$\begin{array}{1 1}(a)\;4367.4A^{\large\circ}&(b)\;3263A^{\large\circ}\\(c)\;3663.6A^{\large\circ}&(d)\;2348.1A^{\large\circ}\end{array}$

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$E_n=-\large\frac{21.7\times 10^{-12}}{n^2}$erg

$\therefore E_2=-\large\frac{21.7\times 10^{-12}}{4}$$=5.425\times 10^{-12}erg$

For removal of electron $E_2=\large\frac{hc}{\lambda}$

$E_2$ should be given to remove electron (i.e) +ve

$\lambda=\large\frac{6.625\times 10^{-27}\times 3.0\times 10^{10}}{5.425\times 10^{-12}}$

$\Rightarrow 3663.6\times 10^{-8}cm$

$\Rightarrow 3663.6A^{\large\circ}$

Hence (c) is the correct answer.

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