Question

Find the quantum number `n` corresponding to the excited state of `He^(o+)` ion if on transition to the ground state that ion emits two photon in succession with wavelength `108.5` and `30.4 nm`

Answer 1

If the wavelength are `lambda_(1),lambda_(2)` then the total energy of the excited start must be
`E_(n)=E_(1)+(2 pi c ħ)/(lambda_(1))+(2 pi c ħ)/(lambda_(2))`
But `E_(1)= -4E_(H)` and `E_(n)= -(4E_(H))/(n^(2))` where we are ignoring reduced mass effects.
Then `4E_(H)=(4E_(H))/(n^(2))+(2pi c ħ)/(lambda_(1))+(2pi c ħ)/(lambda_(2))`
Substituting the values we get `n^(2)= 23`
which we take to mean `n=5`. (The result is sensitive to the values of the various quantities and small difference get multiplied beacuse difference of two large quantities is involved:
`n^(2)=(E_(H))/(E_(H)-(pi c ħ)/(2)((1)/(lambda_(1))+(1)/(lambda_(2))))`.