What transition in the hydrogen spectrum would have the same wavelength as the Balmer transition `n=4` to `n=2` of `He^(+)` spectrum ?

Options

A. n = 4 to n = 3
B. n = 3 to n = 2
C. n = 4 to n = 2
D. n = 2 to n = 1

Answer

D

Explanation

For `He^(+)` ion, `1/lamda=Z^(2)R[1/n_(1)^(2)-1/n_(2)^(2)]`
`(2)^(2)R[1/2^(2)-1/4^(2)]=(3R)/4`
For hydrogen atom, `1/lamda=R[1/n_(1)^(2)-1/n_(2)^(2)]`
`(3R)/4=R[1/n_(1)^(2)-1/n_(2)^(2)]or1/n_(1)^(2)-1/n_(2)^(2)=3/4`
`n_(1)=1andn_(2)=2`.