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 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.