$H_{\alpha }$ line of Balmer series is $6500\mathring{A}$. The wave length of $H\gamma$ is?

The correct option is B $4298\mathring{A}$

In balmer series

when $n_1$ = 2,$n_2$ = 3,it is called alpha,

when $n_1$ = 2, $n_2$ = 5, it is called gamma,

$1/\lambda =109677(1/n_1^2-1/n_2^2)c{m}^{-}1$

$=109677\left(1/2^2\right)-1/5^2$

$=23032.17c{m}^{-1}$

So $\lambda =0.00004298cm=4298A^0$