A photon collides with a stationary hydrogen atom in ground state inelastically. Energy of the colliding photon is 10.2 eV. After a time interval of the order of micro second another photon collides with same hydrogen atom inelastically with an energy of 15eV. What wil be observed by the detector?
(a) 2 photons of energy 10.2 eV
(b) 2 photons of energy 1.4 eV
(c ) One photon of energy 10.2 eV and an electron of energy 1.4 eV
(d) One photon of energy 10.2 eV and another photon of energy 1.4 eV

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.