JEE Main 2020PhysicsNuclear PhysicsMediumMCQ

JEE Main 2020 — Nuclear Physics Question with Solution

JEE Main 2020 (04 Sep Shift 2)

Question

Find the Binding energy per nucleon for Sn50120. Mass of proton mp=1.00783 U, mass of neutron mn=1.00867 U and mass of tin nucleus msn=119.902199 U. (take 1U = 931 MeV)

Choose an option

Show full solutionCorrect option: D
Correct answer
D8.5MeV

Step-by-step explanation

Binding energy =(ΔM)C2=(AM)931

put the value of ∆M 

BE=8.5MeV 

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 2020, 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.