If the longest wavelength of the Balmer series for a hydrogen atom is , then the shortest wavelength of the Brackett series is x . The value of x is

Step-by-step solution

The longest wavelength of the Balmer series corresponds to the transition from n = 3 to n = 2 . Using the Rydberg formula: 1 = R ( 1 2^2 - 1 3^2 ) = R ( 1 4 - 1 9 ) = 5R 36 = 36 5R The shortest wavelength of the Brackett series corresponds to the series limit, i.e., the transition from n = to n = 4 . 1 ' = R ( 1 4^2 - 1 ^2 ) = R 16 ' = 16 R We are given ' = x . x = ' = 16 R 36 5R = 16 5 36 = 80 36 = 20 9 Answer: 20 9