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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 inelastic collision will take place is:

10.2eV

20.4eV

12.1eV

16.8eV

Solution:

n→v(H)Before(n)(H)→v/2Afterloss inK.E.=12mv2−12(2m)(v/2)2=14mv2K.E. lost is used to jump from 1st orbit to 2nd orbitΔK.E.=10.2ev→14mv2=10.212mv2=2×10.2=20.4eV