# The average distance of a 1s hydrogen electron

In the Bohr atom, the Bohr distance, $a_0$, is associated with the average distance of the electron from the nucleus in the ground state of the hydrogen atom. Is this what quantum mechanics also predicts?

From Schrödinger’s wave equation, $\psi$ is called the wave function and its square, $\psi^2$, is properly considered to be a joint probability density function. This is a term used in multivariable calculus courses to represent a probability distribution function over multiple variables (e.g. the Cartesian spacial variables: x, y and z).

We can use the idea of this joint density function to calculate a weighted average analogous to the way you would calculate the average return for a series of gambles with probability, $x_i$, of winning and payouts, $p_i$. This calculation would take the form: return = $\sum_{i=0}^N{x_ip_i}$. Similarly, when we think of calculating a weighted average distance the electron could be found from the center of the nucleus over a region of space, V, we can use the joint probability density function, the square of the wave function multiplied by the distance for each probability element: $\int_V{r\cdot\psi^2dV}$. This recognizes that the integral represents the probability of finding an electron within a particular volume of space. If we write the differential volume element, $dV$, in spherical coordinates, and we integrate over all r, we get: $\int_0^{\pi}\int_0^{2\pi}\int_0^R{r^2}\cdot\sin(\phi)\cdot{r}\cdot\psi^2drd{\theta}d{\phi}$

We can snag the normalized Schrödinger wave function solutions for a hydrogen electron from this website.

And check to see if the wave function is properly defined. That is, $1 = \int_0^{\pi}\int_0^{2\pi}\int_0^R{r^2}\cdot\sin{\phi}\cdot\psi^2drd{\theta}d{\phi}$

Notice that we could have used integration by parts to do this without wolframalpha or as we did in the Calculus of Friendship post where we used differentiation under the integral sign to find this integral exactly $\int_0^\infty{t^n{e^{-at}}dt} = n!(\frac{1}{a})^{n+1}$. Next, we can compute the average distance of the 1s electron:

A value of $\frac{3}{2}a_0$ is in agreement with the value obtained in “Modern physics”, by Serway, Moses, Moyer, Example 8.9, p284. Interestingly, this is larger than the Bohr radius itself, but of the same order of magnitude.

As I tried to extend this sort of calculation to promoted electrons, I had a harder time getting wolframalpha to numerically integrate the equation. This sort of calculation could be used to determine the expected distance required to promote an excited electron. I still haven’t figured out what held up the calculation but it could be due to a problem with round off error with such a small Bohr radius. It would be especially interesting to do this calculation though to compare the classical Coulombic energy difference between the energy levels to the spectral lines observed for the hydrogen atom.

I hope to follow this post up with such a calculation once I am able properly calculate the average distance of the 2s electron.