Question

A photon collides with a stationary hydrogen atom in the ground state inelastically. The energy of the colliding photon is 10.2 eV. After a time interval of the order of microseconds, another photon collides with the same hydrogen atom inelastically with an energy of 15 eV. What will 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

Answer 1

Answer: (C)

One photon of energy 10.2 eV and an electron of energy 1.4 eV

Answer 2

The first photon will excite the hydrogenatom (in ground state) to first excited state (as $E_2-E_1 = 10.2eV).$

• Hence, during de-excitation, due to the 10.2 eV photon, one photon of energy of 10.2 eV will be detected.
• The second photon of energy 15 eV can ionize the atom.
• 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 1.4 eV energy will be released.

$\therefore$ the correct answer is (c).