$(a)\;2\qquad(b)\;3\qquad(c)\;4\qquad(d)\;5$

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For Balmer Series

$\large\frac{1}{\lambda}$=$R_H[\large\frac{1}{n_1^2}-\large\frac{1}{n_2^2}]$

$\Rightarrow\large\frac {1} {4103\times10^{-8} }$= $109678[\large\frac{1}{2^2}-\large\frac{1}{(m+2)^2}]$

$\Rightarrow [\large\frac{1}{2^2}-\frac{1}{(m+2)^2}] = \frac{10^8}{4103 \times 109678}$

$\Rightarrow \large\frac{1}{(m+2)^2} = 0.02778$

$\Rightarrow\large\frac{1}{(m+2)} = 0.1667$

$\Rightarrow m+2 =5.99 \approx 6$

$\therefore m = 4$

$m^{th}$ line represents $4^{th}$ line of balmer series

Hence answer is (c)

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