Given the masses of various atomic particles $m_{p}=1.0072 u, m_{n}=1.0087 u, m_{e}= 0.000548 \mathrm{u}, \mathrm{m}_{\mathrm{v}}^{-}=0, \mathrm{~m}_{\mathrm{d}}=2.0141 \mathrm{u}$, where $\mathrm{p} \equiv$ proton, $\mathrm{n} \equiv$ neutron, $\mathrm{e} \equiv$ electron, $\overline{\mathrm{v}} \equiv$ antineutrino and $\mathrm{d} \equiv$ deuteron. Which of the following process is allowed by momentum and energy conservation?

(C) $\mathrm{n}+\mathrm{p} \rightarrow \mathrm{d}+\gamma$

(A) $n+n \rightarrow$ deuterium atom (electron bound to the nucleus)

(B) $p \rightarrow n+e^{+}+\bar{v}$

(D) $\mathrm{e}^{+}+\mathrm{e}^{-} \rightarrow \gamma$

JEE Main · 2020 · 06 Sep 2020 · Shift 2 · Modern Physics

Given the masses of various atomic particles

m_{p} = 1.0072 u, m_{n} = 1.0087 u, m_{e} = 0.000548 u, m_{v}^{-} = 0, m_{d} = 2.0141 u

, where

p \equiv

proton,

n \equiv

neutron,

e \equiv

electron,

\overline{v} \equiv

antineutrino and

d \equiv

deuteron. Which of the following process is allowed by momentum and energy conservation?

(A) $n + n \to$ deuterium atom (electron bound to the nucleus)
(B) $p \to n + e^{+} + \overset{v}{ˉ}$
(C) $n + p \to d + γ$ ✓ Correct
(D) $e^{+} + e^{-} \to γ$

Answer: (C)

Solution

A process is allowed only if the total rest mass of the reactants is at least that of the products, so that the surplus can appear as released energy. Take

n + p \to d + γ

: the reactant mass is

m_{n} + m_{p} = 1.0087 + 1.0072 = 2.0159 u

, while the product deuteron has mass

2.0141 u

. The reactants are heavier by

Δ m = 0.0018 u

, and that mass defect is carried off as the energy of the emitted

γ

photon, which also lets momentum balance. So it is the capture

n + p \to d + γ

that conservation of energy and momentum permits.

JEE Main · 2020 · 06 Sep 2020 · Shift 2 · Single correct option

Practise more: all JEE Main PYQs on Modern Physics →

JEE Main · 2020 · 06 Sep 2020 · Shift 2 · Modern Physics

Solution

More JEE Main questions on Modern Physics