Find out the quantum number 'n' coresponding to the exited state of `He^(+)` ion if on de-exitation to the ground state that ion emits two photons in sucession with wavelengths `1023.7 and 304 Å (R_(H)=1.097 xx10^(7) m^(-1))`.

Explanation

For two successive transitions say from n to `n_(1)` and then `n_(1)` to 1
`Delta E=(E_(n)-E_(n_(1)))+(E_(n_(1))-E_(1))=E_(n)-E_(1)=(hcR_(H)Z^(2))/(n^(2))+(hcR_(H)Z^(2))/(1^(2))=hcR_(H) Z^(2)[(n^(2)-1)/(n^(2))]" ....(i)"`
Also , `DeltaE_(1)=(hc)/(lamda_(1))+(hc)/(lamda_(2))=hc [(lamda_(1)+lamda_(2))/(lamda_(1)lamda_(2))]" ....(ii)"`
From equations (i) and (ii) `[(n^(2)-1)/(n^(2))]=(1)/(R_(H)Z^(2))[(lamda_(1)+lamda_(2))/(lamda_(1)lamda_(2))]`
`[(n^(2)-1)/(n^(2))]=(1)/(1.097xx10^(7)xx4)[((1026.7+304)xx10^(-10))/(1026.7xx304xx10^(-20))]=0.9716" "therefore n=6`