The wavelength of $H_{\alpha }$ line of the Balmer series is $5994\stackrel{o}{\text{A}}$. The wavelength of the $H_{\beta }$ line of Balmer series is:

The correct option is D $4440\stackrel{o}{\text{A}}$
$H_{\alpha }$ line of Balmer series is produced due to the transition from $n=3$ to $n=2$
$H_{\beta }$ line of Balmer series is produced due to the transition from $n=4$ to $n=2$
Therefore,
$\frac{1}{{\lambda }_{\alpha }}=R_H[\frac{1}{2^2}-\frac{1}{3^2}]...\left(1\right)$

$\frac{1}{{\lambda }_{\beta }}=R_H[\frac{1}{2^2}-\frac{1}{4^2}]...\left(2\right)$
Where $R_H$ is the Rydberg constant.

From the equation $\left(1\right)$ and $\left(2\right)$ we get,
$\frac{{\lambda }_{\beta }}{{\lambda }_{\alpha }}=\frac{5\times 16}{3\times 36}$
$\Rightarrow {\lambda }_{\beta }=\frac{5\times 16}{3\times 36}\times 5994\stackrel{o}{\text{A}}$
$\Rightarrow {\lambda }_{\beta }=4440\stackrel{o}{\text{A}}$