An electron rotates in a circle around a nucleus having positive charge $$Ze$$. Correct relation between total energy $$(E)$$ of electron to its potential energy $$(U)$$ is :

For an electron orbiting a nucleus with charge $$Ze$$:

The potential energy of the electron is given by $$U = -\frac{kZe^2}{r}$$.

Using the centripetal force condition $$\frac{mv^2}{r} = \frac{kZe^2}{r^2}$$, we find the kinetic energy as

$$ KE = \frac{1}{2}mv^2 = \frac{kZe^2}{2r} = -\frac{U}{2} $$

Hence the total energy is

$$ E = KE + U = -\frac{U}{2} + U = \frac{U}{2} $$

It follows that

$$ 2E = U $$

The correct answer is Option (2): \boxed{2E = U}.