An He + ion in the excited state, on the transition to the ground state, emits two photons in succession with wavelengths 108 . 5 and 30 . 4 nm . What is the quantum number n corresponding to the excited state of He + ion.

Step-by-step solution

The value of h c = 1242   nm   eV ∴ The energy of the first photon is Δ E ⁡ 1 = hc λ 1 = 1 2 4 2 n m eV 1 0 8 . 5 n m = 1 1 . 4 5 eV The energy of the second photon is Δ E ⁡ 2 = hc λ 2 = 1 2 4 2 n m eV 3 0 . 4 n m = 4 0 . 8 6 eV The total energy emitted by the electron during jumping from n 2 to n 1 is Δ E = Δ E 1 + Δ E 2   = ( 11 . 45 + 40 . 86 )   eV = 52 . 31   eV or Δ E ⁡ = 1 3 . 6 Z ⁡ 2 1 n 1 2 - 1 n 2 2 eV or &#916