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

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.

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?

• It will depend on the hidden context of the question. If the question is "what is the most probable radius for finding the electron" then you will get a different answer as there is zero probability of the electron being at zero radius - the elemental ring (2d) or elemental shell (3d) has no space at zero radius -the wonders of mathematical imprecision. Further, it gets much worse as the dimensionality increase - everything tends to be at the 'one sigma' radius in high dimensional spaces. – Philip Oakley Oct 25 '17 at 9:52
• The alternate context is that the professor wants to be clear that the electron is 'tightly' bound to the atom. It is always (on average) right there, aligned with the atom. They are together as one. A comparative statistic is say WWII bomb (c.f. electron) aiming - was the bomb cluster on target, and what was the spread of the bomb cluster. The electron is on target, and has spread - they are distinct measures. – Philip Oakley Oct 25 '17 at 10:10
• The most probable point vs most probable distance are two different things – Greg Oct 25 '17 at 11:01

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.

• +1! The claim from the title of the question simply makes no sense at all: the probability of finding an electron at a certaint point is effectively zero. The position of a particle is described by a 3-tuple of real numbers, the set of reals is uncountable, and the probability that a random variable takes a certain value from an uncountable set is 0. – Wildcat Oct 25 '17 at 16:13

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).

• So whats the significance of the graph of the radial distribution function? And isn't it true that when r tends to zero, we get the Origin? And what's the physical significance of the different conclusions we get with probability density and probability? – Anjan Oct 25 '17 at 7:53
• The radial distribution is usually the interesting part since it actually describes the probability of finding the electron somewhere at a certain distance around the nucleus, which is a useful information for the chemist. But from a physics view there's actually no difference, it's just two ways of representing the same thing. And yes, r=0 is the origin which we usually set right at the position o the core. – DSVA Oct 25 '17 at 8:03
• So, can we a take a volume element of radius dr and consider the probability of the electron at the origin? – Anjan Oct 25 '17 at 16:41

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.

• The professor takes a small cubical volume for the origin. I don't know why? – Anjan Oct 25 '17 at 11:35
• That's because to get a probability from the wavefunction you have to use a non-zero volume. The probability of the electron being in a volume of zero size is, naturally, zero. Are you sure the prof is using a cubical volume ? Do you mean spherical, which would be more natural, as the wavefunction is in spherical coordinates ? – StephenG Oct 25 '17 at 12:57
• Just to explain his statement regarding the most probable point at origin he takes a small cubical volume element at the origin. – Anjan Oct 25 '17 at 13:50