The binding energy of an electron in the ground state of hydrogen-like ions in whose spectrum, the t — Atomic Structure Chemistry Question

💡 Solution & Explanation

Third line of Balmer series is $5 \to 2$. $\frac{1}{\lambda} = R Z^2 \left(\frac{1}{4} - \frac{1}{25}\right) = R Z^2 \left(\frac{21}{100}\right)$. Plugging $\lambda = 108.5 \times 10^{-9}$ m and $R = 1.097 \times 10^7$, we find $Z^2 \approx 4$, so $Z=2$ (He$^+$). The binding energy of the ground state is $13.6 \times Z^2 = 13.6 \times 4 = 54.4$ eV. Therefore, correct answer is B.