Q. 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 15n $eV$. What will be observed by the detector?
A) 2 photon of energy 10.2 $eV$.
B) 2 photon of energy of 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$\ eV$
Explanation
(c) 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 ionize the atom.
Hence the balance energy $ie,$
$(15 - 13.6)$ $eV$ = 1.4 $eV$ is retained by the electron.
Therefore, by the second photon an electron of energy 1.4 $eV$ will be released.
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.