The ratio of potential energy of electron in the third orbit of ion to the kinetic energy of electro — Atomic Structure Chemistry Question

💡 Solution & Explanation

P.E. in 3rd orbit of $\text{Li}^{2+}$ ($Z=3, n=3$): $P.E. = -27.2 \frac{3^2}{3^2} = -27.2$ eV. K.E. in 4th orbit of $\text{He}^+$ ($Z=2, n=4$): $K.E. = 13.6 \frac{2^2}{4^2} = 13.6 \times \frac{4}{16} = 3.4$ eV. Ratio = $\frac{-27.2}{3.4} = -8:1$. Therefore, correct answer is B.