The wavelength of the first line in balmer series in the hydrogen spectrum is $\lambda .$ What is the wavelength of the second line-

The correct option is A $\frac{20\lambda }{27}$
For the balmer series wavelength emitted will be,

$\frac{1}{\lambda }=R[\frac{1}{(n_f{)}^2}-\frac{1}{(n_i{)}^2}]$

Here, $n_f=2;n_i=3,4,5,\mathrm{6....}$

For the first balmer line, $n_f=2;n_i=3$

$\Rightarrow {\lambda }_1=R[\frac{1}{(2{)}^2}-\frac{1}{(3{)}^2}]$

$\Rightarrow {\lambda }_1=\frac{36}{5R}=\lambda .....\left(1\right)$

For the second balmer line, $n_f=2;n_i=4$

$\Rightarrow {\lambda }_2=R[\frac{1}{(2{)}^2}-\frac{1}{(4{)}^2}]$

$\Rightarrow {\lambda }_2=\frac{16}{3R}.....\left(2\right)$

From (1) and (2) we get,

$\therefore {\lambda }_2=\frac{20\lambda }{27}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(A\right)$ is the correct answer.