Question

In hydrogen spectrum, the ratio of the longest wavelength in the Lyman series to the longest wavelength in the Balmer series is -

• A. $\frac{27}{5}$
• B. $\frac{5}{9}$
• C. $\frac{5}{3}$
• D. $\frac{5}{27}$

Answer 1

The correct option is D $\frac{5}{27}$

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 Lyman series, $n_1=1$ and $n_2=2$,

$\therefore \frac{1}{{\lambda }_L}=R{Z}^2(\frac{1}{1^2}-\frac{1}{2^2})=\frac{3R{Z}^2}{4}...\left(i\right)$

For Balmer series, $n_1=2$ and $n_2=3$

$\therefore \frac{1}{{\lambda }_B}=R{Z}^2(\frac{1}{2^2}-\frac{1}{3^2})=\frac{5R{Z}^2}{36}...\left(ii\right)$

Dividing equation $\left(ii\right)$ by $\left(i\right)$, we get,

$\frac{{\lambda }_L}{{\lambda }_B}=\frac{\frac{5R{Z}^2}{36}}{\frac{3R{Z}^2}{4}}=\frac{5}{27}$

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