Question

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

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

Answer 1

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

As per spectral lines, wavelength is given by,

$\frac{1}{\lambda }=R{Z}^2(\frac{1}{n_1^2}-\frac{1}{n_2^2})$

For ${\lambda }_1$, $n_1=3$ and $n_2=4$

$\therefore \frac{1}{{\lambda }_1}=R{Z}^2(\frac{1}{3^2}-\frac{1}{4^2})=\frac{7R{Z}^2}{144}$

For ${\lambda }_2$, $n_1=2$ and $n_2=3$

$\therefore \frac{1}{{\lambda }_2}=R{Z}^2(\frac{1}{2^2}-\frac{1}{3^2})=\frac{5R{Z}^2}{36}$

Now,

$\frac{{\lambda }_1}{{\lambda }_2}=\frac{\frac{1}{\frac{7R{Z}^2}{144}}}{\frac{1}{\frac{5R{Z}^2}{36}}}=\frac{20}{7}$

Hence, option $\left(D\right)$ is the correct answer.