Physics · Nuclear Physics

JEE Advanced 2021 — Paper 2 — Question 5

A heavy nucleus $N$ , at rest, undergoes fission $N \rightarrow P+Q$ , where $P$ and Q are two lighter nuclei.

Let $\delta=\mathrm{M}_{\mathrm{N}}$ $-M_{P}-M_{Q}$ , where $M_{P}, M_{Q}$ and $M_{N}$ are the masses of $P, Q$ and $N$ , respectively.

$E_{P}$ and $E_{Q}$ are the kinetic energies of $P$ and $Q$ , respectively. The speeds of $P$ and $Q$ are $v_{P}$ and $v_{Q}$ , respectively.

If $c$ is the speed of light, which of the following statement(s) is(are) correct ?

Option A:
$E_{P}+E_{Q}=c^{2} \delta$
Correct
Option B:
$\mathrm{E}_{\mathrm{P}}=\left(\frac{\mathrm{M}_{\mathrm{P}}}{\mathrm{M}_{\mathrm{P}}+\mathrm{M}_{\mathrm{Q}}}\right) \mathrm{c}^{2} \delta$
Option C:
$\frac{\mathrm{v}_{\mathrm{P}}}{\mathrm{v}_{\mathrm{Q}}}=\frac{\mathrm{M}_{\mathrm{Q}}}{\mathrm{M}_{\mathrm{P}}}$
Correct
Option D:
The magnitude of momentum for P as well as Q is $\mathrm{c} \sqrt{2 \mu \delta}$ , where $\mu=\frac{\mathrm{M}_{\mathrm{P}} \mathrm{M}_{\mathrm{Q}}}{\left(\mathrm{M}_{\mathrm{P}}+\mathrm{M}_{\mathrm{Q}}\right)}$
Correct

Answer: A, C, D

Step-by-step solution

Energy released during process $\begin{gathered} \mathrm{Q}=\delta \mathrm{c}^{2} \end{gathered}$ $\because$

momentum is conserved in process. $\Rightarrow 0=\mathrm{m}_{\mathrm{p}} \mathrm{v}_{\mathrm{p}}-\mathrm{m}_{\mathrm{Q} .} \mathrm{v}_{\mathrm{Q}}$ $\begin{gathered} \Rightarrow \quad \frac{\mathrm{v}_{\mathrm{P}}}{\mathrm{v}_{\mathrm{Q}}}=\frac{\mathrm{m}_{\mathrm{Q}}}{\mathrm{m}_{\mathrm{P}}} \end{gathered}$ $E_{P}=\frac{1}{2} m_{P} v_{P}^{2}$

$\mathrm{E}_{\mathrm{Q}}=\frac{1}{2} \mathrm{~m}_{\mathrm{Q}} \mathrm{v}_{\mathrm{Q}}^{2}$

$\begin{gathered} \Rightarrow \frac{\mathrm{E}_{\mathrm{P}}}{\mathrm{E}_{\mathrm{Q}}}=\frac{\mathrm{m}_{\mathrm{Q}}}{\mathrm{m}_{\mathrm{P}}} \end{gathered}$ Solving (1) and (3), $\mathrm{E}_{\mathrm{P}}=\frac{\mathrm{m}_{\mathrm{Q}}}{\mathrm{m}_{\mathrm{P}}+\mathrm{m}_{\mathrm{Q}}} \delta \mathrm{c}^{2}$ $\mathrm{E}_{\mathrm{Q}}=\frac{\mathrm{m}_{\mathrm{P}}}{\mathrm{m}_{\mathrm{p}}+\mathrm{m}_{\mathrm{Q}}} \delta \mathrm{c}^{2}$ Momentum $\mathrm{P}=\sqrt{2 \mathrm{~m}_{\mathrm{P}} \mathrm{E}_{\mathrm{P}}}=\sqrt{2 \mathrm{~m}_{\mathrm{Q}} \mathrm{E}_{\mathrm{Q}}}$ $=\sqrt{2 \mathrm{~m}_{\mathrm{P}} \frac{\mathrm{m}_{\mathrm{Q}}}{\mathrm{m}_{\mathrm{P}}+\mathrm{m}_{\mathrm{Q}}} . \delta \mathrm{c}^{2}}=\mathrm{c} \sqrt{\frac{2 \mathrm{~m}_{\mathrm{P}} \mathrm{m}_{\mathrm{Q}}}{\mathrm{m}_{\mathrm{P}}+\mathrm{m}_{\mathrm{Q}}} . \delta}$

Answer key and solution verified before publishing.

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Exam: JEE Advanced 2021
Paper: Paper 2
Subject: Physics
Chapter: Nuclear Physics
Topic: Nuclear Fission and Fusion