Find the quantum number ‘n’ corresponding to the excited state of  $H{e}^{+}$ ion, if one transition to the ground state that ion emits two photons in succession with wavelengths 108.5nm and 30.4nm.

Given  ${\lambda }_2=30.4\times {10}^{-7}cm$
 ${\lambda }_1=108.5\times {10}^{-7}cm$
let excited state of  $H{e}^{+}$ be  $n_2.$  it comes from  $n_2.$ and then  $n_1$ to 1 to emit two successive photon
$\frac{1}{{\lambda }_2}=R_H.z^2(\frac{1}{1^2}-\frac{1}{n_{1}2})$

$\frac{1}{30.4\times {10}^{-7}}=109678\times 4[\frac{1}{1^2}-\frac{1}{n_1^2}] n_1 = 2$

Now,  $\frac{1}{{\lambda }_1}=R_H{z}^2[\frac{1}{2^2}-\frac{1}{n_2^2}]$

$\frac{1}{108.5\times {10}^{-7}}=109678\times 4[\frac{1}{2^2}-\frac{1}{n_2^2}]$

$n_2=5$

Thus excited state for He is 5^th orbit.