If $$m$$ and $$e$$ are the mass and charge of the revolving electron in the orbit of radius $$r$$ for hydrogen atom, the total energy of the revolving electron will be:

To find the total energy of the electron revolving in a hydrogen atom, we consider the Bohr model where the electron moves in a circular orbit of radius $$r$$ around the proton. The electron has mass $$m$$ and charge $$e$$, while the proton has charge $$+e$$. The electrostatic force of attraction provides the necessary centripetal force for circular motion.

The electrostatic force between the electron and proton is given by Coulomb's law. Since the charges are opposite, the magnitude is $$\frac{1}{4\pi\epsilon_0} \frac{e \cdot e}{r^2} = \frac{1}{4\pi\epsilon_0} \frac{e^2}{r^2}$$. This force acts as the centripetal force, which is $$\frac{mv^2}{r}$$, where $$v$$ is the orbital speed of the electron.

Setting these equal:

$$\frac{mv^2}{r} = \frac{1}{4\pi\epsilon_0} \frac{e^2}{r^2}$$

Multiply both sides by $$r$$ to simplify:

$$mv^2 = \frac{1}{4\pi\epsilon_0} \frac{e^2}{r}$$

Solve for $$v^2$$:

$$v^2 = \frac{1}{4\pi\epsilon_0} \frac{e^2}{m r}$$

The kinetic energy $$K$$ of the electron is $$\frac{1}{2} m v^2$$. Substitute the expression for $$v^2$$:

$$K = \frac{1}{2} m \left( \frac{1}{4\pi\epsilon_0} \frac{e^2}{m r} \right)$$

The $$m$$ cancels out:

$$K = \frac{1}{2} \cdot \frac{1}{4\pi\epsilon_0} \frac{e^2}{r}$$

The potential energy $$U$$ for the electrostatic attraction is negative and given by $$-\frac{1}{4\pi\epsilon_0} \frac{e^2}{r}$$.

The total energy $$E$$ is the sum of kinetic and potential energy:

$$E = K + U = \left( \frac{1}{2} \cdot \frac{1}{4\pi\epsilon_0} \frac{e^2}{r} \right) + \left( -\frac{1}{4\pi\epsilon_0} \frac{e^2}{r} \right)$$

Factor out $$\frac{1}{4\pi\epsilon_0} \frac{e^2}{r}$$:

$$E = \frac{1}{4\pi\epsilon_0} \frac{e^2}{r} \left( \frac{1}{2} - 1 \right)$$

Simplify the term in parentheses:

$$\frac{1}{2} - 1 = -\frac{1}{2}$$

So:

$$E = \frac{1}{4\pi\epsilon_0} \frac{e^2}{r} \times \left( -\frac{1}{2} \right) = -\frac{1}{2} \cdot \frac{1}{4\pi\epsilon_0} \frac{e^2}{r}$$

In atomic physics, it is common to use units where the constant $$\frac{1}{4\pi\epsilon_0}$$ is set to 1 (as in cgs units). Thus, the expression simplifies to:

$$E = -\frac{1}{2} \frac{e^2}{r}$$

Comparing with the options, this matches option D.

Hence, the correct answer is Option D.