JEE Main 2023PhysicsNuclear PhysicsHardNumerical

JEE Main 2023Nuclear Physics Question with Solution

JEE Main 2023 (24 Jan Shift 2)

Question

The energy released per fission of nucleus of X240 is 200 MeV. The energy released if all the atoms in 120 g of pure X240 undergo fission is _____×1025 MeV.  

(Given NA=6×1023)   

Enter your answer

Show full solutionCorrect answer: 6
Correct answer
6

Step-by-step explanation

Energy released per fission is 200 MeV.

The number of atoms n in 120 g:

n=120240×6×1023 atoms.

Total energy released, E=n×200×106

=6×1025 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 2023, 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.