JEE Advanced 2021 Paper 2 Q05 Physics Modern Physics Nuclear Physics Medium

JEE Advanced 2021 Paper 2 · Q05 · Nuclear Physics

A heavy nucleus $N$, at rest, undergoes fission $N \to P + Q$, where $P$ and $Q$ are two lighter nuclei. Let $\delta = M_N - M_P - M_Q$, where $M_P$, $M_Q$ and $M_N$ are the masses of $P$, $Q$ and $N$ respectively. The speeds of $P$ and $Q$ are $v_P$ and $v_Q$, respectively. $E_P$ and $E_Q$ are the kinetic energies of $P$ and $Q$, respectively. If $c$ is the speed of light, which of the following statement(s) is(are) correct?

  1. A. $E_P + E_Q = c^2 \delta$
  2. B. $E_P = \left(\dfrac{M_P}{M_P + M_Q}\right) c^2 \delta$
  3. C. $\dfrac{v_P}{v_Q} = \dfrac{M_Q}{M_P}$
  4. D. The magnitude of momentum for $P$ as well as $Q$ is $c\sqrt{2\mu\delta}$, where $\mu = \dfrac{M_P M_Q}{M_P + M_Q}$
JEE Advanced multi-correct — pick every correct option, then check.
Reveal answer + step-by-step solution

Correct answer:A, C, D

Solution

Energy released $Q = c^2\delta = E_P + E_Q$ — (A) correct. Momentum conservation (initial momentum zero): $M_P v_P = M_Q v_Q$, so $v_P/v_Q = M_Q/M_P$ — (C) correct. Combined with $E_i = p^2/(2 M_i)$ and total energy $E_P + E_Q = c^2\delta$, momentum magnitude $p^2 = 2\mu c^2 \delta$ where $\mu = M_P M_Q/(M_P + M_Q)$, hence $p = c\sqrt{2\mu\delta}$ — (D) correct. For (B): $E_P = p^2/(2M_P) = (M_Q/(M_P+M_Q)) c^2\delta$, not $M_P/(M_P+M_Q) c^2\delta$ — (B) wrong.

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