Question

What transition in the hydrogen spectrum would have the same wavelength as the Balmer transition n = 6 to n = 2 of He+ spectrum ?

Answer 1

For He+ ion, we have

1/λ = Z2R­H [1/n12 -1/n22]

= (2)2RH [1/(2)2 – 1/(4)2] = RH 3/4 ….(i)

Now for hydrogen atom 1/λ = RH [1/n12-1/n22] ….(ii)

Equating equation (i) and (ii), we get

1/n12 -1/n22 = 3/4

Obviously, n1 = 1 and n2 = 2

Hence, the transition n = 2 to n = 1 in hydrogen atom will have the same wavelength as the transition, n = 4 to n = 2 in He+ species