If m is mass of electron, v its velocity, r the radius of stationary circular orbit around a nucleus with charge Ze, then from Bohr's first postulate, the kinetic energy $K = \frac{1}{2}mv^2$ of the electron in C.G.S. system is equal to
Answer & explanation
Correct answer: option 1
In the revolution of electron, coulomb force provides the necessary centripetal force.
$\Rightarrow \frac{z e^2}{r^2}=\frac{m v^2}{r} \Rightarrow m v^2=\frac{z e^2}{r}$
∴ K.E. = $\frac{1}{2} m v^2=\frac{z e^2}{2 r}$