If the angular momentum of an electron in an orbit is J then the kinetic energy of the electron in that orbit is,
A. \[\dfrac{{{J^2}}}{{2m{r^2}}} \\ \]
B. \[\dfrac{{Jv}}{r} \\ \]
C. \[\dfrac{{{J^2}}}{{2m}} \\ \]
D. \[\dfrac{{{J^2}}}{{2\pi }}\]

Hint:We deduce the linear speed of the electron in the orbit from the given expression for the angular momentum. When we get the linear speed we put in the expression for the kinetic energy to find the kinetic energy of the electron.

Formula used:
\[L = mvr\]
where L is the angular momentum of the particle of mass m in a circular orbit of radius r with linear speed v.
\[K = \dfrac{1}{2}m{v^2}\]
where K is the kinetic energy of the body of mass m moving with speed v.

Complete step by step solution:
When a particle is moving around a circular path then the angular momentum of the particle is the product of the moment of inertia about the axis of rotation and the angular speed.The electron is considered as a point mass.

It is given that the angular speed of an electron in an orbit is J. If the speed of the electron in the orbit is v, then using the formula of angular momentum
\[mvr = J\]
\[\Rightarrow v = \dfrac{J}{{mr}}\]
So, the speed of the electron in the orbit is \[\dfrac{J}{{mr}}\] here r is the radius of the orbit.

Using the kinetic energy formula,
\[K = \dfrac{1}{2}m{v^2} \\ \]
\[\Rightarrow K = \dfrac{1}{2}m{\left( {\dfrac{J}{{mr}}} \right)^2} \\ \]
\[\Rightarrow K = \dfrac{1}{2}m \times \dfrac{{{J^2}}}{{{m^2}{r^2}}} \\ \]
\[\therefore K = \dfrac{{{J^2}}}{{2m{r^2}}}\]
Hence, the kinetic energy of the electron is \[\dfrac{{{J^2}}}{{2m{r^2}}}\].

Therefore, the correct option is A.

Note: Though the speed of the electron in the orbit is very high, we don’t consider the relativistic case to find the momentum or the kinetic energy because the energy of the electron is less than the threshold energy for the relativistic motion.