A neutron moving with a speed v makes a head-on collision with a hydrogen atom in ground state kept at rest. The minimum kinetic energy of the neutron for which inelastic collision will take place is (assume that mass of proton is nearly equal to the mass of neutron)

detailed solution

Correct option is B

Let y = speed of neutron before collision,v1 = speed of neutron after collision,v2 = speed of proton or hydrogen atom after collision,and ∆E = energy of excitationFrom conservation of linear momentum,mv=mv1+mv2.....(i)From conservation of energy,12mv2=12mv12+12mv22+ΔE.......(ii)From Eq. (i),v2=v12+v22+2v1v2From Eq. (ii),v2=v12+v22+2ΔEm ∴ 2v1v2=2ΔEm ∴ v1−v22=v1+v22−4v1v2 ⇒ v1−v22=v2−4ΔEm As v1−v2 must be real, therefore v2−4ΔEm≥0 or  12mv2≥2ΔEThe minimum energy that can be absorbed by hydrogen atom in ground state to go into excited state is 10.2 eV. Therefore,12mvmin2=2×10.2eV=20.4eV