In QM, a particle is described by its wavefunction. Each state of the hydrogen atom has a particular wavefunction and a particular energy. Since you can count the number of different wavefunction, the energy is quantized.

It is important to understand that the wavefunction does not tell you where the electron is, it can only tell the probability to find the electron in a certain region. Thus, if you try to locate the electron, you will always get different results. But, if you measure the average distance to the nucleus, your result will approach the bohr radius.
Because of this, it looks like the electron is present everywhere in space because the wavefunction is defined everywhere.

The wavefunction also contains information about the energy of the system. If the electron is in the ground state, its energy will be definite (the Rydberg energy). Since you do not know where the electron is exactly, you cannot compute its energy classicaly. If you try to measure the location of the particle and compute the energy from the position you get, you will not know its kinetic energy because of the Heisenberg uncertainty principle ($\Delta x\Delta p \geq\hbar$). Indeed, if you locate the particle, the uncertainty of its position will be small, but because of the principle, there will be a great uncertainty on its momentum and thus, its kinetic energy. It is possible to measure the energy of a particle with great precision, but because of this same principle, you won't know exactly where the particle is.

So, to answer your question, the electron's energy is quantized in the hydrogen atom. But, you cannot know the position of the particle and its energy with great precision at the same time.

That 13.6eV is the **difference** in the energy of a bound electron as compared to an electron at rest far from the proton. So it is an energy difference. We don't include the rest mass because that's a constant so it's the same for both the bound and free electron. When we take the difference in the energies the rest mass cancels out.

You may be interested to know the mass of a hydrogen atom is actually less than the mass of an isolated proton + the mass of an isolated electron. The difference in the mass is 13.6eV. That's because to make a hydrogen atom we have to start with an isolated proton and an isolated electron then remove 13.6eV to make them stick together.

## Best Answer

Concerning the factor $\frac{1}{2}$: It seems that OP in his classical reasoning only accounted for the Coulomb potential energy

$$\tag{1}\langle U\rangle ~=~-k_e e^2 \langle \frac{1}{r} \rangle ~=~-\frac{k_e e^2}{a_0} ~<~0.$$

Here $k_e$ is Coulomb's constant and $a_0$ is the Bohr radius.$^1$

However we should also take the kinetic energy $\langle T\rangle>0$ into account! We know from the virial theorem that the kinetic energy

$$\tag{2}\langle T\rangle~=~-\frac{1}{2}\langle U\rangle~>~0$$

is minus half the potential energy for the $1/r^2$ Coulomb force.

Hence the total energy becomes half the Coulomb potential energy:

$$\tag{3} E ~=~ \langle T\rangle+ \langle U\rangle ~=~-\langle T\rangle ~=~\frac{1}{2}\langle U\rangle~<~0,$$

which is (up to sign conventions) the Rydberg energy.

--

$^1$ Here OP's estimate is helped by the fact that the expectation value

$$\tag{4}\langle\frac{1}{r}\rangle ~=~\frac{1}{a_0}$$

in the ground state happens to be the inverse Bohr radius

withoutany non-trivial dimensionless number appearing in eq. (4)! [Note for comparison, that e.g. $\langle r\rangle =\frac{3}{2}a_0$.]