Discussion
Consider the induced nuclear fission reaction
\({}_{92}U^{235} + n \rightarrow {}_{37}Rb^{93} + {}_{55}Cs^{141} + 2n\)
where neutron momenta in both initial and final states are negligible. The ratio of the kinetic energies (KE) of the daughter nuclei, \(\dfrac{KE({}_{37}Rb^{93})}{KE({}_{55}Cs^{141})}\) is ___________
Answer: B
The ratio of kinetic energies is 141/93.
The initial momentum is negligible, and so is the neutrons' momentum afterwards, so the two fragments fly apart with equal and opposite momenta of magnitude p. Non-relativistically KE = p2/2M, so with the same p the kinetic energy is inversely proportional to the mass:
\[ \begin{aligned} \frac{KE_{Rb}}{KE_{Cs}} &= \frac{M_{Cs}}{M_{Rb}} \\ &\approx \frac{141}{93} \end{aligned} \]
The lighter rubidium nucleus carries more energy. 93/141 inverts the ratio, 1 would need equal masses, and 0 would mean one fragment stays at rest, which momentum conservation forbids.
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