What is the ratio of wave number of first line (lowest energy line) of Balmer series of H atomic spectrum to first line of its Brackett series?

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 n₁^2 - 1 n₂^2 ) For the first line (lowest energy) of the Balmer series, the transition is from n₂ = 3 to n₁ = 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 n₂ = 5 to n₁ = 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