Question

Electron in hydrogen atom first jumps from third excited state to second excited state and then from second excited to the first excited state. The ratio of the wavelengths ${\lambda }_1:{\lambda }_2$ emitted in the two cases is

• A. $\frac{27}{5}$
• B. $\frac{7}{5}$
• C. $\frac{20}{7}$
• D. $\frac{27}{20}$

Answer 1

The correct option is C $\frac{20}{7}$

We have,
$\frac{1}{\lambda }=R_H(\frac{1}{n_i^2}-\frac{1}{n_f^2})$
For first trasition from third excited state to second.

$n_i=4,n_f=3$

$\Rightarrow \frac{1}{{\lambda }_1}=R_H(\frac{1}{3^2}-\frac{1}{4^2})$

Similarly, for the second transition from second excited state to first excited state,

$\frac{1}{{\lambda }_2}=R_H(\frac{1}{2^2}-\frac{1}{3^2})$

Ratio of both the wavelength is given by,
$\frac{{\lambda }_1}{{\lambda }_2}=\frac{\frac{1}{4}-\frac{1}{9}}{\frac{1}{9}-\frac{1}{16}}$

$\therefore \frac{{\lambda }_1}{{\lambda }_2}=\frac{20}{7}$

Hence, option $\left(B\right)$ is correct.