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. `2` photon of energy `10.2 eV`
B. `2` photon 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`

Answer

C

Explanation

The first photon will excite the hydrogen atom (in ground state) in first excited state (as `E_(2)-E_(1)=10.2 eV`). Hence during de-excitation a photon of `10.2 eV` will be released. The second photon of energy 15 eV can ionise the atom. Hence the balance energy i.e., `(15-13.6)eV=1.4 eV` is retained by the electron. therefore, by the second photon an electron of energy `1.4 eV` willl be released.