Find the quantu number 'n' corresponding to the excited state of He^(+) ion if on transition to the ground state that ion emits two photons in succession with wavelength 108.5 and 30.4 nm respectively.

Find the quantu number 'n' corresponding to the excited state of He^(+

1.	Find the quantu number 'n' corresponding to the excited state of He^(+) ion if on transition to the ground state that ion emits two photons in succession with wavelength 108.5 and 30.4 nm respectively.
Answer» Solution :Suppose the electron in the excited state is present in the shell `n_(2)`. First it falls from `n_(1) " to " n_(1)` and then from `n_(1)` to ground state (for which n =1). Thus, the two transitions INVOLVED and the corresponding wavelengths emitted are (i) `n_(2) rarr n_(1), lamda_(1) = 108.5 nm = 108.5 xx 10^(-7)cm` (ii) `n_(1) rarr 1, lamda_(2) = 30.4 nm = 30.4 xx 10^(-7) cm` Applying Rydberg's formula first to case (ii), we get `bar(V) = (1)/(lamda) = RZ^(2) ((1)/(1^(2)) - (1)/(n_(1)^(2))), " i.e., " (1)/(30.4 xx 10^(-7)) = 109677 xx 2^(2) xx ((1)/(1^(2)) - (1)/(n_(1)^(2)))` or `(1)/(n_(1)^(2)) = 1 - (1)/(30.4 xx 10^(-7) xx 4 xx 109677) = 1 - 0.75 = 0.25 or n_(1)^(2) = (1)/(0.25) = 4 or n_(1) = 2` Applying Rydberg's formula now to case (i), we get `(1)/(108.5 xx 10^(-7)) = RZ^(2) ((1)/(2^(2)) - (1)/(n_(2)^(2))) = 109677 xx 2^(2) xx ((1)/(4) - (1)/(n_(2)^(2)))` or `(1)/(n_(2)^(2)) = (1)/(4) - (1)/(108.5 xx 10^(-7) xx 4 xx 109677) = 0.25 - 0.21 = 0.04` or `n_(2)^(2) = (1)/(0.04) = 25 or n_(2)=5`