Physics · Nuclear Physics

JEE Main 2024 — 8 April, Shift 1 — Question 36

Binding energy of a certain nucleus is $18 \times 10^{8} \mathrm{~J}$ . How much is the difference between total mass of all the nucleons and nuclear mass of the given nucleus:

Option A:
$0.2 \mu \mathrm{~g}$
Option B:
$20 \mu \mathrm{~g}$
Correct
Option C:
$2 \mu \mathrm{~g}$
Option D:
$10 \mu \mathrm{~g}$

Answer: B

Step-by-step solution

$\Delta \mathrm{mc}^{2}=18 \times 10^{8}$

$\Delta \mathrm{m} \times 9 \times 10^{16}=18 \times 10^{8}$

$\Delta \mathrm{m}=2 \times 10^{-8} \mathrm{~kg}=20 \mu \mathrm{~g}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 8 April, Shift 1
Subject: Physics
Chapter: Nuclear Physics
Topic: Mass Defect, Binding Energy and Q-Value of Nuclear Reaction