Question

An excited $H{e}^{+}$ ion emits two photons in succession, with wavelengths $108.5\text{nm}$ and $30.4\text{nm}$, in making a transition to ground state. The quantum number $n$, corresponding to its initial excited is (for photon of wavelength $\lambda$ energy $E=\frac{1240\text{eV}}{\lambda \text{(in nm)}}$

• A. $n=5$
• B. $n=4$
• C. $n=7$
• D. $n=6$

Answer 1

The correct option is A $n=5$

When an electron transition takes place from $n^{th}$ to $m^{th}$ level , a photon of wavelngth $\lambda$ is released.

Let us consider, electron makes a transition from $n^{th}$ state to ground state via an intermediary state $m$

For first transition:

$E_n=+13.6(Z{)}^2(\frac{1}{m^2}-\frac{1}{n^2})=\frac{hc}{\lambda }$ $(n>m)$

Substituting the given data, $E_n=13.6\times 4(\frac{1}{m^2}-\frac{1}{n^2})=\frac{1240}{108.5}..\left(1\right)$

For second transition:

$E_m=+13.6(Z{)}^2(1-\frac{1}{m^2})=\frac{hc}{\lambda }$

Susbtituting the given data,

$E_m=+13.6\times 4(1-\frac{1}{m^2})=\frac{1240}{30.4}.....\left(2\right)$

Adding $\left(1\right)$ and $\left(2\right)$ we get,

$E_n+E_m=13.6\times 4(1-\frac{1}{n^2})=1240\times (\frac{1}{108.5}+\frac{1}{30.4})$

$\Rightarrow n=5$

Hence, option (B) is the correct answer.