JEE Main 2026 — Chemistry Physical Chemistry

2026mcqmedium

Consider the following spectral lines for atomic hydrogen :

A. First line of Paschen series

B. Second line of Balmer series

C. Third line of Paschen series

D. Fourth line of Bracket series.

The correct arrangement of the above lines in ascending order of energy is :

Official previous-year question

Held on 21 Jan 2026 · Verified 6 Jul 2026.

Options

A
$\mathrm{D}<\mathrm{C}<\mathrm{A}<\mathrm{B}$
B
$\mathrm{D}<\mathrm{A}<\mathrm{C}<\mathrm{B}$
C
$\mathrm{C}<\mathrm{D}<\mathrm{B}<\mathrm{A}$
D
$\mathrm{A}<\mathrm{B}<\mathrm{C}<\mathrm{D}$

Did you get this right?

Solution

The energy of a spectral line in the hydrogen atom is given by $E = 13.6 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$ eV.

A. First line of Paschen series: $n_1 = 3, n_2 = 4$ .

$E_A = 13.6 \left( \frac{1}{3^2} - \frac{1}{4^2} \right) = 13.6 \left( \frac{1}{9} - \frac{1}{16} \right) = 13.6 \times \frac{7}{144} \approx 0.66$ eV.

B. Second line of Balmer series: $n_1 = 2, n_2 = 4$ .

$E_B = 13.6 \left( \frac{1}{2^2} - \frac{1}{4^2} \right) = 13.6 \left( \frac{1}{4} - \frac{1}{16} \right) = 13.6 \times \frac{3}{16} = 2.55$ eV.

C. Third line of Paschen series: $n_1 = 3, n_2 = 6$ .

$E_C = 13.6 \left( \frac{1}{3^2} - \frac{1}{6^2} \right) = 13.6 \left( \frac{1}{9} - \frac{1}{36} \right) = 13.6 \times \frac{3}{36} = 13.6 \times \frac{1}{12} \approx 1.13$ eV.

D. Fourth line of Brackett series: $n_1 = 4, n_2 = 8$ .

$E_D = 13.6 \left( \frac{1}{4^2} - \frac{1}{8^2} \right) = 13.6 \left( \frac{1}{16} - \frac{1}{64} \right) = 13.6 \times \frac{3}{64} \approx 0.6375$ eV.

Comparing the values: $E_D (0.6375) < E_A (0.66) < E_C (1.13) < E_B (2.55)$ .

The ascending order of energy is $D < A < C < B$ .

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Solution

The energy of a spectral line in the hydrogen atom is given by $E = 13.6 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$ eV.

A. First line of Paschen series: $n_1 = 3, n_2 = 4$ .

$E_A = 13.6 \left( \frac{1}{3^2} - \frac{1}{4^2} \right) = 13.6 \left( \frac{1}{9} - \frac{1}{16} \right) = 13.6 \times \frac{7}{144} \approx 0.66$ eV.

B. Second line of Balmer series: $n_1 = 2, n_2 = 4$ .

$E_B = 13.6 \left( \frac{1}{2^2} - \frac{1}{4^2} \right) = 13.6 \left( \frac{1}{4} - \frac{1}{16} \right) = 13.6 \times \frac{3}{16} = 2.55$ eV.

C. Third line of Paschen series: $n_1 = 3, n_2 = 6$ .

$E_C = 13.6 \left( \frac{1}{3^2} - \frac{1}{6^2} \right) = 13.6 \left( \frac{1}{9} - \frac{1}{36} \right) = 13.6 \times \frac{3}{36} = 13.6 \times \frac{1}{12} \approx 1.13$ eV.

D. Fourth line of Brackett series: $n_1 = 4, n_2 = 8$ .

$E_D = 13.6 \left( \frac{1}{4^2} - \frac{1}{8^2} \right) = 13.6 \left( \frac{1}{16} - \frac{1}{64} \right) = 13.6 \times \frac{3}{64} \approx 0.6375$ eV.

Comparing the values: $E_D (0.6375) < E_A (0.66) < E_C (1.13) < E_B (2.55)$ .

The ascending order of energy is $D < A < C < B$ .

JEE Main 2026 — Chemistry Physical Chemistry

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Solution

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JEE Main 2026 — Chemistry Physical Chemistry

Options

Did you get this right?

Solution

Your note

JEE Main Physical Chemistry in other years

More JEE Main Physical Chemistry questions

Other Chemistry topics for JEE Main

Organic Chemistry

Inorganic Chemistry

Keep practicing Physical Chemistry