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

For an atom, wave number = 1/λ = RHZ2 [1/n12-1/n22]

For He+ spectrum Z = 2, n2=4, n1=2


Therefore [-v] = 1/λ = RH× {4× [1/22-1/42]}


= RH× {4× [1/4 – 1/16]}


= RH× {[4× 3] / 16}


= 3RH / 4


For hydrogen spectrum, wave number = 3RH/4, Z = 1


Therefore, wave number = 1/λ = RH× 1[1/n12-1/n22]


Or


RH [1/n12-1/n22] = 3RH/4


Or


1/n12 – 1/n22 = 3/4


Which can be so for n1 = 1 & n2 = 2, i.e. the transition is from n = 2 to n = 1


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