Chemistry · Structure of Atom

JEE Main 2026 — 21 January, Evening Shift — Question 48

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 sereis

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

Answer: D

Energy of emitted radiation is proportional to, $\Delta E \propto \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$

A. First Paschen line (4→3)

\frac{1}{9} - \frac{1}{16} = \frac{7}{144}$$ B. Second Balmer line (4→2) $$\Delta_B = \frac{1}{4} - \frac{1}{16} = \frac{3}{16}$$ C. Third Paschen line (6→3) $$\Delta_C = \frac{1}{9} - \frac{1}{36} = \frac{1}{12}$$ D. Fourth Brackett line (8→4) $$\Delta_D = \frac{1}{16} - \frac{1}{64} = \frac{3}{64}$$ Numerical comparison: $$\frac{3}{64} < \frac{7}{144} < \frac{1}{12} < \frac{3}{16}$$ Hence, $D < A < C < B$

Answer key and solution verified before publishing.

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