Question

Fission of 235U occurs when the nucleus absorbs a single neutron and decays to smaller nuclei, liberating additional neutrons in the process. Each of these neutrons stimulates fission of other 235U nuclei in the next cycle of the chain reaction. If the average number of neutrons produced in each fission event is 1.9, how many cycles of the chain reaction are required to consume 2.3 kg of 235U?

Answer 1

mass of U atom = 235*(1.673*10^-27) Kg
= 3.931*10^-25 Kg

Number of fission required for 2.3 Kg U = 2.3 Kg / (3.931*10^-25)
=5.85*10^24

In 1st chain reaction, number of fission = 1
In 2nd chain reaction, number of fission = 1.9
In 3rd chain reaction, number of fission = 1.9^2
In 4th chain reaction, number of fission = 1.9^3
and so on

so,
5.85*10^24 = 1+1.9 + 1.9^2 + 1.9^3 + .....1.9^n

we need to find the volume of 1
RHS is GP
with a = 1
r = 1.9

sum of GP = a*(r^n - 1) / (r-1)
5.85*10^24 = 1*(1.9^n -1)/(1.9-1)
5.265*10^24 = (1.9^n -1)
1.9^n = 5.265*10^24
take log on both sides
log (1.9^n) = log (5.265*10^24)
n*log (1.9)=24.72
n*0.278 = 24.72
n = 88.7
Answer: 89