The total energy of the electron in the orbit is — Atomic Structure Chemistry Question

💡 Solution & Explanation

Kinetic energy $K.E. = \frac{1}{2}mv^2 = \frac{n^4 h^4}{128 \pi^4 m^2 k}$. Potential energy $P.E. = -\frac{k}{r^4} = -k \left(\frac{nh}{4\pi\sqrt{km}}\right)^4 = -\frac{n^4 h^4}{256 \pi^4 m^2 k}$. Total Energy = $K.E. + P.E. = \frac{2n^4 h^4}{256 \pi^4 m^2 k} - \frac{n^4 h^4}{256 \pi^4 m^2 k} = \frac{n^4 h^4}{256 k \pi^4 m^2}$. Therefore, correct answer is C.