Arrange the range of frequencies of the following series of hydrogen atom in ascending order

(A) Balmer
(B) Lyman
(C) Paschen
(D) Brackett

Choose the correct answer from the options given below:

(D), (C), (A), (B)

The correct answer is Option (4) → (D), (C), (A), (B)

The frequency of spectral lines depends on the energy difference between levels:

$E = 13.6 \left( \frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}} \right) \, eV$

Series classification:

(B) Lyman: transitions to $n=1$ → highest energy difference → highest frequency.

(A) Balmer: transitions to $n=2$ → lower than Lyman.

(C) Paschen: transitions to $n=3$ → lower than Balmer.

(D) Brackett: transitions to $n=4$ → lowest frequency.

Ascending order of frequency:

Brackett < Paschen < Balmer < Lyman