The angular momentum of electron in Bohr's orbit is J. What will be the K.E. of electron in that Boh — Atomic Structure Chemistry Question

💡 Solution & Explanation

Angular momentum $J = mvr$, which implies $m = \frac{J}{vr}$. Kinetic energy is $K.E. = \frac{1}{2}mv^2 = \frac{1}{2}\left(\frac{J}{vr}\right)v^2 = \frac{1}{2}\frac{Jv}{r}$. Therefore, correct answer is A.