JEE Main 2020PhysicsNuclear PhysicsEasyMCQ

JEE Main 2020 — Nuclear Physics Question with Solution

JEE Main 2020 (06 Sep Shift 2)

Question

Given the masses of various atomic particles mP=1,0072 u, mn=1,0087 u, me=0.000548 u, mv¯=0, md=2.0141 u, where p=proton, n≡neutron, e≡electron, v¯≡antineutrino and d¯≡deuteron. Which of the following process is allowed by momentum and energy conservation :

Choose an option

Show full solutionCorrect option: C
Correct answer
Cn+p→d+γ

Step-by-step explanation

For n+p→d+γ,

The mass defect, ∆m=mp+mn-md

⇒∆m=1.0072+1.0087-2.0141

⇒∆m=0.0018

⇒∆m>0

∴only n+p→d+γ is possible

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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.