Atoms

Step: Find the relation between the energy levels corresponding to the hydrogen atom and \(He^+\)– ion.
\(E_{n,H}=-\frac{13.6}{n^2}~\text{eV}\)
\(E_{n,He^+}=-\frac{13.6\times 2^2}{n^2}~\text{eV}\\ E_{2n,He^+}=-\frac{13.6\times 2^2}{(2n)^2}~\text{eV}=-\frac{13.6}{n^2}~\text{eV}=E_{n,H}\)
The energy level corresponding to the  \(n^{\text{th}}\) quantum number of the  \(H\)-atom has the same value as the  \(2n^{\text{th}}\) level of the  \(He^{+}\)-ion. Therefore, the energy of the photon causing the \(n=2\rightarrow4\) transition in the  \(He^+\)-ion is equal to the one causing the  \(n=1\rightarrow2\) transition in the  \(H\) -atom.
 Hence, option (1) is the correct answer.