Question

The photon radiated from hydrogen corresponding to $2^{nd}$ line of Lyman series is absorbed by a hydrogen like atom $X$ in $2^{nd}$ excited state. As a result, the hydrogen like atom ${}^{\prime }{X}^{\prime }$ makes a transition to $n^{th}$ orbit. Then

• A. $X=H{e}^{+}$,$n=4$
• B. $X=L{i}^{++}$,$n=6$
• C. $X=H{e}^{+}$,$n=6$
• D. $X=L{i}^{++}$,$n=9$

Answer 1

The correct option is D $X=L{i}^{++}$,$n=9$

Energy of $n^{th}$ state in $H\text{-atom}$ is same a energy of $3n^{th}$ state in $L{i}^{++}$.
$\because 2^{nd}$ line of Lyman series in $H\text{- atom}$ is obtained by transition from $n=3$ to $n=1$
$\therefore$ Energy is obtained by
$E=-13.6Z^2[\frac{1}{n_1^2}-\frac{1}{n_2^2}]$
$E_1=-13.6\times \frac{8}{9}\text{eV}.$
So,
This same amount of energy is obtained in $L{i}^{++}$ by transition from $n=9$ to $n=3$
By the formula
$E_2=-13.6\times (3{)}^2[\frac{1}{9}-\frac{1}{81}]$
$E_2=-13.6\times \frac{8}{9}eV.$