Question

Find the quantum number ${}^{\prime }{n}^{\prime }$ corresponding to the state of $H{e}^{+}$ ion, if on transition to the ground state that ion emits two photons in succession with wavelength $108.5$ and $30.4nm$ (take $R_H=1.09\times {10}^{5}c{m}^{-1}$)

Answer 1

Given, $D_2=30.4\times {10}^{-7}cm$
$D_1=108.5\times {10}^{-7}cm$
Let, excited state of $H{e}^{+}$ be $n_2$. It comes from $n_2$ to $n_1$ and then $n_1$ to $1$ to emit two succesive photon
$\frac{1}{D_2}=R_H.Z^2[\frac{1}{1^2}-\frac{1}{n_1^2}]\frac{1}{30.4\times {10}^{-7}}=1.09\times {10}^5\times 4[\frac{1}{1^2}-\frac{1}{n_1^2}]\therefore n_1=2\text{Now for}{D}_1:n_1=2and{n}_2:?\frac{1}{D_1}=R_H{Z}^2[\frac{1}{2^2}-\frac{1}{n_2^2}]\frac{1}{108.5\times {10}^{-7}}=1.09\times {10}^5\times 4\times [\frac{1}{2^2}-\frac{1}{n_2^2}]\therefore n_2=5$
Thus, excited state for $He$ is $5^{th}$ orbit.