Exam

JEE Main

Year

2025

Date

7 April

Shift

Morning Shift

Subject

Physics

Chapter

Atomic Physics

JEE Main 2025 · 7 April · Morning Shift · Physics · Atomic Physics

For a hydrogen atom, the ratio of the largest wavelength of Lyman...

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

Step-by-step solution

$\mathrm {Lyman}$

$\frac{1}{\lambda_1}=\mathrm{R}\left[\frac{1}{1}-\frac{1}{4}\right]=\frac{3 \mathrm{R}}{4}$
$\lambda_1=\frac{4}{3 R}$ ________...(1)
and $\mathrm {Balmer}$
$\frac{1}{\lambda_2}=\mathrm{R}\left[\frac{1}{4}-\frac{1}{9}\right]=\frac{5 \mathrm{R}}{36}$
$\lambda_2=\frac{36}{5 R}$
Then, $\frac{\lambda_1}{\lambda_2}=\frac{5}{27}$