A spectral line in the Balmer series of a hydrogen atom is observed. Its wavelength is given to be 20 27 times the wavelength of the first line of the Balmer series. The principal quantum number ( n ) of the initial higher energy state for this unknown spectral line is

Step-by-step solution

For the Balmer series, the transitions terminate at n₁ = 2 . For the first line of the Balmer series ( n = 3 n = 2 ): 1 ₁ = R ( 1 2^2 - 1 3^2 ) = R ( 1 4 - 1 9 ) = 5R 36 For the unknown spectral line from state n to n=2 : 1 _n = R ( 1 2^2 - 1 n^2 ) = R ( 1 4 - 1 n^2 ) It is given that _n = 20 27 ₁ . Therefore, taking the reciprocal: 1 _n = 27 20 ( 1 ₁ ) Substituting the value of 1 ₁ : 1 _n = 27 20 5R 36 = 3R 16 Equating the two expressions for 1 _n : R ( 1 4 - 1 n^2 ) = 3R 16 1 4 - 1 n^2 = 3 16 1 n^2 = 1 4 - 3 16 =