In the hydrogen spectrum, the longest wavelength in the Lyman series is 120 nm and the shortest wave — Atomic Structure Chemistry Question

💡 Solution & Explanation

The longest wavelength in the Lyman series corresponds to the $n=2 \to 1$ transition: $E_{2\to 1} = \frac{hc}{120}$. The shortest wavelength in the Balmer series corresponds to the $n=\infty \to 2$ transition: $E_{\infty \to 2} = \frac{hc}{360}$. The ionization energy corresponds to the $n=\infty \to 1$ transition, which is $E_{ion} = E_{\infty\to 2} + E_{2\to 1} = \frac{hc}{360} + \frac{hc}{120} = hc \left(\frac{1}{360} + \frac{3}{360}\right) = \frac{4hc}{360} = \frac{hc}{90}$. Thus, the wavelength required to ionize the atom is 90 nm. Therefore, correct answer is 0090.