JEE Main 2019PhysicsNuclear PhysicsMediumMCQ

JEE Main 2019 — Nuclear Physics Question with Solution

JEE Main 2019 (10 Jan Shift 2)

Question

Consider the nuclear fission, Ne20→2He4+C12. Given that the binding energy/nucleon of Ne20, He4 and C12 are 8.03 MeV, 7.86 MeV, respectively. Identify the correct statement:

Choose an option

Show full solutionCorrect option: B
Correct answer
BEnergy of 9.72 MeV has to be supplied.

Step-by-step explanation

Q=Ef-Ei
=12×7.86+4×7.07+4×7.07-(20×8.03)
Q=-9.72 MeV

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About this question

This is a previous-year question from JEE Main 2019, 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.