According to Wikipedia, the Bohr radius (already described here) is:
a physical constant, approximately equal to the most probable distance between the nucleus and the electron in a hydrogen atom in its ground state.
However, I tried to calculate the mean value of the radius $\langle r\rangle_\psi$ between an electron and its proton in the ground state $1s$ (considering a particle in a spherically symmetric potential in the spherical coordinates $r, \theta, \varphi$). The wave function is hence given by:
\begin{equation}
\psi_{nlm} = Y_l^m(\theta, \varphi).R_{nl}(r)
\label{1}
\end{equation}
\begin{equation}
\Longrightarrow \psi_{1s0} = Y_0^0(\theta, \varphi).R_{10}(r) = \sqrt{\frac{1}{\pi}}.\left(\frac{Z}{a_0}\right)^{3/2}.\exp\left({\frac{-Zr}{a_0}}\right)
\label{2}
\end{equation}
Considering that the nucleus has a charge of $+e$, the formula of the mean distance of the electron from the proton is given by:
$$\langle r\rangle_\psi \equiv \langle\psi|r|\psi\rangle = \int_{\mathbb{R}^3} r|\psi(x)|^2 \ d\vec{r} = {\color{red}{\frac{3}{2}a_0 \neq a_0}}$$
How to explain that $\langle r\rangle_\psi \neq a_0$, which is by definition the most probable (so mean) value of $r$ ?
Best Answer
Assuming you are talking about the Hydrogen Ground State Radius, both your calculations and what Wikipedia stated are truth. Its average value is indeed $\frac{3}{2} a_0$ as you've calculated. The most probable value however is $a_0$ (see image below). I hope this clears up.