Which transition in the hydrogen spectrum would have the same wavelength as the Balmer type transition from n = 4 to n = 2 of He + spectrum

Step-by-step solution

For He + ion, the wave number associated with the Balmer transition, n   =   4 to n   =   2 is given by: 1 λ   =   RZ 2   1 n 1 2   -   1 n 2 2 Where, n 1   =   2 n 2   =   4 Z = atomic number of helium 1 λ = R ( 2 ) 2 1 2 2 - 1 4 2 ⇒ 1 λ = 4 R 4 - 1 16 ⇒ 1 λ =   3 R 4 ⇒ λ   =   4 3 R According to the question, the desired transition for hydrogen will have the same wavelength as that of He + .