The transition from state `n = 4` to `n = 3` in a `He^(o+)` ion result in ultraviolet radition Intrated radiation will be abtained in the transiion from

Options

A. `n = 2 rarr n = 1`
B. `n = 3 rarr n = 2`
C. `n = 5 rarr n = 4`
D. `n = 8 rarr n = 6`

Explanation

In case of H-spectrum UV spectrum appose from `n_(1) = 1.to n_(2) = 2,3….`
Here in case of `He^(theta)` spectrum UV spectrum is from `n_(1) to n_(2) = 4`
`v_(UV)= RZ^(2)(1/3^(2) -(1)/(4^(2))) = R xx 4((1)/(3^(2)) - (1)/(4^(2)))`
in case of H-spectrum IR appose from `n_(1) = 3,n_(2) = 4,5....`
`v_(m) for H_(2)^(o+) = R xx I^(2)((1)/(3^(2)) - (1)/(4^(2)))`.....(i)
`v_(m) for H^(o+) = R xx 4((1)/((n)^(2)) - (1)/(n_(2)^(2))) = R((4)/(n_(1)^(2)) - (4)/(n_(2)^(2)))`......(iii)
Comparing the coefficient of equation (i) and (ii) we get
`(4)/(n_(1)^(2)) = (1)/(9) ,n_(1) =6`
`(4)/(n_(2)^(2)) = (1)/(16) ,n_(2) =8`
The transition for `He^(o+)` in IR is `n_(1) rarr 6` to`n_(2) rarr 8`