Let be the longest wavelength observed in the Lyman series and be the shortest wavelength observed i — Atomic Structure Chemistry Question

💡 Solution & Explanation

The longest wavelength in the Lyman series corresponds to the lowest energy transition ($n_2=2 \to n_1=1$). $\frac{1}{\lambda_1} = R_H(\frac{1}{1^2} - \frac{1}{2^2}) = \frac{3 R_H}{4} \implies \lambda_1 = \frac{4}{3 R_H}$. The shortest wavelength in the Balmer series corresponds to the series limit ($n_2=\infty \to n_1=2$). $\frac{1}{\lambda_2} = R_H(\frac{1}{2^2} - 0) = \frac{R_H}{4} \implies \lambda_2 = \frac{4}{R_H}$. The ratio $\frac{\lambda_1}{\lambda_2} = \frac{4 / (3 R_H)}{4 / R_H} = \frac{1}{3}$.