Q3 / 75 · JEE Main 2026 (21 January Shift 2) · Chemistry
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 :
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 :
- A$\mathrm{D}<\mathrm{C}<\mathrm{A}<\mathrm{B}$
- Correct answerB$\mathrm{D}<\mathrm{A}<\mathrm{C}<\mathrm{B}$
- C$\mathrm{C}<\mathrm{D}<\mathrm{B}<\mathrm{A}$
- D$\mathrm{A}<\mathrm{B}<\mathrm{C}<\mathrm{D}$
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 $D < A < C < B$.
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 $D < A < C < B$.
Hence, the correct answer is option: 'B'.