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

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`