Question

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

Answer 1

For He + ion, the wave number associated with the Balmer transition, n = 4 to n = 2 is given by :
v- = $\frac{1}{\lambda}$ = RZ2 ($\frac{1}{n_1^2}-\frac{1}{n_2^2}$)
Where, n1 = 2 ,n2 = 4 , Z = atomic number of helium
v- = $\frac{1}{\lambda}$ = R (2)2 ($\frac{1}{4}-\frac{1}{6}$) = 4R ($\frac{4-1}{16}$)
v- = $\frac{1}{\lambda}$ = $\frac{3R}{4}$ ⇒ λ = $\frac{4}{3R}$
According to the question, the desired transition for hydrogen will have the same wavelength as that of He+.
⇒ R(1)2[$\frac{1}{n_1^2}-\frac{1}{n_2^2}$] = $\frac{3R}{4}$
By hit and trail method, the equality given by equation (1) is true only when n1 = 1 and n2 = 2.
∴ The transition for n2 = 2 to n = 1 in the hydrogen spectrum would have the same wavelength as the Balmer transition n = 4 to n= 2 of He+ spectrum.