Confusion regarding Bohr Radius The bohr radius is derived by setting the centripetal force of electron to equal the coulomb force.
$$\frac{mv^2}{r}=\frac{kq_1q_2}{r^2}.$$
The electron is not a particle orbiting the electron according to quantum mechanics. Its position defined by its wave function. So, does the bohr radius make any sense quantum mechanically?
 A: In hindsight the Bohr model may seem obsolete,
because it was still partially based on classical mechanics
(the concept of well-defined electron orbits).
But at least it correctly predicted the essential features
of the hydrogen atom.
One of these features is the predicted radius of the orbits
$$r_n=a_0n^2 \quad \text{with }n=1,2,3,...  \tag{1}$$
where
$$a_0 = \frac{4\pi\epsilon_0\hbar^2}{e^2m_e}$$
is the Bohr radius.
This result is still qualitatively correct when
the hydrogen atom is treated with modern quantum
mechanics (based on Schrödinger's equation).
Here the wave functions $\psi_{nlm}(r,\theta,\phi)$
of the orbitals have an average radius
(see Expectation powers of $r$ for hydrogen)
$$\langle r\rangle_{nlm} = a_0 \frac{3n^2 - l(l+1)}{2} \tag{2}.$$
This is kind of a similar result as (1) from Bohr's model.
In both cases (1) and (2) the Bohr radius $a_0$
sets the order of magnitude for the radial sizes.
A: As OP correctly observes, Bohr's model is semiclassical. In the case of hydrogen, the potential is an attractive Coulomb potential
$$V(r)=-\frac{e^2}{r}\tag{1}\label{1}.$$
Solving the time-independent Schrödinger equation (TISE) for the bound states i.e.
$$-\frac{\hbar^2}{2m}\nabla^2\psi-\frac{e^2}{r}\psi=E\psi\qquad E<0\tag{2}\label{2}$$
yields the hydrogen eigenfunctions
$$\psi_{n\ell m}(\vec{r})=R_{n\ell}(r)Y^{m}_\ell(\hat{r}).\tag{3}\label{3}$$
The corresponding eigenenergies are labeled by $n$ and are exactly those in Bohr's model.
It turns out that the radial part in \eqref{3} contains a decreasing exponential factor multiplying a polynomial part, so the asymptotic behaviour is
$$R_{n\ell}(r)\sim e^{-\frac{r}{na}}\tag{4}\label{4}$$
where $a$ is Bohr's radius, which clearly appears as an exponential damping lenght-scale.
More specifically, in the case of the ground state $\psi_{100}$, modulo a normalization factor, \eqref{4} is exactly the radial part and it turns out that the quantum average of the radial distance in this state is proportional the Bohr radius.
$$\langle r\rangle_{100}=\frac{3}{2}a\tag{5}\label{5}$$
For a generic eigenstate the quantum average is just proportional to $a$,
$$\langle r\rangle_{nlm}\propto a\tag{6}\label{6}$$
the proportionality constants depending on the quantum numbers.
Finally, in the case of the ground state, the radial probability density
$$\rho(r)=r^2\lvert R_{10}(r)\rvert^2\tag{7}\label{7}$$
is maximum at $r=a$, i.e. it is the most probable value.
A: The following discussion is informed by the discussion by Michael Fowler titled the Bohr Atom I recommend it.

The concept of the Bohr atom hinges on assuming that the state of an electron bound to a hydrogen nucleus has a definable kinetic energy. A kinetic energy in the sense that there is opportunity for interconversion of potential energy and kinetic energy. A smaller radial distance of the electron with respect to the nucleus corresponds to lower potential.
As to kinetic energy: the assumption is one of correspondence. A correspondence between the kinetic energy that a particle would have on one hand, and the kinetic energy in terms of the de Broglie relationship on the other hand.

The Coulomb force and gravity are both inverse square laws, so I will use gravity as example.
Take the case of a planet in an slightly eccentric orbit around the Sun. From aphelion to perihelion the orbital altitude of the planet is decreasing, with corresponding increase in velocity, with the opposite process happening from perihelion to aphelion.
For a given force law - here an inverse square law - there is at every radial distance a corresponding rate of conversion between potential energy and kinetic energy.
In the case of an inverse square law: while it is not the case that for a given orbit the amount of kinetic energy is  equal to the amount of potential energy, it is the case that at each point in time the rate of change of kinetic energy is equal to the rate of change of potential energy.
In the case of orbiting motion:
-The derivative of the (Coulomb) potential with respect to radial distance recovers the Coulomb force
-The derivative of kinetic energy of the orbiting motion with respect to the radial distance results in the expression for the centripetal force.
Equating Coulomb force to the required centripetal force corresponds to identifying the state such that the rate of change of kinetic energy will match the rate of change of potential energy.

Micheal Fowler describes:
Bohr offered the hypothesis that the state of an electron bound to a hydrogen nucleus has a minimum energy in order for that bound state to exist at all. A candidate for that minimum energy is Planck's constant times the frequency of the electron's motion.
Micheal Fowler quotes Bohr:
Bohr said later: "As soon as I saw Balmer's formula, the whole thing was immediately clear to me."
The energy differences between the spectral lines of the Balmer series provide a wealth of information.
If the spectral lines correspond to energy states according to successive multiples of angular momentum, then the energy level of the state with lowest possible angular momentum can be inferred.
A: This is to be understood as a treatment before the current probabilistic model emerged.
This is still taught in schools and college to bridge the understanding, the syllabus is often laid out in a chronological order. Teaching the wrong treatment and later show its limitation and later bring out the newer models.
A: Yes.
The reduced Compton wavelength of the electron is:
$$ \bar{\lambda}_C = \frac{\hbar}{m_ec} $$
where the Compton wavelength is the quantum/relativistic limit on localizing and electron. Any more local, and there is enough $\Delta p$ to provide energy to make a positron/electron pair.
The Bohr radius is:
$$ a_0 =\bar{\lambda}_C / \alpha \approx 137\bar{\lambda}_C$$
where $\alpha$ represents the strength of QED. This mean at $Z\ge 137$, it gets weird.
tl;dr It's totally quantum.
A: After multiplying both side by $r$, one recognizes the non-relativistic virial theorem. It does by itself not fix the Bohr radius. https://en.wikipedia.org/wiki/Virial_theorem
