$(a)\;2\qquad(b)\;3\qquad(c)\;1\qquad(d)\;4$

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For II line of Balmer series $n_1=2,n_2=4$

$\therefore \large\frac{1}{\lambda_{2B}}=$$R_H\times\big[\large\frac{1}{2^2}-\frac{1}{4^2}\big]$

$\Rightarrow R_H\times \large\frac{3}{16}$-----(1)

For I line of Balmer series $n_1=1,n_2=2$

$\therefore \large\frac{1}{\lambda_{1L}}$$=R_H\times \big[\large\frac{1}{1^2}-\frac{1}{2^2}\big]$

$\Rightarrow R_H \times \large\frac{3}{4}$-------(2)

By equ (1) & (2)

$\large\frac{\lambda_{2B}}{\lambda_{1L}}=\frac{R_H\times 3\times 16}{4\times 3\times R_H}$$=4$

Hence (d) is the correct answer.

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