A specific spectral line in the Balmer series of a He ⁺ ion is found to have the exact same wavelength as the first line of the Lyman series of a hydrogen atom. The principal quantum number of the upper state for this transition in the He ⁺ ion is _______.

Step-by-step solution

For the first line of the Lyman series of a hydrogen atom ( Z=1 ), the transition is from n₂ = 2 to n₁ = 1 . Using the Rydberg formula: 1 _ H = R_ H (1)^2 [ 1 1^2 - 1 2^2 ] = R_ H ( 1 - 1 4 ) = 3R_ H 4 For a line in the Balmer series of a He ⁺ ion ( Z=2 ), the transition is from an unknown upper state n to n₁ = 2 . 1 _ He ^+ = R_ H (2)^2 [ 1 2^2 - 1 n^2 ] = 4R_ H ( 1 4 - 1 n^2 ) = R_ H ( 1 - 4 n^2 ) Given that the wavelengths are equal, their reciprocals are also equal: 3R_ H 4 = R_ H ( 1 - 4 n^2 ) 3 4 = 1 - 4 n^2