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 photons of energy 1.4 eV
C. Two photons of energy 10.2 eV
D. One photon of energy 10.2 eV and another photon of 1.4 eV

Answer

B

Explanation

For hydrogen atom,
`E_(1) = -13.6 eV, E_(2) - E_(1) = -3.4 - (-13.6)`
or `(E_(2)-E_(1))=10.2 eV`
Due to inelastic collision of photon with stationary hydrogen atom, the photon will be absorbed with its energy 10.2 eV. The electron will jump from ground state (first orbit) to second orbit.
The electron falls back to its original state in less than a microsecond and releases a photon of energy 10.2 eV. Another photon collides with same hydrogenatom inelastically with an energy of 15 eV. This photon on absorption will knock out an electron and will ionise the atom as the ionisation energy is only 13.6 eV.
`therefore` Balance energy =15-13.6
or Balance energy = 1.4 eV
The knocked out electron will retain energy =1.4 eV with it .
The detector will observe one photon of energy 10.2eV and an electron of energy 1.4 eV.
Option (b) represents the anwer .