To what series does the spectral line of atomic hydrogen belong if its wave number is equal to the d — Atomic Structure Chemistry Question

💡 Solution & Explanation

486.1 nm corresponds to $4 \to 2$. 410.2 nm corresponds to $6 \to 2$. $\bar{\nu} = \bar{\nu}_{6\to 2} - \bar{\nu}_{4\to 2} = R\left(\frac{1}{2^2} - \frac{1}{6^2}\right) - R\left(\frac{1}{2^2} - \frac{1}{4^2}\right) = R\left(\frac{1}{4^2} - \frac{1}{6^2}\right)$. This identifies a transition from $n=6$ to $n=4$, which belongs to the Brackett series. Therefore, correct answer is D.