A `He^+` ion in the excited state, on transition to the ground state, emits two photons in succession with wavelengths 108.5 nm and 30.4 nm. What is the quantum number n corresponding to the excited state of `He^+` ion ?

Answer

21

Explanation

The value of `hc = 1242 nm eV`
The energy of the first photon is,
`DeltaE_(1) = (hc)/lambda_(1) = (1242 nm eV)(108.5 nm) 11.45 eV`
The energy of the second photon is,
`DeltaE_(2) = (hc)/lambda_(2) = (242 nm eV)/(30.4 nm) 40.86 eV`
The total energy emitted by the electron while jumping from n2 to `n_1` is,
`DeltaE = DeltaE_(1) + DeltaE_(2)`
`= (11.45 + 40.86)eV = 52.31 eV`
or , `DeltaE = 13.6 Z^(2) (1/n_(1)^(2) + 1/n^(2)) eV`
or, `DeltaE 13.6 Z^(2) (1/n_(1)^(2) - 1/n^2)` eV
or `52.31 eV = 13.6 Z^(2) (1/n_(1)^(2) -1/n^(2)) eV`
But the final state of the electron is the ground state,
`therefore n_(1) = 1`
`therefore 52.31 eV = 13.6 Z^(2) (1-1/2^(2)) eV`
`1-1/n^(2) = 3.85/2^(2) = 3.85/4 = 0.96` (for `He^(+), Z = 2`)
or `n^(2) = 1/0.04 = 25`
`therefore n = 5`