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 :

We need to arrange the given spectral lines in ascending order of energy. The energy of a spectral line transition is given by:

$$E = 13.6 \left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right) \text{ eV}$$ where $$n_f$$ is the lower energy level and $$n_i$$ is the upper energy level.

For the first line of the Paschen series (A) with $$n_f = 3$$ and $$n_i = 4$$,

$$E_A = 13.6\left(\frac{1}{9} - \frac{1}{16}\right) = 13.6 \times \frac{7}{144} = 0.6611 \text{ eV}.$$

For the second line of the Balmer series (B) with $$n_f = 2$$ and $$n_i = 4$$,

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

For the third line of the Paschen series (C) with $$n_f = 3$$ and $$n_i = 6$$,

$$E_C = 13.6\left(\frac{1}{9} - \frac{1}{36}\right) = 13.6 \times \frac{3}{36} = 13.6 \times \frac{1}{12} = 1.1333 \text{ eV}.$$

For the fourth line of the Brackett series (D) with $$n_f = 4$$ and $$n_i = 8$$,

$$E_D = 13.6\left(\frac{1}{16} - \frac{1}{64}\right) = 13.6 \times \frac{3}{64} = 0.6375 \text{ eV}.$$

Arranging these energies in ascending order gives

$$E_D (0.6375) < E_A (0.6611) < E_C (1.1333) < E_B (2.55),$$

so the order of the spectral lines is D < A < C < B. The correct answer is Option B.