Suppose the neutron and hydrogen atom move at speed `v_(1)` and `v_(2)` after the collision. The collision on will be inelastic if a part of the kinetic energy is used to excite the atom. Suppose an energy `DeltaE` is used in this way. Using conservation of linear momentum and energy.
`mv=mv_(1)+mv_(2)`.....(i)
and `(1)/(2)mv^(2)=(1)/(2)mv_(1)^(2)+(1)/(2)mv_(2)^(2)+DeltaE`......(ii)
From (i), `v^(2)=v_(1)^(2)+v_(2)^(2)+2v_(1)v_(2)`,
From (ii) `v^(2)=v_(1)^(2)+v_(2)^(2)+(2DeltaE)/(m)`
thus,`2v_(1)v_(2)=(2DeltaE)/(m)`
Here, `(v_(1)-v_(2))^(2)-4v_(1)v_(2)=v^(2)-(4DeltaE)/(m)`
As `v_(1)-v_(2)` must be real, `v^(2)-(4DeltaE)/(m) gt 0`
or `(1)/(2)mv^(2) gt 2DeltaE`
The minimum energy that can be absorbed by the hydrogen atom in ground state to go in an excited state is `10.2 eV`. Thus, the minimum kinetic energy of the nerutron needed for an inelastic collision is
`(1)/(2)mv_(min)^(2)=2xx10.2eV=20.4 eV`