JEE Main 2024PhysicsNuclear PhysicsEasyMCQ

JEE Main 2024 — Nuclear Physics Question with Solution

JEE Main 2024 (27 Jan Shift 2)

Question

The atomic mass of C126 is 12.000000 u and that of C136 is 13.003354 u. The required energy to remove a neutron from C136, if mass of neutron is 1.008665 u, will be:

Choose an option

Show full solutionCorrect option: C
Correct answer
C4.95MeV

Step-by-step explanation

Given the nuclear reaction is

C136+Energy→C126+n10

The mass defect for the above reaction can be calculated as follows:

Δm=(12.000000+1.008665)-13.003354 u=-0.00531 u

Hence, the energy required is given by

E=0.00531×931.5 MeV=4.95 MeV

Practice this on the real CBT interface

Solve this JEE Main question (and the rest of the Nuclear Physics chapter) on PrepSharp's TCS iON-style CBT player — with timer, bookmarks and session analytics.

Solve interactively →

About this question

This is a previous-year question from JEE Main 2024, covering the Nuclear Physics chapter of Physics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.