Question

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

• A. $2$ photons of energy $10.2eV$
• B. $2$ photons of energy $1.4eV$
• C. One photon of energy $10.2eV$ and an electron of energy $1.4eV$
• D. One photon of energy $10.2eV$ and another photon of energy $1.4eV$

Answer 1

Answer: (C)

The first photon will excite the hydrogen atom (in ground state) to first excited state (as $E_2-E_1=\text{10.2}eV$ ). Hence, during de-excitation a photon of $\text{10.2}eV$ will be released.
The second photon of energy $15eV$ can ionise the atom. Hence the balance energy i. e., $\left(15-\text{13.6}\right)eV=\text{1.4}eV$ is retained by the electron. Therefore, by the second photon an electron of energy $\text{1.4}eV$ will be released.