(a) Example 38.6 explains that the number of fission events in a chain reaction increases by a factor \(k\) with each generation. Show that the total number of fission events in \(n\) generations is \(N=\left(k^{n+1}-1\right) /(k-1) \cdot(b)\) In a typical nuclear explosive, \(k\) is about 1.5 and the generation time is about \(10 \mathrm{ns}\). Use the result from (a) to calculate the time for all the nuclei in a 10 -kg mass \(^{235} \mathrm{U}\) to fission. (Hint: Sum a series in part (a), and neglect 1 compared with \(N\) in part (b).)

To start, we know that the number of fission events after the initial one is given by \(k^n\), because each generation of fissions produces \(k\) times as many events as the previous generation. This is a geometric series, and the sum to \(n\) terms of a geometric series is given by the formula \(N=\frac{k^{n+1}-1}{k-1}\). Thus, the total number of fission events, including the initial one, after \(n\) generations will follow this formula as given in the problem statement.

For a nuclear explosive, we have \(k \approx 1.5 \) and the generation time is \(10 ns\); also, we have a 10 kg mass of \({}^{235} U\). Completed fission of 1 mole of \({}^{235} U\) will involve \({N_a} = 6.023 × 10^{23}\) atoms – Avogadro's number. The number of moles in 10 kg of \({}^{235} U\) is given by \(\frac{10 × 10^3}{235}\), thus the total number of uranium atoms, \(N\), in 10 kg of \({}^{235} U\) is \(\frac{10 × 10^3 × N_a}{235} \) . This number is much larger than 1, so we can neglect 1 compared to \(N\) in total number of fission events.

We want to find the time for all these uranium atoms to undergo fission. Since these fissions take place in 'generations', we want to find the number of generations, \(n\), such that the total number of fissions, as given by our formula, is equal to the total number of uranium atoms in 10 kg of \({}^{235} U\). That is, we solve the equation \(\frac{1.5^{n+1} - 1}{1.5 - 1} = \frac{10 × 10^3 × N_a}{235} \) for \(n\). However, since the number of uranium atoms is so large, we can further simplify this by keeping only the highest power of \(1.5\) in the numerator, neglecting 1. So, our equation becomes \(1.5^{n+1} = \frac{10 × 10^3 × N_a}{235} \). Solving this equation gives us the value of \(n\). Then, since each generation lasts for \(10 ns\), the total time for all fissions is given by \(10n ns\).