The shortest wavelength of the line in hydrogen atomic spectrum of Lyman series when $R_H=109678c{m}^{-1}$ is:

The correct option is D $911.7\mathring{A}$
Rydberg's formula is:
$\frac{1}{\lambda }=R_H{Z}^2(\frac{1}{n_1^2}-\frac{1}{n_2^2})$

For hydrogen,
$Z=1$ and for Lyman series, $n_1=1$ and $n_2=\infty$

(for shortest wavelength)

On substituting values, we get-

$\frac{1}{\lambda }=109678\times (1{)}^2\times [\frac{1}{(1{)}^2}-\frac{1}{(\infty {)}^2}]$

$\frac{1}{\lambda }=109678c{m}^{-1}$

or $\lambda =\frac{1}{109678}c{m}^{-1}$

$=9.117\times {10}^{-6}cm$

$=911.7\times {10}^{-8}cm$

$=911.7\stackrel{\circ }{A}$.

Hence, option D is correct.