What transition in the hydrogen spectrum would have the same wavelength as the Balmer transition to — Atomic Structure Chemistry Question

💡 Solution & Explanation

For $He^+$ ($Z=2$), $\frac{1}{\lambda} = R(2)^2 \left(\frac{1}{2^2} - \frac{1}{4^2}\right) = 4R\left(\frac{1}{4} - \frac{1}{16}\right) = R\left(\frac{4}{4} - \frac{4}{16}\right) = R\left(1 - \frac{1}{4}\right) = R\left(\frac{1}{1^2} - \frac{1}{2^2}\right)$. This precisely matches the Rydberg equation for hydrogen ($Z=1$) undergoing a transition from $n=2$ to $n=1$. Therefore, correct answer is C.