Question

A gas of identical hydrogen like atoms has some atoms in the lowest (ground) energy level $A$ and some atoms in a particular upper (excited) energy level $B$ and there are no atoms in any other energy level. The atoms of the gas make transition to a higher energy level by absorbing monochromatic light photons having energy of $2.7eV$. Subsequently, the atoms emit radiation of only six different photon energies. Some of the emitted photons have an energy of $2.7eV$,some have more energy and some have less energy than $2.7eV$.
Find the principal quantum number of the initially exited level $B$.
Find the ionization energy for the atoms.
Find maximum and the minimum energies of the emitted photons.

Answer 1

Level $C$ must corresponds to $n=4$ since the total number of possible transitions from level $4$ to lower levels is $1+2+3=6$
The level $B$ corresponds either to $n=2$ or, $n=3$.
The energies of all the photons involved in the transitions are (in $eV$):
$E_{4,2}=13.6Z^2(\frac{1}{2^2}-\frac{1}{4^2})=2.55Z^2$
$E_{4,3}=0.66Z^2$
$E_{4,1}=12.75Z^2$
$E_{3,2}=1.89Z^2$
$E_{3,1}=12.1Z^2$
$E_{2,1}=10.2Z^2$
The only possible choices are:
$Z=1$ and $E_{4,2}=2.55eV\approx 2.7eV$
$Z=2$ and $E_{4,3}=2.64eV\approx 2.7eV$
The choice $Z=2$ however is inconsistent with the fact that some photons have energy less then $2.7eV$
Thus, $Z=1$ and the quantum number of level $B$ is $n=2$
The ionisation energy is, $13.6eV\times 1^2=13.6eV$
The maximum energy of the photons is $12.75eV$ ($E_{4,1}$) while the minimum is $0.66eV$ ($E_{4,3}$).