A hydrogen atom in a state with a binding energy of 0.85 eV undergoes a transition to a state having an excitation energy of 10.2 eV . Determine the quantum numbers n of the upper and the lower energy states participating in this transition.

Step-by-step solution

The binding energy of a hydrogen atom in state n is given by B.E. = 13.6 n^2 eV . For the upper state, the binding energy is 0.85 eV . 13.6 n₂^2 = 0.85 n₂^2 = 13.6 0.85 = 16 n₂ = 4 The excitation energy of a state n is the energy required to excite the electron from the ground state ( n=1 ) to that state. E = 13.6 (1 - 1 n₁^2 ) eV For the lower state, the excitation energy is 10.2 eV . 13.6 (1 - 1 n₁^2 ) = 10.2 1 - 1 n₁^2 = 10.2 13.6 = 0.75 1 n₁^2 = 0.25 n₁^2 = 4 n₁ = 2 The quantum numbers of the upper and lower en