JEE Advanced 2021 · Paper 2 · Physics Q5

← JEE Advanced 2021 · Paper 2

PhysicsMultiple correct+4 / -2Question 5 of 57

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. $E_P$ 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?

E_P + E_Q = c^2\delta

E_P = \left(\dfrac{M_P}{M_P + M_Q}\right)c^2\delta

\dfrac{v_P}{v_Q} = \dfrac{M_Q}{M_P}

The magnitude of momentum for

P

as well as

Q

c\sqrt{2\mu\delta}

, where

\mu = \dfrac{M_PM_Q}{(M_P + M_Q)}

Answer and worked solution

Official answer: A, C, D

Energy. The mass defect appears as kinetic energy: $E_P+E_Q=\delta c^2$ . (A) is true.

Momentum. The nucleus is at rest, so P and Q have equal and opposite momenta $p$ . Then
$\frac{v_P}{v_Q}=\frac{M_Q}{M_P},$
so (C) is true.

Sharing the energy. Since $E=\frac{p^2}{2M}$ , $E_P:E_Q=M_Q:M_P$ , which gives
$E_P=\frac{M_Q}{M_P+M_Q}\delta c^2.$
(B) has the masses the wrong way round. False.

Size of the momentum. $p^2=2M_PE_P=\dfrac{2M_PM_Q}{M_P+M_Q}\delta c^2=2\mu\delta c^2$ , so $p=c\sqrt{2\mu\delta}$ . (D) is true.

(A), (C), (D)

Topics:Nuclear fission and fusion Mass defect and binding energy Conservation of linear momentum

← Previous question Next question →

JEE Advanced 2021 · Paper 2, Physics question 5