$\begin{array}{1 1}(a)\;2.5\times 10^{-19}J&(b)\;1.95\times 10^{-17}J\\(c)\;2.5\times 10^{-17}J&(d)\;1.95\times 10^{-19}J\end{array}$

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$\bar {V}=Rz^2(\large\frac{1}{n_1^2}-\frac{1}{n_2^2})$

Putting the values $n_1=1,n_2=3,z=3$

We get $\bar{V}=8.77\times 10^7m^{-1}$

$\Rightarrow \lambda=\large\frac{1}{\bar V}$$=113.9A^{\circ}$

Corresponding transition responsible is $1\to \infty$

(i.e) $\Delta E=13.6\times 3^2\big(\large\frac{1}{1^2}-\frac{1}{\infty^2})$

$\Delta E=122.4eV$

$\Rightarrow 1.95\times 10^{-17}J$

Hence (b) is the correct answer.

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