The ratio of longest to shortest wavelength emitted in Paschen series of hydrogen atom is

Question

Aicharya · Accepted Answer

For Paschen series, Longest wavelength corresponds to $\begin{aligned} & \mathrm{n}_1=3, \mathrm{n}_2=\mathrm{n}_1+1=4 \ & \lambda_{\max }=\frac{144}{7 \mathrm{R}} \end{aligned}$ Shortest wavelength corresponds to $\mathrm{n}_1=3, \mathrm{n}_2=\infty$ $\begin{aligned} \lambda_{\min } & =\frac{9}{R} \ 	herefore \quad & \frac{\lambda_{\max }}{\lambda_{\min }}=\frac{\left(\frac{144}{7 R}ight)}{\left(\frac{9}{R}ight)}=\frac{144}{63} \end{aligned}$

The ratio of longest to shortest wavelength emitted in Paschen series of hydrogen atom is

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