ y − λ sr
[1 − exp(λ sr − λ y )t]
( 3 )
λ sr = 0.693/(28 × 365 × 24) = 2.825 × 10
−6 h
−1
λ y = 0.693/64 = 0.0108 h
−1
(a) For t = 1 h and using the values for the decay constants N sr /N y = 3.56×
10
5
(b) For t = 10 years, N sr /N y = 3,823
7.98 If N is the number of atoms of each component at t = 0, then at time t the
number of atoms of the two components will be
N 235 = N 0 exp(−λ 235 t)
( 1 )
N 238 = N 0 exp(−λ 238 t)
( 2 )
Dividing the two equations
N 235 /N 238 = (0.7/100) = exp −(λ 235 − λ 238 ) t
(3)