The frequency of light emitted for the transition $n=4$ to $n=2$ of $H{e}^{+}$ is equal to the transition in $H$-atom corresponding to which of the following:

The correct option is B $n=2$ to $n=1$
We know that,
$\text{Wave Number,}\overline{\nu }=\frac{1}{\lambda }=R_H{Z}^2(\frac{1}{n_1^2}-\frac{1}{n_2^2})$
where,
$R_H=\text{Rydberg constant}=109677{\text{cm}}^{-1}$

Given: $n_1=2$
$n_2=4$
The frequency of light emitted for the transition in $H{e}^{+}=$ The frequency of light emitted in the transition for $H$ atom

∴ We can say that their wavelengths would also be the same
(since $v=\frac{c}{\lambda }$)

For $H{e}^{+}$ atom, $Z=2$
$\overline{\nu }=\frac{1}{\lambda }=R_H{2}^2(\frac{1}{n_1^2}-\frac{1}{n_2^2})$

For $H$ atom, $Z=1$
$\overline{\nu }=\frac{1}{\lambda }=R_H{1}^2(\frac{1}{n_1^2}-\frac{1}{n_2^2})$

As their wavelengths are the same, the wave numbers would also be the same.
$\therefore R_H{1}^2(\frac{1}{n_1^2}-\frac{1}{n_2^2})=R_H{2}^2(\frac{1}{2^2}-\frac{1}{4^2})$
$=R_H(\frac{1}{1^2}-\frac{1}{2^2})$

Hence we can say that the frequency of light emitted for the transition $n=4$ to $n=2$ in helium ion is equal to the transition in $H$ atom corresponding to $n_1=2$ to $n_2=1$.