# Bohr radius: does the mean distance between the proton et the electron in $^1H$ equals to $a_0$ or $\frac{3}{2}a_0$?

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:

$$$$\psi_{nlm} = Y_l^m(\theta, \varphi).R_{nl}(r) \label{1}$$$$

$$$$\Longrightarrow \psi_{1s0} = Y_0^0(\theta, \varphi).R_{10}(r) = \sqrt{\frac{1}{\pi}}.\left(\frac{Z}{a_0}\right)^{3/2}.\exp\left({\frac{-Zr}{a_0}}\right) \label{2}$$$$ Considering that the nucleus has a charge of $$+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$$ ?

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.