$\begin{array}{1 1}(a)\;1.8\times 10^{-2}cm&(b)\;1.8\times 10^{-4}cm\\(c)\;2.8\times 10^{-4}cm&(d)\;1.7\times 10^{-2}cm\end{array}$

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$mur=\large\frac{nh}{2\pi}$

$\therefore \large\frac{nh}{2\pi}$$=4.2178\times 10^{-34}$

Or $n=\large\frac{4.2178\times 10^{-34}\times 2\times 3.14}{6.625\times 10^{-34}}$

$\Rightarrow 4$

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

The transition spectral line for $4^{th}$ to $3^{rd}$ shell is

$\large\frac{1}{\lambda}$$=109678\big[\large\frac{1}{3^2}-\frac{1}{4^2}\big]$

$\Rightarrow \lambda=1.8\times 10^{-4}cm$

Hence (b) is the correct answer.

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