# Electron orbital velocity in hydrogen atom?

1. Jul 30, 2014

### carrz

Are there any actual numbers for electron orbital velocity in hydrogen atom?

2. Jul 30, 2014

### sophiecentaur

No, because you cannot model an electron, bound to a nucleus as an orbiting object. The Bohr atom model is not a valid one - which is why it was replaced some time ago by the model which treats the electron, when bound to an atom, as a wave or a probability density function. That is why the energy states are nowadays referred to as 'orbitals' and not 'orbits'. The term 'orbital' should. perhaps, not even be used but, for the sake of Chemists, it seems to be.

3. Jul 30, 2014

### carrz

Is there some velocity associated with electron in Bohr atom model?

But if we can measure electron orbital magnetic moment then I should be able to work out some velocity from Biot-Savart law, right? Can we measure electron orbital magnetic moment in hydrogen atom?

4. Jul 30, 2014

### sophiecentaur

You can get a number but what meaning would it have if the electron is not in one particular place at any time? You can only have a velocity if there is a displacement / position.

5. Jul 30, 2014

### carrz

I think it can only mean "rate of change in position", in units distance per unit time. But whatever the meaning it should be related and proportional to its momentum. Wouldn't you agree?

I know we can not quite measure it, but are you sure QM actually forbids electrons to move in continuous trajectories? And why is that again, uncertainty principle, radiation maybe? It was radiation problem Bohr atom model suffered mostly, wasn't it?

Isn't that true for a magnetic moment as well? Both the magnitude and orientation of a magnetic field is defined by the charge velocity vector, isn't it?

6. Jul 30, 2014

### sophiecentaur

My take on this is:
When an electron is in a bound state, it doesn't have a defined position so what meaning can velocity (= rate of change of position) have? If you stick the electron in a vacuum and fire it in a direction, so it is not bound, then you can assign it a position and momentum (subject to Heisenberg) and predict when it will hit a target, for instance. When it's bound, the momentum is known accurately, because of its defined energy state so its position is not knowable to finer than a fuzzy region, given by the wave equation solution.
I don't think there's much point in trying to use the Bohr model for anything else but interesting History discussions. We know it's flawed.
As for magnetic moment, there are a whole range of orbits and velocities for a classical orbiting charge which will produce the same field.

7. Jul 30, 2014

### Staff: Mentor

For bound states, yes, QM says that the electron does not have a continuous trajectory. This is neither the uncertainty principle nor the answer to the radiation problem (which appears in the Rutherford model not the Bohr model), although both of these results fall out of the solution of the Schrodinger equation for the bound states of the hydrogen atom, along with the lack of a continuous trajectory.

Not in QM. This is most clear in the case of the intrinsic magnetic moment of subatomic particles, which is associated with the (sadly misnamed - nothing is spinning) quantum mechanical property of spin, but it's also the case for the magnetic moment associated with the (equally misnamed) orbital angular momentum of the electron.

There are a number of sites (including wikipedia) that have pretty decent explanations of what the solutions to the Schrodinger equation for hydrogen look like.

8. Jul 30, 2014

### carrz

That's even better, I'll try to find that momentum number of an electron in hydrogen atom and then I'll work out some velocity from there.

Not knowable position doesn't really mean "cannot move in continuous trajectory". What precisely is the equation, theory, or experiment that is supposed to forbid electrons to move in continuous trajectories when bound to an atom?

9. Jul 30, 2014

### carrz

Are you saying it is the Schrodinger equation which somehow implies electrons can not move in continuous trajectory when bound to an atom? Can you explain?

I don't see how a probability cloud exclude continuous trajectories. It seems to me Schrodinger equation can be applied to classical systems, like solar system, just the same.

10. Jul 30, 2014

### Staff: Mentor

Look at the solutions of the Schrodinger equation for the bound electron in a hydrogen atom. Calculate the expectation value of the position as a function of time in these solutions. Does this describe a continuous trajectory? No.

Now look at the solutions of the Schrodinger equation for an unbound particle moving through a cloud chamber or similar device. Calculate the expectation value of the position as a function of time for these solutions (this is the "Mott problem" - google for it). Does this describe a continuous trajectory? Yes.

It can be. However the solution for a collection of 1050 atoms traveling in close formation (that's what a planet is) in the gravity well of a mass 1012 meters distant doesn't look anything like the solution for a bound electron. Instead, it predicts pretty much exactly the same thing as classical mechanics does.

11. Jul 31, 2014

### Matterwave

You can easily calculate the "velocity" in the Bohr model.

In the Bohr model, the angular momentum is quantized $L=n\hbar$, classically (the Bohr model is a semi classical model) the velocity is related to the angular momentum $L=mrv$. Therefore very simply we have:

$$v=\frac{n\hbar}{mr}$$

So for the ground state, the radius is the Bohr radius $r=a_0$ and n=1 so:

$$v=\frac{\hbar}{ma_0}\approx 2*10^6 m/s$$

What this value means...probably nothing...

12. Jul 31, 2014

### sophiecentaur

From your remarks here and elsewhere, I can see that you are too much wedded to the Classical model to make any significant progress with this. You cannot ignore QM and, to get on with QM you just have to 'think different'. This was a problem everyone had a hundred years ago but there was a sea change in opinions and they managed to bring us to where we are today.
If you insist on a classical approach, you may well get a numerical answer but it will have no significance. You may as well include the base dimensions of the Great Pyramid in your calculations. It just cannot be as simple as you are making out or there would never have been the need to go to QM and beyond.

13. Jul 31, 2014

### carrz

You can replace Coulombs potential with gravity potential and Schrodinger equation will describe the same thing for planetary orbits, and you can also get their harmonic oscillator or wave function, even though planets move in continuous trajectories, or maybe better to say, because of it.

14. Jul 31, 2014

### sophiecentaur

You can do lots of Maths and get lots of different, mathematically correct, answers but do they mean anything in the real world? I am not sure what you hope to get out of this. You asked a question at the start yet you seem to be telling us 'your' answer, which, afaics, is counter to accepted Physics. All the replies are telling you this, it seems.

15. Jul 31, 2014

### Staff: Mentor

You can indeed make that replacement and solve for the energy eigenstates of an orbiting planet under the assumption the planet represents a single quantum object in the same way that an electron does. However, unlike the electron, the planet consists of an enormous number of individual particles with no particular coherency relationship between them. Thus, this assumption is an unrealistic oversimplification; it's only useful for showing that the energy eigenstates are so close to one another that we can treat them as continuous (again, unlike the electron). The classical trajectory that we observe for classical objects like planets emerges when we do not make this oversimplification.

16. Jul 31, 2014

### carrz

I was thinking the same thing. I know what F=ma or v=s/t mean, there is a reason or causality implicit in those equations, but what does Schrodinger's equation mean? That electrons vanish and then pop up into existence someplace else, for no reason, without causality? I'm not convinced, and I don't think Schrodinger's equation carries any such meaning.

It's like an equation to tells us the odds of getting a 6 out of 10 dice rolls. But it can not tell us which side will turn up on each try, and similarly it is unreasonable to expect of Schrodinger's equation to be anything more than just statistical. It tells us general odds, but nothing about specific time or specific location, and so I conclude there is plenty of room for any kind of trajectories in that equation, including continuous one.

I was thinking to calculate orbital magnetic moment or electron's momentum and see how it compares with experimental measurements. Would you be surprised if got some velocity from measured orbital magnetic moment via Biot-Savart law, then used that velocity to calculate momentum, and it turns out it matches experimental measurements?

17. Jul 31, 2014

### carrz

I'd say that's a reasonable number. By the way, do you know in what units would electron orbital magnetic moment be measured?

Last edited: Jul 31, 2014
18. Jul 31, 2014

### carrz

Ok. Do you know if electron orbital magnetic moment can actually be measured? I imagine it would be very difficult to decouple from its spin magnetic moment and that of the proton. Also momentum, can we actually measure it, and how?

19. Jul 31, 2014

### Staff: Mentor

You may not have noticed, but you've subtly changed the subject. It was "What does the Schrodinger equation and the formalism of QM say?" and now you're suggesting that whatever they say, there's more to the story than that.

You are describing a hypothetical ("there is room for" means "cannot be excluded") hidden variable theory in which the position of the particle is a hidden variable.

If you're not already familiar with the DeBroglie-Bohm pilot wave interpretation, you may want to spend some time reading about it.

However, you'll find yourself in the dead end that all discussions of interpretations lead to: There is no way of extracting from a non-local hidden variable model (Bell/Aspect experiments have rejected the local hidden variable possibility) any prediction that will be different from the statistical predictions of quantum mechanics.

20. Jul 31, 2014

### Staff: Mentor

It's a reasonable number in the same way that 75 kilograms is a reasonable number for the weight of a large unicorn - unicorns don't exist, but if they did I'd expect them to weigh within a decimal order of magnitude or so of what a good-sized antelope weighs. However, just as I have no idea what good an estimate of the weight of a non-existent unicorn is, I'm not sure what you can do with that reasonable number.

As for the units of magnetic moment... This is one of the things that wikipedia wil get right: http://en.wikipedia.org/wiki/Magnetic_moment#Units