The kinetic energy of an electron in the second Bohr orbit of a hydrogen atom is ( is the Bohr radiu — Atomic Structure Chemistry Question

💡 Solution & Explanation

Velocity $v_n = \frac{nh}{2\pi m r_n}$. For $n=2$, $r_2 = 4a_o$, so $v_2 = \frac{2h}{2\pi m (4a_o)} = \frac{h}{4\pi m a_o}$. Kinetic energy $K.E. = \frac{1}{2} m v_2^2 = \frac{1}{2} m \left(\frac{h}{4\pi m a_o}\right)^2 = \frac{h^2}{32\pi^2 m a_o^2}$. Therefore, correct answer is C.