JEE Main 2022PhysicsNuclear PhysicsEasyMCQ

JEE Main 2022Nuclear Physics Question with Solution

JEE Main 2022 (24 Jun Shift 1)

Question

Nucleus A is having mass number 220 and its binding energy per nucleon is 5.6 MeV. It splits in two fragments B and C of mass numbers 105 and 115. The binding energy of nucleons in B and C is 6.4 MeV per nucleon. The energy Q released per fission will be:

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Show full solutionCorrect option: D
Correct answer
D176MeV

Step-by-step explanation

The total binding energy of nucleus A is,

EA=220×5.6=1232 MeV

The total binding energy of B and C combined is,

EB+EC=105×6.4+115×6.4=1408 MeV

Therefore, the energy released per fission is,

Q=EB+EC-EA=1408-1232=176 MeV

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About this question

This is a previous-year question from JEE Main 2022, 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.