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Chemistry · Structure of Atom

JEE Main 2026 — 24 January, Evening Shift — Question 60

The wavelength of spectral line obtained in the spectrum of Li2+\mathrm{Li^{2+}} ion, when the transition takes place between two levels whose sum is 4 and difference is 2, is

  1. Option A:

    2.28×10−7 cm2.28 \times 10^{-7} \mathrm{~cm}

  2. Option B:

    2.28×10−6 cm2.28 \times 10^{-6} \mathrm{~cm}

  3. Option C:

    1.14×10−7 cm1.14 \times 10^{-7} \mathrm{~cm}

  4. Option D:

    1.14×10−6 cm1.14 \times 10^{-6} \mathrm{~cm}

    Correct

Answer: D

Step-by-step solution

Given:

n2+n1=4n_2 + n_1 = 4 n2−n1=2n_2 - n_1 = 2

On solving, we have

n2=3,  n1=1n_2 = 3,\; n_1 = 1

For Li2+\mathrm{Li^{2+}}, Z=3Z = 3, thus we have

= Z^2 R \left(1 - \frac{1}{9}\right)$$ $$\frac{1}{\lambda} = 9R \cdot \frac{8}{9} = 8R$$ $$\frac{1}{\lambda} = 8 \times 1.097 \times 10^5$$ $$\frac{1}{\lambda} = 8.776 \times 10^5$$

\lambda = 1.14 \times 10^{-6},\mathrm{cm}

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Paper
24 January, Evening Shift
Subject
Chemistry
Chapter
Structure of Atom
Topic
Analysis of Spectra of H-like species
The wavelength of spectral line obtained in the spectrum of Li 2+… | JEE Main 2026 PYQ with Solution · DhiX AI