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JEE Main 202622 JanuaryEvening Shift
Exam
JEE Main
Year
2026
Date
22 January
Shift
Evening Shift
Subject
Chemistry
Chapter
Structure of Atom

JEE Main 2026 · 22 January · Evening Shift · Chemistry · Structure of Atom

The energy of first (lowest) Balmer line of H atom is x ~J . The...

The energy of first (lowest) Balmer line of H atom is x Jx \mathrm{~J}. The energy (in J) of second Balmer line of H atom is :

Options

  1. A1.35x1.35 x
  2. Bx2x^{2}
  3. C2x2 x
  4. Dx1.35\frac{x}{1.35}

Correct answer

A. 1.35x1.35 x

Step-by-step solution

First Balmer line:
transition n=3 to n=2, energy E1=13.6(14−19)=13.6×536=xE_1 = 13.6(\frac{1}{4} - \frac{1}{9}) = 13.6 \times \frac{5}{36} = x J.
Second Balmer line:
transition n=4 to n=2, energy E2=13.6(14−116)=13.6×316E_2 = 13.6(\frac{1}{4} - \frac{1}{16}) = 13.6 \times \frac{3}{16} J.
Ratio: E2E1=316536=316×365=2720=1.35\frac{E_2}{E_1} = \frac{\frac{3}{16}}{\frac{5}{36}} = \frac{3}{16} \times \frac{36}{5} = \frac{27}{20} = 1.35.
Thus second Balmer line energy is 1.35x1.35x J.
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