The shortest wavelength in the Paschen series of the Hydrogen spectrum is Rydberg constant of hydrogen $=1…
The shortest wavelength in the Paschen series of the Hydrogen spectrum is Rydberg constant of hydrogen $=1.097 \times 10^7 \mathrm{~m}^{-1}$.
- $91.2 \mathrm{~nm}$
- $364.6 \mathrm{~nm}$
- $820.4 \mathrm{~nm}$
- $2278.9 \mathrm{~nm}$
Solution
For shortest wavelength of Paschen series $\left(n_1=3\right.$,
$\begin{aligned}
& \left.\mathrm{n}_2=\infty\right) \\
& \frac{1}{\lambda_{\mathrm{S}}}=\mathrm{R}_{\mathrm{H}}\left(\frac{1}{3^2}-\frac{1}{\infty}\right) \\
& \lambda_{\mathrm{S}}=\frac{9}{\mathrm{R}_{\mathrm{H}}}=\frac{9}{1.097 \times 0^7} \mathrm{~m} \\
& =820.4 \mathrm{~nm}
\end{aligned}$
Asked in: AP EAMCET 2022 (06 Jul Shift 1)
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