Most probable point for finding an electron in the 1s orbital of a Hydrogen atom

Question

My professor says that the most probable point for finding an electron in a 1s orbital of a hydrogen atom is at its origin.

He explains this by citing the fact that the square of the wave function which gives the probability density is maximum at the origin.

At the same time, we all agree that the Bohr radius is the distance at which probability of finding the electron is maximum for 1s orbital.

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I can understand the reasoning he gave about the origin being the highest probable point for finding an electron, but then why are the graphs of the radial distribution function so?

Answer 1

I provide here a few supplementary comments; the major part of your question has been answered by DSVA.

• For the electron's position in a 1s orbital, the probability density $\rho(r)$ is maximal at the origin, whereas the radial probability density $p(r)$ is maximal at the Bohr radius.

• probability versus probability density: The probability density integrated over a volume yields a probability. We cannot talk about probabilities at a point for a continuous system because a point has measure zero---that is, if we make a measurement of a random variable for a continuous system, we will never get exactly the value represented by the point. Instead, we need to consider a higher-dimensional object, which has non-negligible measure. A natural choice is a shell centered around the origin with radius $R$ and thickness $\Delta R$, assumed small. We can find the probability that the electron is located within the shell as $$P(R < r < R+\Delta R) = \int_\text{shell}\mathrm{d}\mathbf{r}\,\rho(r) = \int_R^{R+\Delta R}\mathrm{d}r\,4\pi r^2\rho(r)\approx 4\pi R^2 \rho(R)\,\Delta R.$$ Motivated by this calculation, we define the radial probability density $p(r) := 4\pi r^2\rho(r)$, which is a more important measure of where the electron actually is, because of the radial symmetry in the system---we are more interested in the distance between the electron and the origin than in the position vector of the electron.

• probability density versus radial probability density: The probability density takes as its fundamental unit a point---imagine a cube of infinitesimal volume, as your professor uses. The radial probability density takes as its fundamental unit a sphere---imagine a shell of infinitesimal width, as before. We will restate the contents of the first bullet point in this context: The probability of finding the electron in a given infinitesimal cube is maximized for a cube centered about the origin. The probability of finding the electron in a given infinitesimal shell is maximized for a shell at the Bohr radius. In the latter case, although the probability density is lowered for large $r$, the increased surface area means that there is more space we can hope to find the electron in, and the competition of these two factors result in a maximum for the radial probability density at the Bohr radius.

Answer 2

It makes a difference if you look in a point or in a volume element.

For the radial distribution we sum up the probabilities within (very thin) shells at different distances from the core. If the radius get's bigger the volume in this shell gets bigger. Remember that the surface of a sphere goes by $4 \cdot \pi \cdot r^2$ so the volume of the shell is essentially for all practical purposes $4 \cdot \pi \cdot r^2 \cdot \Delta r$. At the core the radius is zero which means the volume is zero and thus the probability is zero too.

At the same time the probability of finding the electron in a point at a certain distance goes down, but the volume goes up. Because of the scaling of the volume and the "decay" of the probability to find it in a point further away it turns out that there's a maximum of finding the electron at a certain distance away from the core (but not a single point at that distance).

Answer 3

He explains this by citing the fact that the square of the wave function which gives the probability density is maximum at the origin.

Not exactly.

The answer by @dsva explains why this is wrong and I'll just expand that in a minute, but first note that it's easy to see why this is wrong.

If the electron was most likely to be at the origin it would be likely to interact with the nucleus, something we expect not to happen. In fact there is a small chance of this because the nucleus is a non-zero size.

Essentially he's missing the need to sum over the shell, and the shell's volume is zero at the origin because of the volume element $dV = 4\pi r^2dr$ with $r=0$.

The probability of finding the electron in a region of zero radial size is zero, but we can evaluate the relative probability for two radii. For 1-s orbitals that is :

$$\frac {r_1^2 e^{-{r_1}/a}}{r_1^2 e^{-{r_1}/a}+r_2^2 e^{-{r_2}/a}}$$

and

$$\frac {r_2^2 e^{-{r_2}/a}}{r_1^2 e^{-{r_1}/a}+r_2^2 e^{-{r_2}/a}}$$

For $r_1=0$ we clearly get a relative probability of zero compared to any $r_2\neq 0$.

At the same time, we all agree that the Bohr radius is the distance at which probability of finding the electron is maximum for 1s orbital.

This is because of what an expectation value is and how it is calculated in QM.

An expectation value is an average over the entire space of the effect of an operator on a wavefunction. It is not simply the operator multiplied by the probability density.

In short an expectation value can be considered an average over all space, whereas the probability density is not an average over all space.