JEE Main Chemistry Question (2026) — Solution

Step-by-step solution

The wave number of a spectral line in the hydrogen emission spectrum is given by the Rydberg formula: = R_H ( 1 n_1^2 - 1 n_2^2 ) For the first line (lowest energy) of the Balmer series, the transition is from n_2 = 3 to n_1 = 2: _ Balmer = R_H ( 1 2^2 - 1 3^2 ) = R_H ( 1 4 - 1 9 ) = 5R_H 36 For the first line of the Brackett series, the transition is from n_2 = 5 to n_1 = 4: _ Brackett = R_H ( 1 4^2 - 1 5^2 ) = R_H ( 1 16 - 1 25 ) = 9R_H 400 Taking the ratio of the two wave numbers: _ Balmer _ Brackett = 5R_H 36 9R_H 400 = 5 36 400 9 = 500 81 To express this ratio in the form 5 : x, we divide the numerator and the denominator by 100: 500 81 = 5 0.81 The ratio is 5 : 0.81. Answer: 5:0.81