The wavelengths of the lines in this series are given by , in which and is a constant. What is the v — Atomic Structure Chemistry Question

💡 Solution & Explanation

For He$^+$ ($Z=2$) ending at $n=4$: $\frac{1}{\lambda} = R(2^2)\left(\frac{1}{4^2} - \frac{1}{n^2}\right) = 4R\left(\frac{n^2 - 16}{16n^2}\right) = \frac{R(n^2 - 16)}{4n^2}$. Rearranging yields $\lambda = \frac{4n^2}{R(n^2 - 16)}$. Comparing with the given formula, $C = \frac{4}{R}$. Given $R \approx 1.09 \times 10^7$ m$^{-1}$, $C = \frac{4}{1.09 \times 10^7} \approx 366.97 \times 10^{-9}$ m $= 366.97$ nm. Therefore, correct answer is A.