Find the quantum 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 wavelengths state that ion emits two photons in succession with wavelengths`1026.7 and 304Å. (R = 1.096 xx 10^(7)m^(_1)`

Explanation

Given ` (1)/(lambda_(1)) + (1)/(lambda_(2)) = RZ^(2) (1 - (1)/(n^(2)))`
`(1)/(n^(2)) = 1 - [(lambda_(1) +lambda_(2))/(lambda_(1) lambda_(2)) xx (1)/(RZ^(2))]`
`= 1 - (1330.7 xx 10^(-10))/(1026.7 xx 304 xx 10^(-20) xx 4 xx 1.096 xx 10^(7))`
Thus `(1)/(n^(2)) = 0.0275 implies 6.03`
Hence the quantum number`= 6`