$(a)\;3246A^o\qquad(b)\;3985A^o\qquad(c)\;4862A^o\qquad(d)\;4005A^o$

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The transition responsible for second Balmer ($\beta$-line) line is $4\to 2$

$\Delta E=13.6(1)^2(\large\frac{1}{2^2}-\frac{1}{4^2}) = 2.55eV$

Now, $\lambda=\large\frac{hc}{\Delta E}$

$\Rightarrow \lambda=4.862\times 10^{-7}m$

$\Rightarrow \lambda=4862\;\overset{\circ}{A}$

Hence (c) is the correct answer.

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