Is the Bohr radius deprecated?
Solution 1
Perhaps a more recent theory regarding the atomic radius that you might be interested in is Schrödinger's quantum mechanical model. The wave function, represented by $\psi$, is pretty useful in judging the probability of finding an electron at any particular point.
When the wavefunction, $\psi$, is squared the result is a number that is directly proportional to the probability of finding and electron at specific coordinate in 3D space. The radial portion of the wavefunction really only tells us if there is high or low probability at various distances from the nucleus (possible radii for the electrons). Multiplying this probability by the area available at that distance will give us the Radial Distribution Function for the given electron. The concentric spherical shells have areas equal to the surface area of a sphere which is $4\pi r^2$.
So, essentially the radial probability distribution function is $\psi^2\cdot 4\pi r^2$
Since your question is regarding a hydrogen atom, we can look at the wave function of the $1s$ orbital.
$$\psi_{1s}=\frac{1}{\sqrt{\pi}a_{0}^\frac32}\cdot e^{\frac{r}{a_{0}}}$$
So the radial distribution function is
$$f(r) = \frac{1}{\pi a_{0}^3}\cdot e^{\frac{2r}{a_{0}}}\cdot 4\pi r^2$$
Here, $a_{0}$ is the Bohr radius, and $r$ is the distance from the nucleus.
This represents the probability of finding an electron at a given distance $r$. So, on differentiating the function with respect to $r$, and equating it to zero, and then solving for $r$, you get the distance at which $f(r)$ is maximum, which turns out to be equal to $a_{0}$, the Bohr radius.
So according to Schrödinger's theory, the Bohr radius that was estimated in Bohr's model is actually the distance at which the radial probability distribution function is maximum. And that's what the 'atomic radius' was defined as later. (Note, however, that according to Schrödinger, the electron can actually be found anywhere between $r=0$ and $r=\infty$.)
Here's a small extract from wikipedia:
It turns out that this is a maximum at $r=a_{0}$. That is, the Bohr picture of an electron orbiting the nucleus at radius $a_{0}$ is recovered as a statistically valid result.
Ref.: Radial distribution (utexas)
Solution 2
In a word: No. The Bohr radius is a key concept and it is not deprecated.
In the modern outlook, the Bohr radius is the length unit of the atomic system of units, i.e., it is the natural length scale that comes out as a combination of the reduced Planck constant $\hbar$, the electrostatic interaction constant $\frac{e^2}{4\pi\epsilon_0}$, and the electron mass $m_e$. When doing atomic and molecular physics as well as quantum chemistry, all calculations are done in multiples of the Bohr radius.
And yes, it does not have a simple definition as "the radius of the groundstate orbit" that it does in the (deprecated) Bohr model, but that does not mean it is not useful.
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Derek Seabrooke
Updated on June 21, 2020Comments

Derek Seabrooke over 2 years
The Bohr radius ($a_0$ or $r_{\text{Bohr}}$) is a physical constant, equal to the most probable distance between the nucleus and the electron in a hydrogen atom in its ground state. It is named after Niels Bohr, due to its role in the Bohr model of an atom. Its value is $5.29177210903(80)×10^{−11}\ \mathrm m$.
Source: https://en.wikipedia.org/wiki/Bohr_radius
The Bohr model itself is generally considered to be deprecated. Is this still considered to be a more or less accurate approximation of the radius of a hydrogen atom? If not, what is the current thought?

Derek Seabrooke over 2 yearsThanks for the answer. I am not calling it useless. What I really want to know is, if the Bohr radius is not the radius of the groundstate orbit what is the radius of the groundstate orbit to the best of your reckoning?

Cosmas Zachos over 2 years@Derek Seabrooke . There is no "orbit". The ground state is a spherically symmetric orbital, with vanishing angular momentum, and the Bohr radius is some sort of an average size thereof, as Nikhil reviews. There are technical arguments mapping Bohr's picture to the proper modern one, but they are too technical to shed unqualified light on the issue.

Derek Seabrooke over 1 yearIs the e in the wave function the Euler constant rather than elementary charge?