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

Answer

B

Explanation

`barv=1/(lamda)=(1/(2^(2))-1/(4^(2)))RZ^(2)=3/4R`
In H- spectrum for the same `barv` or `lamda` as `Z=1,n=1, n_(2)-2`
Hence D is the correct answer.