Which specific transition in the hydrogen atom spectrum would emit a photon of the exact same wavele — Atomic Structure Chemistry Question

💡 Solution & Explanation

The wavenumber is given by $\bar{ u} = R Z^2 (\frac{1}{n_1^2} - \frac{1}{n_2^2})$. For $He^+$ ($Z=2$) transitioning from $4 \to 2$, $\bar{ u} = R \times 2^2 \times (\frac{1}{2^2} - \frac{1}{4^2}) = 4R(\frac{1}{4} - \frac{1}{16}) = 4R(\frac{3}{16}) = \frac{3R}{4}$. For Hydrogen ($Z=1$), the transition matching this energy requires $\frac{1}{n_1^2} - \frac{1}{n_2^2} = \frac{3}{4}$. This is satisfied when $n_1=1$ and $n_2=2$ ($\frac{1}{1} - \frac{1}{4} = \frac{3}{4}$).