Question

A gas of identical H-like atom 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 makes transition to a higher energy level by absorbing monochromatic light of photon energy $2.7eV$. Some have more and some have less than $2.7eV$. Find the minimum energy of the emitted photon in eV.

(Multiply your answer by 10)

Answer 1

The electrons being present in I shell and another shell $n_1$. These are excited to higher level $n_2$ by absorbing $2.7eV$ and on de-excitation emits six $\lambda$ and thus, excited state $n_2$ comes to be $4$.
$[6=\sum {\Delta }_H=\sum (n_2-1)\therefore n_2=4]$
Now, $E_1=\frac{R_H\cdot c\cdot h}{1^2};E_{n_1}=\frac{R_H\cdot c\cdot h}{n_1^2};E_4=-\frac{R_H\cdot c\cdot h}{4^2}$
Since, de-excitation leads to different $\lambda$ having photon energy $<2.7eV$ and thus, absorption of $2.7eV$ energy causing excitation to $4th$ shell and then re-emitting photons of $>2.7eV$ is possible only when $n_1=2$ (the de-excitation from $4th$ shell occurs in I, II and III shell).
$E_4-E_2=2.7eV$
$E_4-E_3=<2.7eV$
$E_4-E_1>2.7eV$
$\therefore E_{n_1}=E_2=\frac{R_H\times c\times h}{2}=\frac{E_1}{2^2}$ Since $n_1=2$
(as obtained by disscussion)
Also, $E_4-E_2=2.7eV$
$\therefore [\frac{-E}{4^2}+\frac{E_1}{2^2}]=2.7eV$
$\therefore E_1=-14.4eV$
$IE=14.4eV$
$E_{max}=E_4-E_1=\frac{E_1}{4^2}+E_1$
$=-\frac{14.4}{16}+14.4=13.5eV$
$E_{min}=E_4-E_3=[\frac{-E_1}{4^2}+\frac{E_1}{3^2}]$
$=0.7eV$
Note: It is ${}_1{\text{H}}^2$ atom.

When it is multiplied by 10, we get $0.7\times 10=7$

Hence, the answer is 7.