Difference between average position of electron and average separation

[happyparticle]

Homework Statement

Difference between average position of electron and average separation

Relevant Equations

$\langle r \rangle = \int_0^{\infty} \int_0^{\pi} \int_0^{2 \pi} \Psi^* \Psi r^2 sin \theta dr d\theta d \phi$

Hi,
I asked this question elsewhere, but I didn't understand the answer. It seems to be easy to understand, but for some reason I'm really confuse.
I'm not sure how to find the average position of an electron and the average separation of an electron and his proton in a hydrogen atom.
To be precise, I can find one but I'm not sure if this is the position or the distance between the electron and the proton.I thought that the average position of an electron was $\langle r \rangle = \int_0^{\infty} \int_0^{\pi} \int_0^{2 \pi} \Psi^* \Psi r^2 sin \theta dr d\theta d \phi$.
Where $\Psi = R(r)Y(\theta, \phi)$
However, after looking on the internet it seems like this is the average distance between de proton and the electron (the expected value of r). I thought it was the position of the electron since we cover all the space from 0 to $\infty$ and over all the angles.
If the above is the average separation between the electron and the proton what is the average position of the electron?

 

[PeroK]

If the above is the average separation between the electron and the proton what is the average position of the electron?

That depends on where the proton is.

 

[happyparticle]

I misread
The proton is at the origin.

 

[happyparticle]

The average value of$R(r)$ seems to be the average distance between the proton and electron, but I may be wrong.

 

[FactChecker]

I don't know if this makes a difference to you, but the proton stays within the nucleus, which means that its position can only change about one 10,000th of the size of the atom and its electron orbits. So the difference between the two interpretations that you are wondering about is relatively tiny.

 

[hutchphd]

Homework Statement:: Difference between average position of electron and average separation
Relevant Equations:: $\langle r \rangle = \int_0^{\infty} \int_0^{\pi} \int_0^{2 \pi} \Psi^* \Psi r^2 sin \theta dr d\theta d \phi$

I thought that the average position of an electron was ⟨r⟩=∫0∞∫0π∫02πΨ∗Ψr2sinθdrdθdϕ.

What you have written down is not <r>
EDIT: OK I see you do recognize the need for the r operator. Never Mind

 

[PeroK]

The average value of$R(r)$ seems to be the average distance between the proton and electron, but I may be wrong.

That's explicitly what it is. The atom itself would have a spatial wavefunction that is approximately the position of the nucleus, which is much more massive than the electron.

 

[bob012345]

Homework Statement:: Difference between average position of electron and average separation
Relevant Equations:: $\langle r \rangle = \int_0^{\infty} \int_0^{\pi} \int_0^{2 \pi} \Psi^* \Psi r^2 sin \theta dr d\theta d \phi$

Hi,
I asked this question elsewhere, but I didn't understand the answer. It seems to be easy to understand, but for some reason I'm really confuse.
I'm not sure how to find the average position of an electron and the average separation of an electron and his proton in a hydrogen atom.
To be precise, I can find one but I'm not sure if this is the position or the distance between the electron and the proton.I thought that the average position of an electron was $\langle r \rangle = \int_0^{\infty} \int_0^{\pi} \int_0^{2 \pi} \Psi^* \Psi r^2 sin \theta dr d\theta d \phi$.
Where $\Psi = R(r)Y(\theta, \phi)$
However, after looking on the internet it seems like this is the average distance between de proton and the electron (the expected value of r). I thought it was the position of the electron since we cover all the space from 0 to $\infty$ and over all the angles.
If the above is the average separation between the electron and the proton what is the average position of the electron?

Depending on the state of the electron there is the peak of the radial probability density curve and there is the expectation value of r. For a 1s electron in a Hydrogen atom, the peak of the radial probability density is 1 Bohr radius $a_0$ which is the most probable position but the expectation value $<r>= \large\frac{3a_0}{2}$ which is the average position. Is this what you mean?

http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/hydr.html

 

[songanh]

When you say "average position of electron", you have to specify which reference point that you're referring to, i.e. choose your origin. Otherwise, it doesn't make sense to talk about position of something.
In this case, the most natural choice is to make the position of the proton to be the reference point. So if we are more careful with the wording, $\braket{r}$ is "the average position of electron about the proton".
But you see, when you say that, it is the same thing as "the separation between proton and electron".
So in this context, those two things are, approximately, the same.
P.S. I believe your expression for $\braket{r}$ is missing an r

 

[hutchphd]

Homework Statement:: Difference between average position of electron and average separation
Relevant Equations:: $\langle r \rangle = \int_0^{\infty} \int_0^{\pi} \int_0^{2 \pi} \Psi^* \Psi r^2 sin \theta dr d\theta d \phi$

I thought that the average position of an electron was ⟨r⟩=∫0∞∫0π∫02πΨ∗Ψr2sinθdrdθdϕ.

This is a bad mistake. Keep track of units
|ψ|² has units of 1/volume in 3D space.
Always look at the units. Always.

 

[mjc123]

Am I understanding it differently from everyone else, but I assume "position" refers to a specific point in space with coordinates (x,y,z) or (r,θ,φ). If you consider a spherical shell of radius r centred on the nucleus, every point on the surface of the sphere corresponds to a separation r, but each point is a distinct "position". By the symmetry, for every position (x,y,z), the diametrically opposite position (-x, -y, -z) has equal probability density. Therefore the average position is (0,0,0), i.e. at the nucleus. But that does not mean that the average separation is 0.

 

[bob012345]

Am I understanding it differently from everyone else, but I assume "position" refers to a specific point in space with coordinates (x,y,z) or (r,θ,φ). If you consider a spherical shell of radius r centred on the nucleus, every point on the surface of the sphere corresponds to a separation r, but each point is a distinct "position". By the symmetry, for every position (x,y,z), the diametrically opposite position (-x, -y, -z) has equal probability density. Therefore the average position is (0,0,0), i.e. at the nucleus. But that does not mean that the average separation is 0.

Well, that's kind of like saying the average position (location) of the human race is at the center of the earth. Perhaps there is some small mathematical truth to it but it's misleading. That's probably not what was meant by this problem.

Edit: Or maybe it is as @hutchphd suggests in post #13.

 

[hutchphd]

Therefore the average position is (0,0,0), i.e. at the nucleus. But that does not mean that the average separation is 0.

I think this is a very good answer and perhaps the intent of the question.

 

[bob012345]

I think this is a very good answer and perhaps the intent of the question.

Perhaps, but we don't have the exact wording of the problem.

 

[hutchphd]

We have what we have.