Most probable electron location in Hydrogen ground state

[jaydnul]

From the ground state wave equation the most probable outcome of anyone measurement will be in the center of the atom, at the nucleus. The expectation value is found to be the Bohr radius.

So does this mean that if you measure the position of an electron in a hydrogen atom in the ground state it will most likely annihilate itself?

 

[Orodruin]

No, why would it? The electron is not an anti-proton.

 

[Dale]

No. Protons and electrons are not antiparticles.

EDIT: scooped by Orodruin!

 

[jaydnul]

So what happens?

 

[Orodruin]

Nothing really. There is not enough energy for electron capture. The state where the electron is found within the nucleus is not an energy eigenstate and the atom may be ionized or end up in a different energy level.

No. Protons and electrons are not antiparticles.

EDIT: scooped by Orodruin!

I'm two for two tonight - I should probably quit while I am ahead.

 

[jaydnul]

I get it. The only way quarks and electrons interact is through EM, right? (And Gravity)

So can it be in the space between two quarks?

 

[Nugatory]

So can [the electron] be in the space between two quarks?

That question makes sense only if you have also have definite positions for the quarks - but Orodruin's comment above about position eigenstates not being bound energy eigenstates applies to the quarks as well. A state in which the positions of the electron and the quarks (three of them, in the most common hydrogen case) are all known is not a state in which you have a hydrogen atom.

You are, I think, trying to think of a particle with a known position as if it were a little tiny ball located at that position. That's not going to work, because that's just not what particles are.

 

[jaydnul]

Sorry to resurrect this thread but I think I was calculating the probability density wrong. The most probable location for one measurement would be the Bohr radius, not the nucleus, right?

Or is my first post right? The most probable outcome of ONE measurement is the nucleus, and the expectation value (average) is the Bohr radius.

 

[Orodruin]

This would depend on how you define "most probable". The nucleus has the highest probability density, but the Bohr radius has the highest probability per radial distance.

 

[jaydnul]

I'm a little confused. If the nucleus has the highest probability density, then wouldn't the radius of the nucleus (as opposed to the Bohr radius) have the highest probability per radial distance?

 

[mfb]

The highest probability density is calculated "per volume", while the other value is "per radial distance".
You get much more volume per radial distance if the radius is larger (it is growing like the surface of a sphere, with the radius squared).

There is much more volume between 0.95 and 1.05 times the Bohr radius than between 0 and 0.1 times the Bohr radius.

 

[jaydnul]

Oh, OK. So if you added up all the infinitesimal probability densities per volume at different radial intervals, the Bohr radius has the highest magnitude compared to other radial intervals.

But nature doesn't know we think in "radial distances". What I mean is nature doesn't care that two volume elements are at the same radial distance, it just cares about each individual volume element. So really, if we actually do the measurement, the nucleus we be the most probable, yes?

 

[Orodruin]

Oh, OK. So if you added up all the infinitesimal probability densities per volume at different radial intervals, the Bohr radius has the highest magnitude compared to other radial intervals.

Correct.

But nature doesn't know we think in "radial distances". What I mean is nature doesn't care that two volume elements are at the same radial distance, it just cares about each individual volume element. So really, if we actually do the measurement, the nucleus we be the most probable, yes?

If by "most probable" you mean the highest probability per volume, then yes.

 

[jaydnul]

Correct.
If by "most probable" you mean the highest probability per volume, then yes.

If the experiment was done and you had to place a bet on where you think the electron would be found, would you put your money on the nucleus or Bohr radius?

 

[Orodruin]

That would depend on whether the bet is for a fixed volume or a fixed radial interval.

Edit: And on the odds I am given ...

 

[jaydnul]

Sorry if this is excruciating but it's just not clicking.

You're saying that if I asked you to bet on the radius the electron will be found, you would say the Bohr radius. But if I asked you to bet on the volume the electron will be at, you would say the nucleus? That would mean our betting conditions will effect the outcome of the experiment.

 

[jtbell]

Let's try it this way.

Divide up the space in the vicinity of the atom into a lot of small equal-size volume elements. Make them all cubes, if you like.

The probability of finding the electron in a volume element that is located somewhere at the Bohr radius is smaller than the probability of finding it in the volume element that is located at r=0.

However, there are many more of those volume elements at the Bohr radius (located at different directions from the nucleus, but the same distance) than there are at the nucleus (where there is precisely one such volume element).

Therefore it turns out that the total probability of finding the electron at the Bohr radius is larger than the probability of finding it at r=0.

 

[jaydnul]

Ahh, completely understand now. Thanks

One more question. When you say that finding the electron at the nucleus is not an energy eigenstate, what do you mean?

When in the ground state (which has the energy value 13.6 eV), there is a probability of finding the electron at the nucleus. So why wouldn't that be an energy eigenstate?

 

[Orodruin]

If you measure the electron to be within the nucleus, then it is no longer in the ground state. Your measurement has affected the system and the electron is now in whatever linear combination of states that make up a wave function where the electron is inside the nucleus. As this wave function evolves in time, the electron probability density will spread accordingly.

 

[USeptim]

The electron wave function tells that it's mostly inside the Bohr radius and it's also at the nucleus with an small overall probability but with a high density of probability. However the electron wave function exists at both places at the same time (and also it spreads out of the Bohr radius).

In order to find the electron mostly inside the radius you would need a high energy interaction and that would probably free the electron.

In order to have a particle in a definite position it would need infinite energy and since the nucleus is not a point but a very small region, you would have to transfer a lot of energy to the electron.

 

[jaydnul]

K, thanks guys.