1000 Solved Problems in Modern Physics

7.3 Solutions 421

7.95 Let the proportion of^235 U and^238 Uatt=0bex:(1−x). Present day

radioactive atoms for the two components after timetwill be

N 235 =xN 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 × 108 years

λ 238 = 0. 693 /T 1 / 2 ,T 1 / 2 = 4. 5 × 109 years

t= 3 × 109 years

Substituting these values in (3) and solving forxwe getx= 1 / 22

Therefore, the ratio of^235 U and^238 Uatoms3× 109 years ago was

1/22:21/22 or 1:21

7.96|dN/dt|=Nλ

N= 1. 0 × 10 −^6 × 6. 02 × 1023 / 242 = 2. 487 × 1015

|dN/dt|=Nλ

80 + 3 / 3 , 600 = 2. 487 × 1015 × 0. 693 /T 1 / 2

Solving forT 1 / 2 , we findT 1 / 2 = 2. 15 × 1013 sor6. 8 × 105 years

Forα-decay: 80=Nλα= 2. 487 × 105 λα

λα= 1. 01 × 10 −^6 year−^1

For fission 3/ 3 , 600 = 2. 487 × 1015 λf

λf= 1. 05 × 10 −^11 year−^1

7.97 Nsr=Nsr^0 exp(−λsrt)(1)

Ny=

λsrNsr^0

λy−λsr

[exp(λsrt)−exp(−λyt)] (2)

Dividing the two equations

Ny

Nsr

=

λsr

λy−λsr

[1−exp(λsr−λy)t](3)

λsr= 0. 693 /(28× 365 ×24)= 2. 825 × 10 −^6 h−^1

λy= 0. 693 / 64 = 0 .0108 h−^1

(a) Fort=1 h and using the values for the decay constantsNsr/Ny= 3. 56 ×

105

(b) Fort=10 years,Nsr/Ny= 3 , 823

7.98 IfNis the number of atoms of each component att=0, then at timetthe

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)