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

Answer

n=2 to n=1

Explanation

For a H or H like atom, the energy of the nth shell is given by
`E_(n)=-(2.178 xx 10^(-18) Z^2)/(n^2) J//atom `
For `He^(+)` ion (Z=2), the energy emitted in a transition from n=4, to n=2 i.e,
`=-(2.178 xx 10^(-18) xx 4)/(4^2)-(-(2.178 xx 10^(-8) xx 4)/(2^2))`
`=2.178 xx 10^(-18) xx 4 (1/4-1/(16))`
`=1.633 xx 10^(-18)"J/atom"`
The wavelength corresponding to this energy is given by `lambda=(hc)/(triangle E)=(6.626 xx 10^(-34) xx 3 xx 10^(8))/(1.633 xx 10^(-18)) (therefore triangleE=(hc)/(lambda))`
`=1.217 xx 10^(-7)m`
This wavelength falls in the ultraviolet region where Lyman series in the spectrum of hydrogen appears. For Lyman series `n_1=1`. Thus a transition from a higher level to `n_1=1 ` level in H may give this wave length. Suppose the higher level is `n_2`. For H spectrum we have `bar=1/lambda=R (1/n_(1)^2-1/n_2^2)`
In the present case, `n_1=1`
`R=1.0974 xx 10^(7)m^(-1), lambda=1.217 xx 10^(-7)m`
`(1)/(1.217 xx 10^(-7))=1.0974 xx 10^(7) (1/1^2-1/n_2^2)`
or `n_2=1.9951=2`
Hence, transition n=2 to n=1 in hydrogen will have the same wavelength as n=4 to n=2 transition in `He^(+)` spectrum.