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?

Answer:-

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.