Question 139: Atoms & Nuclei

Question

For a hydrogen atom, the ratio of the largest wavelength of Lyman series to that of the Balmer series is

Accepted Answer

Lyman
$$\begin{aligned}
& \frac{1}{\lambda_1}=\mathrm{R}\left[\frac{1}{1}-\frac{1}{4}ight]=\frac{3 \mathrm{R}}{4} \
& \lambda_1=\frac{4}{3 \mathrm{R}}\quad	ext{..... (1)}
\end{aligned}$$
and Balmer
$$\begin{aligned}
&\begin{aligned}
& \frac{1}{\lambda_2}=\mathrm{R}\left[\frac{1}{4}-\frac{1}{9}ight]=\frac{5 \mathrm{R}}{36} \
& \lambda_2=\frac{36}{5 \mathrm{R}}
\end{aligned}\
&	ext { Then, } \frac{\lambda_1}{\lambda_2}=\frac{5}{27}
\end{aligned}$$

Answer

$5: 27$

Answer

$27: 5$

Answer

$3: 4$

Answer

$5: 36$

Atoms & Nuclei

For a hydrogen atom, the ratio of the largest wavelength of Lyman series to that of the Balmer series is

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