Q.33:- What transition in the hydrogen spectrum would have the same wavelength as the Balmer transition n = 4 to n = 2 of He+ spectrum?
For H-like particles, ṽ = (2π2mZ2e4/ch3)×(1/n12 – 1/n22) = RZ2(1/n12 – 1/n22)
∴ For He+ spectrum, Balmer transition, n=4 to n=2
ṽ = 1/λ = RZ2(1/22 – 1/42) = R×4×3/16 = 3R/4
For hydrogen spectrum ,
ṽ = 1/λ = R(1/n12 – 1/n22) = 3R/4 ⇒ (1/n12 – 1/n22) = 3/4
which can be true only for n1=1 and n2=2 i.e. transition from n=2 to n=1.