A sample of hydrogen atoms in an excited state de-excites to the ground state, emitting exactly 6 distinct spectral lines. The longest wavelength among these emitted lines matches an unknown transition in the He ⁺ ion. The principal quantum numbers of this transition in He ⁺ are:

Step-by-step solution

The number of spectral lines emitted when electrons de-excite to the ground state is given by n(n-1) 2 . Given that 6 lines are emitted: n(n-1) 2 = 6 n^2 - n - 12 = 0 (n-4)(n+3) = 0 Thus, the initial excited state of the hydrogen atoms is n=4 . The longest wavelength corresponds to the smallest energy gap, which is the transition from n=4 to n=3 . The wavenumber for this transition in hydrogen ( Z=1 ) is: 1 = R_H (1)^2 ( 1 3^2 - 1 4^2 ) For the He ⁺ ion ( Z=2 ), let the transition be from n₂ to n₁ . The wavenumber