JEE Main 2020PhysicsNuclear PhysicsHardMCQ

JEE Main 2020 — Nuclear Physics Question with Solution

JEE Main 2020 (06 Sep Shift 1)

Question

You are given that  Li37=7.0160u,Mass of Mass of He24=4.0026u and Mass of  He11=1.0079H When 20g of Li37 is converted into 24 He by proton capture, the energy liberated, (in kWh ), is : [Mass of nucleon =1GeV/c2]

Choose an option

Show full solutionCorrect option: D
Correct answer
D1.33×106

Step-by-step explanation

E=ΔmC2

E=(1.0079+7.0160-2(4.0026)×931

=1.33×106

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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.