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 transitions to a higher energy level by absorbing monochromatic light of photon of energy $2.7 cV$.

Subsequently, the atoms emit radiations of only six different photon energies. Some of the emitted photons have energy $2.7 cV$, some have energy more or some have energy less than $2.7 cV$.

(i) Find the principal quantum number of the initially excited level B.

(ii) Find the ionization energy for the gas atoms.

(iii) Find the maximum and minimum energy of emitted photons.

Answer 1

(i) Let the higher energy level attained by atoms by absorbing $2.7 cV$ be $C$. It has been given that when atoms at $C$ emit photons by transition at lower levels, six photons are cmitted. We know that number of photons emitted (or number of spectral lines formed)

$$=\frac{n(n-1)}{2}$$

where $n$ is the highest level from which transitions occur at lower levels.

$$\therefore \quad 6=\frac{n(n-1)}{2}, \quad \therefore n=4$$

The higher level $C$ attained $=4$.

Between levels $n=1$ and $n=4$, the level $B$ cannot be $n=3$ because it is mentioned that some of the photons subsequently emitted have energy less than the value that caused excitation from $B$ to $C$ by absorbing $2 \cdot 7 cV$. Therefore, the level of $B$ i.c. $n=2$.

(ii) $\quad E_{C}-E_{B}=\frac{13 \cdot 6}{4} Z^{2}-\frac{13 \cdot 6 Z^{2}}{16}$

or, $\quad 2 \cdot 7=13 \cdot 6 \times \frac{3 Z^{2}}{16} \quad$ or, $\quad Z^{2}=\frac{2 \cdot 7 \times 16}{3 \times 13 \cdot 6}=\frac{18}{17}=1, \quad \therefore \quad Z=1$

Ionization energy $=13 \cdot 6 Z^{2}=13 \cdot 6 eV$

(iii) $E_{\max }=E_{4}-E_{1}=(13.6-0.85) cV =12.75 eV$

$E_{\min }=E_{4}-E_{3}=\frac{13.6}{9}-\frac{13.6}{16}=\frac{13.6 \times 7}{9 \times 16} eV =0.66 eV \text {. }$