The shortest wavelength in hydrogen spectrum approximately is
The shortest wavelength in hydrogen spectrum approximately is
$121 \mathrm{~nm}$
$91.2 \mathrm{~nm}$
$182 \mathrm{~nm}$
$89.2 \mathrm{~nm}$
Solution
Sequence of series in hydrogen spectrum are Lyman series $(n=1)$, Balmer series $(n=2)$, Paschen series $(n=3)$, Bracket series $(n=4)$ and Pfund series $(n=5)$.
For Lyman series, $n_1=1$
For shortest wavelength in Lyman series i.e., the energy difference in two states showing transition should be maximum i.e., $n_2=\infty$
Using formula, $\frac{1}{\lambda}=R_{\mathrm{H}}\left[\frac{1}{n_1^2}-\frac{1}{n_2^2}\right]$ here
$
\begin{aligned}
R_{\mathrm{H}} & =\text { Rydberg constant } \\
\frac{1}{\lambda} & =R_{\mathrm{H}}\left[\frac{1}{1^2}-\frac{1}{\infty^2}\right] \Rightarrow \frac{1}{\lambda}=R_{\mathrm{H}} \\
\lambda & =\frac{1}{R_{\mathrm{H}}}=\frac{1}{109678} \\
& =9.117 \times 10^{-6} \mathrm{~cm}=91.2 \mathrm{~nm}
\end{aligned}
$
Hence, Lyman series is a spectral series of transitions and resulting ultraviolet emission lines of $\mathrm{H}$-atom