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.

Answer

`T_(min)=(T)/(2)=10.2=20.4 eV`

Explanation

The loss in kinetic energy should be `10.2 eV` so that the hydrogen atom may be excited to first excited state. For minimum kinetic energy case, the two atoms will move together.
`mv_(0)=(m+m)v rArr v=v_(0)//2`
loss in kinetic energy `=(1)/(2) mv_(0)^(2)-(1)/(2)(2m)((v_(0))/(2))^(2)`
`10.2V=(1)/(4).mv_(0)=(1)/(2).T :. T=2xx10.2eV=20.4eV`