Chemistry · Structure of Atom

JEE Main 2026 — 24 January, Morning Shift — Question 65

The hydrogen spectrum consists of several spectral lines in Lyman series $(L_1, L_2, L_3,\ldots;\, L_1$ has lowest energy among Lyman series $)$ . Similarly, it consists of several spectral lines in Balmer series $(B_1, B_2, B_3,\ldots;\, B_1$ has lowest energy among Balmer lines $)$ . The energy of $L_1$ is $x$ times the energy of $B_1$ . The value of $x$ is $\_\_\_\_ \times 10^{-1}$ (Nearest integer).

Answer: 54

Numerical answer — enter this value.

Step-by-step solution

Lowest energy Lyman line:

L_1 : 2 \to 1

E_{L_1} \propto \left(1 - \frac{1}{4}\right) = \frac{3}{4}

Lowest energy Balmer line:

B_1 : 3 \to 2

E_{B_1} \propto \left(\frac{1}{4} - \frac{1}{9}\right) = \frac{5}{36}

Ratio:

= \frac{3/4}{5/36} = \frac{27}{5} = 5.4$$

x = 5.4 = 54 \times 10^{-1}

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 24 January, Morning Shift
Subject: Chemistry
Chapter: Structure of Atom
Topic: Analysis of Spectra of H-like species