Assume that under specific conditions, hydrogen atoms are permitted to undergo only those transitions where the principal quantum number n changes by 2 . Determine the wavelength(s) emitted by hydrogen that fall within the visible range ( 380 nm to 780 nm ).

Step-by-step solution

The visible range of the hydrogen emission spectrum corresponds to the Balmer series, where the final state is n_f = 2 . Given the condition that the principal quantum number changes by 2 ( n = 2 ), the initial state must be n_i = 2 + 2 = 4 . The energy difference for the transition from n_i = 4 to n_f = 2 is given by: E = 13.6 ( 1 n_f^2 - 1 n_i^2 ) eV E = 13.6 ( 1 2^2 - 1 4^2 ) eV = 13.6 ( 1 4 - 1 16 ) eV E = 13.6 3 16 eV = 2.55 eV The corresponding wavelength can be calculated using the relation = hc E : 1242 eV