Physics · Nuclear Physics

JEE Main 2024 — 27 January, Shift 2 — Question 36

The atomic mass of ${ }_{6} \mathrm{C}^{12}$ is 12.000000 u and that of ${ }_{6} \mathrm{C}^{13}$ is 13.003354 u . The required energy to remove a neutron from ${ }_{6} \mathrm{C}^{13}$ , if mass of neutron is 1.008665 u , will be :

Option A:
62.5 MeV
Option B:
6.25 MeV
Option C:
4.95 MeV
Correct
Option D:
49.5 MeV

Answer: C

Step-by-step solution

${ }_{6} \mathrm{C}^{13}+$ Energy $\rightarrow{ }_{6} \mathrm{C}^{12}+{ }_{0} \mathrm{n}^{1}$ $\Delta \mathrm{m}=(12.000000+1.008665)-13.003354$

$=-0.00531 \mathrm{u}$

$\therefore$ Energy required $=0.00531 \times 931.5 \mathrm{MeV}$

$=4.95 \mathrm{MeV}$

Answer key and solution verified before publishing.

Practise Nuclear Physics

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.

Solve a similar one free

Exam: JEE Main 2024
Paper: 27 January, Shift 2
Subject: Physics
Chapter: Nuclear Physics
Topic: Mass Defect, Binding Energy and Q-Value of Nuclear Reaction