A photon of `10.2 eV`. Energy collides with a hydrogen atom in ground state inelastically. After few microseconds one more photon of energy `15 e V` collides with the same hydrogen atom. Then what can be detected by a suitable detector.

Options

A. one photon of `10.2 eV` and an electron of energy `1.4 eV`
B. `2` photon of energy `10.2 eV`
C. `2` photons of energy `3.4 eV`
D. `1` photon of `3.4 eV` and one electron of `.4 eV`

Answer

A

Explanation

First photon will excite the atom to `I` excited state, which when returining to ground state will emit a photon of energy `10.2 eV` second photon will ionize the atom (`13.6 eV` will be used up in this process). The extra energy `(=15013.6=1.4 eV)` will be carried by electron as its kinetic energy. So a photon of energy `13.6 eV` and an electron of energy `1.4 eV` will be emitted.