A hydrogen atom emits ultraviolet radiation having a wavelength of 102.5 nm . Determine the quantum numbers of the states involved in the transition.

Step-by-step solution

The wavelength of the emitted photon is given by the Rydberg formula: 1 = R_H ( 1 n₁^2 - 1 n₂^2 ) Since the emitted radiation is in the ultraviolet region, it belongs to the Lyman series, which corresponds to transitions to the ground state ( n₁ = 1 ). Substituting = 102.5 nm = 102.5 10⁻⁹ m and R_H = 1.097 10^7 m ⁻¹ : 1 102.5 10⁻⁹ = 1.097 10^7 ( 1 - 1 n₂^2 ) 9.756 10^6 = 1.097 10^7 ( 1 - 1 n₂^2 ) 1 - 1 n₂^2 = 9.756 10^6 1.097 10^7 0.889 1 n₂^2 = 1 - 0.889 = 0.111 1 9 n₂^2 = 9 n₂ = 3 Thus, the quantum numbers of the