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.