Question

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

Answer 1

For an atom, `bar(v) =(1)/(lamda)=R_(H)Z^(2)((1)/(n_(1)^(2))-(1)/(n_(2)^(2)))`
For `He^(+)` spectrum : Z = 4, `n_(2) = 4, n_(1) = 2`.
`:. " " bar(v) =(1)/(lambda)=R_(H) xx4((1)/(2^(2))-(1)/(4^(2)))=(3 R_(H))/(4)`
For hydrogen spectrum : `bar(v) =(3 R_(H))/(4) ` and z= 1
` :. " " bar(v) =(1)/(lambda)=R_(H) xx 1((1)/(n_(1)^(2))-(1)/(n_(2)^(2)))`
or `" " R_(H)((1)/(n_(1)^(2))-(1)/(n_(2)^(2))) =(3R_(H))/(4) " " or " " (1)/(n_(1)^(2))-(1)/(n_(2)^(2))=(3)/(4)`
This corresponds to `n_(1) = 1 , n_(2) = 2` and means that the transition has place in the Lyman series from n = 2 to n= 1.
Thus, the transiton is from `n_(2)` to `n_(1)` in case of hydrogen spectrum.