Question

A photon of energy 10.2 $eV$ collides inelastically with $H-$atom in ground state. After a certain time interval of few $\mu s$ another photon of energy 15 $eV$ collides inelastically with the same $H$ atom, the observation made by a suitable detector is

• A. 1 photon with energy 10.2 $eV$ and an electron with 1.4 $eV$.
• B. Two photons with 10.2 $eV$
• C. Two photons with 1.4 $eV$
• D. One photon with 3.4 $eV$ and 1 electron with 1.4 $eV$

Answer 1

The correct option is A 1 photon with energy 10.2 $eV$ and an electron with 1.4 $eV$.

The energy of the H atom is given by:
$E_n=-\frac{13.6}{n^2}$, where $n$ is the principal quantum number.

Now the $10.2eV$ electron will excite the ground state electron to the 2nd orbit.

After few microsecond when the another photon of energy $15eV$collides inelastically with the same H atom the previously excited electron will come down to ground state with the emission of a photon of energy $10.2eV$ and we will see another electron with energy $15-13.6=1.4eV$ due to that $15eV$ photon.