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 $r$ ?


1 Answer 1


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.

The Expectation Value for Radius
Hydrogen Ground State

  • $\begingroup$ so the most probable radius is not the mean radius ? Curious... $\endgroup$ May 31, 2022 at 5:58

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