GATE 2024 PH – Question 17
For $^{235}_{92}\mathrm U+n\to{}^{93}_{37}\mathrm{Rb}+{}^{141}_{55}\mathrm{Cs}+2n$, neglect initial and final neutron momenta. The daughter kinetic-energy ratio $\mathrm{KE}(\mathrm{Rb})/\mathrm{KE}(\mathrm{Cs})$ is
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Correct answer: (B) $141/93$
Explanation
In the fission reaction the incoming neutron has negligible momentum, so the **total momentum is zero** before the reaction. The two heavy fragments (Rb and Cs) fly apart with equal and opposite momenta (neglecting the neutron momenta):
$$p_{Rb}=p_{Cs}=p .$$
**Kinetic energy** in terms of momentum: $KE=\dfrac{p^2}{2m}$, so for the same $p$, $KE\propto\dfrac1m$.
$$\frac{KE(\text{Rb})}{KE(\text{Cs})}=\frac{m_{Cs}}{m_{Rb}}=\frac{141}{93}\quad(\text{option B}).$$
The lighter fragment carries more of the kinetic energy.