The electronic transition in the `He^(+)` spectrum from n=4 to n=2 corresponds to which transition in the H-spectrum?

Options

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

Answer

A

Explanation

For transition in `He^(+)(Z=2)`
`1/(lamda)=RZ^(2)(1/(n_(1)^(2))-1/(n_(2)^(2)))`
`=Rxx4(1/4-1/16)=3/4R`
For H atom (Z=1) `1/(lamda)=R(1/(n_(1)^(2))-1/(n_(2)^(2)))`
`R(1/(n_(1)^(2))-1/(n_(2)^(2)))=3/4R`
`1/(n_(1)^(2))-1/(n_(2)^(2))=3/4:.n=2,n=1`