JEE MAIN - Physics (2025 - 7th April Morning Shift - No. 17)
For a hydrogen atom, the ratio of the largest wavelength of Lyman series to that of the Balmer series is
$5: 27$
$27: 5$
$3: 4$
$5: 36$
Explanation
Lyman
_7th_April_Morning_Shift_en_17_1.png)
$$\begin{aligned} & \frac{1}{\lambda_1}=\mathrm{R}\left[\frac{1}{1}-\frac{1}{4}\right]=\frac{3 \mathrm{R}}{4} \\ & \lambda_1=\frac{4}{3 \mathrm{R}}\quad\text{..... (1)} \end{aligned}$$
and Balmer
_7th_April_Morning_Shift_en_17_2.png)
$$\begin{aligned} &\begin{aligned} & \frac{1}{\lambda_2}=\mathrm{R}\left[\frac{1}{4}-\frac{1}{9}\right]=\frac{5 \mathrm{R}}{36} \\ & \lambda_2=\frac{36}{5 \mathrm{R}} \end{aligned}\\ &\text { Then, } \frac{\lambda_1}{\lambda_2}=\frac{5}{27} \end{aligned}$$
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