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Chemistry · Structure of Atom

JEE Main 2026 — 6 April, Morning Shift — Question 47

If shortest wavelength of hydrogen atom in Lyman series is x , then longest wavelength in Balmer series of He+\mathrm{He}^{+}is :

  1. Option A:

    9x5\frac{9 x}{5}

    Correct
  2. Option B:

    36x5\frac{36 x}{5}

  3. Option C:

    x4\frac{x}{4}

  4. Option D:

    5x9\frac{5 x}{9}

Answer: A

Step-by-step solution

1x=R×12×(112−1∞2)\frac{1}{\mathrm{x}}=\mathrm{R} \times 1^{2} \times\left(\frac{1}{1^{2}}-\frac{1}{\infty^{2}}\right)

\frac{1}{\lambda}=\mathrm{R} \times 2^{2} \times\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right) \end{gathered}$$ $\frac{\mathrm{x}}{\lambda}=\mathrm{R} \times 4 \times \frac{5}{4 \times 9}$ $\frac{1}{\lambda}=\frac{1}{x} \times \frac{5}{9}$ $\lambda=\frac{9 \mathrm{x}}{5}$

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Paper
6 April, Morning Shift
Subject
Chemistry
Chapter
Structure of Atom
Topic
Analysis of Spectra of H-like species

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