Question 488: Atoms & Nuclei

Question

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

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 $$\mathrm{n}_1=3, \mathrm{n}_2=\infty$$
$$\begin{aligned}
\lambda_{\min } & =\frac{9}{R} \
	herefore \quad \frac{\lambda_{\max }}{\lambda_{\min }} & =\frac{\left(\frac{144}{7 R}ight)}{\left(\frac{9}{R}ight)}=\frac{144}{63}
\end{aligned}$$

Answer

$$\frac{144}{63}$$

Answer

$$\frac{25}{9}$$

Answer

$$\frac{9}{25}$$

Answer

$$\frac{63}{144}$$

Atoms & Nuclei

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

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