The ratio of potential energy of electron in the third orbit of Li^2+ ion to the kinetic energy of e — Atomic Structure Chemistry Question

💡 Solution & Explanation

Potential Energy is given by P.E. = -2 * 13.6 * Z^2 / n^2 eV. For Li^2+ (Z = 3) in the third orbit (n = 3): P.E. = -2 * 13.6 * 3^2 / 3^2 = -27.2 eV. Kinetic Energy is given by K.E. = 13.6 * Z^2 / n^2 eV. For He^+ (Z = 2) in the fourth orbit (n = 4): K.E. = 13.6 * 2^2 / 4^2 = 3.4 eV. The ratio P.E. / K.E. = -27.2 / 3.4 = -8:1.