At what minimum kinetic energy must a hydrogen atom move for it's inelastic headon collision with another stationary hydrogen atom so that one of them emits a photon? Both atoms are supposed to be in the ground state prior to the collision.

Explanation

Photon can be emitted in `H-H` collision only if one of the `H` is excited to an `n=2` state which then dexcites to `n=1` state by emitting a photon. Let `v_(1)` and `v_(2)` be the velocites of two Hydrogen atoms after the collision and `M` their masses. Then, energy momentum conservation
`Mv_(1)+Mv_(2)=sqrt(2MT)`
( in the frame of the stationary `H` atom)
`(1)/(2)Mv_(1)^(2)+(1)/(2)Mv_(2)^(2)+(3)/(4) ħR=T`
`(3)/(4) ħR=VR(1-(1)/(4))` is the excitation energy of the `n=2` state from the ground state.
Eliminating ` v_(2)(1)/(2)M{2v_(1)^(2)-2sqrt((2T)/(M))v_(1)+(2T)/(M)}+(3)/(4) ħR=T`
`M{(v_(1)-(1)/(2)sqrt((2T)/(M))^(2))}+(1)/(2)T+(3)/(4) ħR=T`
or `M{(v_(1)-(1)/(2)sqrt((2T)/(M)))^(2)}+(3)/(4) ħR=(1)/(2)T`
For minimum`T`, the square on the left should vanish. Thus `T=(3)/(2) ħR= 20.4 eV`