In the hydrogen spectrum, the longest wavelength in the Lyman series is 120 nm and the shortest wave — Atomic Structure Chemistry Question
Question
In the hydrogen spectrum, the longest wavelength in the Lyman series is 120 nm and the shortest wavelength in the Balmer series is 360 nm. From this data, the longest wavelength (in nm) of light that could ionize hydrogen atom is
💡 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.