Physics · Atomic Physics

JEE Main 2025 — 7 April, Morning Shift — Question 47

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

Option A:
$5: 27$
Correct
Option B:
$5: 36$
Option C:
$3: 4$
Option D:
$27: 5$

Answer: A

Step-by-step solution

$\frac{\lambda_{C}}{\lambda_{L}}=E_{0}\left[1-\frac{1}{4}\right]=\frac{E_{0} 3}{4}$

\begin{gathered} \frac{\lambda_{C}}{\lambda_{B}}=E_{0}\left[\frac{1}{4}-\frac{1}{9}\right]=\frac{E_{0} 5}{4 \times 9} \end{gathered}

So, $\frac{\lambda_{C} \lambda_{L}}{\lambda_{B} \times \lambda_{C}}=\frac{5 E_{0} \times 4}{4 \times 9 \times 3 E_{0}}=\frac{5}{27}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 7 April, Morning Shift
Subject: Physics
Chapter: Atomic Physics
Topic: Hydrogen Spectrum

In a hydrogen like ion, the energy difference between the 2^ nd excitation energy state and ground is 108.8 eV . The atomic number of the…

For a hydrogen atom, the ratio of the largest wavelength of Lyman… | JEE Main 2025 PYQ with Solution · DhiX AI