Question

Find the quantum number $n$ corresponding to the excited state of $H{e}^{+}$ ion if on a transition to the ground state that ion emits two photons in succession with wavelength $108.5$ nm and $30.4$ nm respectively.

• A. 7
• B. 5
• C. 3
• D. 4

Answer 1

The correct option is B 5

Given ${\lambda }_2=30.4\times {10}^{-7}cm$

${\lambda }_1=108.5\times {10}^{-7}cm$

Suppose excited state of $H{e}^{+}$ is $n_2$. The electron falls from $n_1$ to $n_1$ and then $n_1$ to $1$ energy level to emit two successive photons.

$\therefore \frac{1}{{\lambda }_2}=R_H\times Z^2[\frac{1}{1^2}-\frac{1}{n_1^2}]$

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

$\therefore n_1=2$

Now for ${\lambda }_1:n_1=2$ and $n_2=n$

$\therefore \frac{1}{{\lambda }_1}=R_H\times Z^2[\frac{1}{2^2}-\frac{1}{n^3}]$

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

$\therefore n=5$

Thus, the excited state of $H{e}^{+}$ is $5th$ orbit.

Hence, the correct option is $B$