JEE Main 2016PhysicsAtomic PhysicsMediumMCQ

JEE Main 2016 — Atomic Physics Question with Solution

JEE Main 2016 (10 Apr Online)

Question

A neutron moving with a speed 'v' makes a head on collision with a stationary hydrogen atom in ground state. The minimum kinetic energy of the neutron for which  perfactly inelastic collision will take place is :

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Show full solutionCorrect option: A
Correct answer
A20.4 eV

Step-by-step explanation

Let velocity before collision v from momentum conservation mv=( m+m ) v 1

v 1 = v 2

Loss in K.E. =12mv2-122mv22

=14 mv2

K.E. lost is used to jump from 1st orbit to 2nd orbit ∆ K.E.=10.2 eV

⇒  14 mv2=10.2

12 mv2=2×10.2=20.4 eV

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About this question

This is a previous-year question from JEE Main 2016, covering the Atomic 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.