An electron in a hydrogen atom is initially in the fourth excited state and emits a photon of energy 2.86 eV while making a transition to the nth energy level. Find n.

Question

G1 Physics · Accepted Answer

The fourth excited state corresponds to principal quantum number n i n_i n i ​ = 5. Hydrogen energy levels are E n E_n E n ​ = -13.6/n² eV. Photon energy emitted is E p h o t o n E_photon E p ​ h o t o n = E i n i t i a l E_initial E i ​ ni t ia l - E f i n a l E_final E f ​ ina l in magnitude: 2.86 = 13.6(1/ n f 2 n_f² n f 2 ​ - 1/5²). Thus, 2.86/13.6 = 1/ n f 2 n_f² n f 2 ​ - 1/25. Since 2.86/13.6 ≈ 0.2103, we get 1/ n f 2 n_f² n f 2 ​ ≈ 0.2103 + 0.04 ≈ 0.2503 ≈ 1/4. Therefore, n f n_f n f ​…