A photon collides with a stationary hydrogen atom in ground state inelastically . Energy of the colliding photon is 10.2eV. After a time interval of order of micro second another photon collides with same hydrogen atom inelastically with an energy of `15 eV` what will be observed by the detector?

Options

A. one photon of energy `10.2 eV` and an electron of energy `1.4 eV`.
B. Two photon of energy `1.4 eV`
C. Two photon of energy `10.2 eV`
D. one photon of energy `10.2 eV` and another photon of `1.4 eV`.

Answer

A, D

Explanation

Initially, a photon of energy `10.2 eV` colides inelastically with a hydrogen atom in ground state.
For hydrogen atom
`E_(1) = - 13.6 eV: E_(2) = - (13.6)/(4) = - 3.4 eV`
`:. E_(2) - E_(1) =10.2 eV`
The electron atom will be jumps to second orbit after obserbing the photon of energy `10.2 eV`. The electron jumps back to its original state in less than a microsecond and releases a photon of energy `10.2 eV`.Another photon of energy `15 eV` strikes the hydrogen atom inelastic. Thuis energy is `13.6 eV`. The remaining energy of `1.4 eV` is left with the electron as its kinetic energy.