In the context of an electron orbiting around a nucleus with a positive charge of $$\mathrm{Ze}$$, we are dealing with classical physics approximations and the electrostatic force between the electron and the nucleus. In such a setup, the electron's potential energy (U) is due to electrostatic interaction, and it is given by Coulomb's law:

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

Where:

• $U$ is the potential energy of the electron,
• $k$ is Coulomb's constant,
• $Z$ is the atomic number (number of protons in the nucleus),
• $e$ is the charge of an electron, and
• $r$ is the radius of the orbit of the electron around the nucleus.

The negative sign indicates that the potential energy is negative because the electron and nucleus attract each other.

The total energy (E) of the electron in orbit is the sum of its kinetic energy (K) and its potential energy (U). Since the electron is in a stable orbit, its kinetic energy can be shown to be exactly half the magnitude of its potential energy but positive:

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

Therefore,

$$E = K + U = -\frac{1}{2}U + U = \frac{1}{2}U$$

To find a relation between total energy (E) and potential energy (U), we rearrange the equation as follows:

$$2E = U$$

This is to say, the total energy (E) is half the magnitude of potential energy (U) but negative, and the correct relationship between them, when looking for a positive proportionality, yields to $2E = U$. Hence, the correct option is:

Option C: $$2 \mathrm{E} = \mathrm{U}$$