A neutron moving with a speed `v` makes a head-on collision with a hydrogen in ground state kept at rest which inelastic collision will be take place is (assume that mass of photon is nearly equal to the mass of neutron)

Explanation

Suppose the neutron and the hydrogen atom move at speeds `upsilon_(1) and upsilon_(2)` after the collision .The collision will be inelstic if a part of the kinetic energy is used to excited the atom .Soppose an energy `DeltaE` is used in this way .Using conservation of linear momentum and energy
`m upsilon = m upsilon_(1) + m upsilon_(2)`...(i)
`and (1)/(2) m upsilon^(2) = (1)/(2)m upsilon_(1)^(2) + (1)/(2) m upsilon_(2)^(2) + DeltaE`...(ii)
From (i), `upsilon^(2)= upsilon_(1)^(2)+ upsilon_(2)^(2) +2 upsilon_(1)upsilon_(2)`
From (ii),`upsilon^(2)= upsilon_(1)^(2)+ upsilon_(2)^(2) +(2Delta E)/(m)`
Thus `2 upsilon_(1) upsilon_(2) = (2Delta E)/(m)`
Hence,`(upsilon_(1)- upsilon_(2))^(2) =(upsilon_(1)+ upsilon_(2))^(2)- 4 (2Delta E)/(m)upsilon_(1) upsilon_(2)= upsilon^(2)- (4Delta E)/(m)`
As `upsilon_(1)- upsilon_(2)` must be real,
`upsilon^(2)-(4Delta E)/(m) ge 0`
`or, `(1)/(2) m upsilon^(2) gt 2 delta E`
The minimum energy that the can be obserbed by the hydrogen atom in ground state to in an excited state is `10.2eV` Thus the minimum kinetic energy of the neutron needed for an inslastic collision is
`(1)/(2) m upsilon_(min)^(2) = 2 xx 10.2eV = 20.4 eV`