$\begin{array}{1 1}(a)\;4814.8A^{\large\circ}&(b)\;3426.7A^{\large\circ}\\(c)\;5234A^{\large\circ}&(d)\;3672.6A^{\large\circ}\end{array}$

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For $H_{\alpha}$ line of Balmer series $n_1=2,n_2=3$

For $H_{\beta}$ line of Balmer series $n_1=2,n_2=4$

$\large\frac{1}{\lambda_{H_{\large\alpha}}}$$=R_H\big[\large\frac{1}{2^2}-\frac{1}{3^2}\big]$------(1)

$\large\frac{1}{\lambda_{H_{\large\beta}}}$$=R_H\big[\large\frac{1}{2^2}-\frac{1}{4^2}\big]$------(2)

By equations (1) & (2)

$\large\frac{\lambda_{H_{\large\alpha}}}{\lambda_{H_{\large\beta}}}=\frac{\Large\frac{1}{4}-\Large\frac{1}{9}}{\Large\frac{1}{4}-\Large\frac{1}{16}}$

$\lambda_{\beta}=\lambda_{\alpha}\times \big[\large\frac{80}{108}\big]$

$\Rightarrow 6500\times \large\frac{80}{108}$

$\Rightarrow 4814.8A^{\large\circ}$

Hence (a) is the correct answer.

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