Which of the following transitions of $\mathrm{He}^{+}$ion will give rise to spectral line which has same wavelength as the spectral line in hydrogen atom?

Let the transition of electron in hydrogen atom $(Z=1)$ gives wavelength $\lambda_1$ for transition from $n_2$ to $n_1$, then $\frac{1}{\lambda}=R Z^2\left(\frac{1}{m^2}-\frac{1}{n^2}\right)$ Here, $\lambda=\lambda_1, Z=1, m=n_1, n=n_2$ $\frac{1}{\lambda_1}=R(1)^2\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)=R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$ Now, if electron makes transition in $\mathrm{He}^{+}$ion from $n_4$ to $n_3$ and gives wavelength $\lambda_2$, then $\begin{aligned} \frac{1}{\lambda_2} & =R(2)^2\left(\frac{1}{n_3^2}-\frac{1}{n_4^4}\right) \\ & =R\left[\frac{1}{\left(\frac{n_3}{2}\right)^2}-\frac{1}{\left(\frac{n_4}{2}\right)^2}\right] \end{aligned}$ Given that, $\quad \lambda_1=\lambda_2$ $\Rightarrow R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)=R\left[\frac{1}{\left(\frac{n_3}{2}\right)^2}-\frac{1}{\left(\frac{n_4}{2}\right)^2}\right]$ On comparing beth sides, we get $n_1=\frac{n_3}{2} \text { and } n_2=\frac{n_4}{2}$ The value of $n_1, n_2, n_3$ and $n_4$ must be an integer, as they denote principal quantum number. If $n_3=2$ and $n_4=4$, then the given condition is satisfied.