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

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Let us calculate $\Delta E$ first

$\Delta E=21.7\times 10^{-19}z^2(\large\frac{1}{n_1^2}-\frac{1}{n_2^2})$

Substituting $n_1=2$ and $n_2=4$ $z=2$

$\Delta E=1.632\times 10^{-18}J$

Now $\lambda=\large\frac{hC}{\Delta E}$

$\Rightarrow 1.218\times 10^{-17}m$

$\Rightarrow 1218A^{\large\circ}$

It means H-atom this transition would lie in Lyman series.Hence our aim is how to find the transition $m\to 1$

$\Rightarrow \Delta E=21.7\times 10^{-19}\times 1^2(\large\frac{1}{n_1^2}-\frac{1}{n_2^2})$

$\Rightarrow 1.632\times 10^{-18}=21.7\times 10^{-19}(1-\large\frac{1}{n_2^2})$

$\Rightarrow n_2=2$

Hence the corresponding transition in H-atom is $2\to 1$

Hence (a) is the correct answer.

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