Which of the following transitions of He^ + ion will give rise to...

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,$\quad \lambda_1=\lambda_2\Rightarrow R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)=R \left[\frac{1}{ \left(\frac{n_3}{2}\rright)^2}-\f frac{1}{\left(\frac{n_4}{2}\right)^2}\right]On comparing beth sides, we getn_1=$\frac{n_3}{2} \text{\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.