Given the masses of various atomic particles m _ P = 1, 0072u , m _ n = 1, 0087u , m _ e = 0 .000548 u , m ¯ v = 0 , m _ d = 2 .0141 u , where p = proton, n ≡ neutron, e ≡ electron, ¯ v ≡ antineutrino and ¯ d ≡

Correct answer: C. n + p → d + γ For n + p → d + γ , The mass defect, ∆ m = m _ p + m _ n - m _ d ⇒ ∆ m = 1 . 0072 + 1 . 0087 - 2 . 0141 ⇒ ∆ m = 0 . 0018 ⇒ ∆ m > 0 ∴ onlyn + p → d + γ ispossible

JEE Main 20206 SeptemberEvening Shift

Exam: JEE Main
Year: 2020
Date: 6 September
Shift: Evening Shift
Subject: Physics
Chapter: Nuclear Physics

JEE Main 2020 · 6 September · Evening Shift · Physics · Nuclear Physics

Given the masses of various atomic particles m _ P = 1, 0072u , m _...

Given the masses of various atomic particles

{m}_{P} = 1, 0072u

{m}_{n} = 1, 0087u

{m}_{e} = 0 .000548 u

m\overset{¯}{v} = 0

{m}_{d} = 2 .0141 u

, where

p =

proton,

n ≡

neutron,

e ≡

electron,

\overset{¯}{v} ≡

antineutrino and

\overset{¯}{d} ≡

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

Options

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

Correct answer

C. $n + p \to d + \gamma$

Step-by-step solution

For $n + p \to d + \gamma$ ,

The mass defect, $∆ m = {m}_{p} + {m}_{n} - {m}_{d}$

$⇒ ∆ m = 1 . 0072 + 1 . 0087 - 2 . 0141$

$⇒ ∆ m = 0 . 0018$

$⇒ ∆ m > 0$

$∴ onlyn + p \to d + \gamma ispossible$

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