Chemistry · Structure of Atom

JEE Main 2026 — 4 April, Morning Shift — Question 49

What is the ratio of wave number of first line (lowest energy line) of Balmer series of H atomic spectrum to first line of its Brackett series?

  1. Option A:

    5:15: 1

  2. Option B:

    5:0.81

    Correct
  3. Option C:

    5:1.755: 1.75

  4. Option D:

    5:27

Answer: B

Step-by-step solution

V‾1=RH(z)2[122−132]⇒Ist\quad \overline{\mathrm{V}}_{1}=\mathrm{R}_{\mathrm{H}}(\mathrm{z})^{2}\left[\frac{1}{2^{2}}-\frac{1}{3^{2}}\right] \Rightarrow \mathrm{I}^{\mathrm{st}} line of Balmer series V‾2=RH(z)2[142−152]⇒Is\overline{\mathrm{V}}_{2}=\mathrm{R}_{\mathrm{H}}(\mathrm{z})^{2}\left[\frac{1}{4^{2}}-\frac{1}{5^{2}}\right] \Rightarrow \mathrm{I}^{\mathrm{s}} line of Brackett series V‾1 V‾2=50081⇒5:0.81\frac{\overline{\mathrm{V}}_{1}}{\overline{\mathrm{~V}}_{2}}=\frac{500}{81} \Rightarrow 5: 0.81

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Subject
Chemistry
Chapter
Structure of Atom
Topic
Analysis of Spectra of H-like species