Q:

I have a few q's and I've numbered them so they aren't missed.
1. Do electrons actually move around the proton, in particular consider the hydrogen atom (1 atom & 1 proton). Does the electron move around the proton or not? When I mean move around, I mean like literally a kenetic motion in the shape of an orbit around the proton.
2. If yes to a, what experiments or data have proved this? Like how do you know?
3. If yes, what is the magnitude of the velocity? is it correct to use m*v^2/r=K*q^2/r^2. Given we know mass of electron, charge and atomic radius, is this a correct model to solve for v?
4. okay last one and its qualitative:
so usually an electron in motion produces a B-field. if it is true that electrons move with some v around the proton, why isn't a b-field produced?

- Sera S. (age 19)

St. Paul MN USA

- Sera S. (age 19)

St. Paul MN USA

A:

1. Generally atoms' electrons do not move in anything like the classical sense. In particular, for the lowest energy state of the hydrogen atom, the electron cloud goes absolutely nowhere, keeping a fixed distribution in time. There's some kinetic energy, associated with a distribution of purely radial velocities. Quantum states generally have distributions of both position and velocity. That doesn't mean sometimes one value and sometimes another. It just plain means distributed, like the way a water wave is just plain spread out..

2. The answer to 1. was no.

3. Those models give something like the right numbers, but the models are completely wrong. For example, the tangential (orbital) velocity in that ground state is exactly zero.

4. For the ground state, it isn't true- hence no such B. There are higher-energy states in which the electron cloud does actually rotate. These do have associated B fields. You can even have states combining ones with different energies so that the actual positions of the cloud rotate in something like the way you picture classically. There is also some B from the intrinsic electron spin, quite apart from any orbital effects.

Mike W.

2. The answer to 1. was no.

3. Those models give something like the right numbers, but the models are completely wrong. For example, the tangential (orbital) velocity in that ground state is exactly zero.

4. For the ground state, it isn't true- hence no such B. There are higher-energy states in which the electron cloud does actually rotate. These do have associated B fields. You can even have states combining ones with different energies so that the actual positions of the cloud rotate in something like the way you picture classically. There is also some B from the intrinsic electron spin, quite apart from any orbital effects.

Mike W.

*(published on 10/18/2011)*

Q:

cool thank you for the answer. I did not expect the electron to move around the proton but that is picture highschool chem teachers have. What elements have such B-fields caused by the whole electron cloud moving? Also I am a bit confused with your comment on electron spin. Is it still correct to think that a charge in motion creates a b-field?

- Sera S (age 19)

St. Paul MN USA

- Sera S (age 19)

St. Paul MN USA

A:

On the second question, charge in motion definitely creates a B-field. However, spins also create B-fields, even though it's not quite right to picture them as charges in motion.

The orbital magnetic contribution depends very strongly on whether the atoms exist separately in a gas or interacting in a crystal. I'm guessing you're interested in the separate-atom case. The first four elements in the periodic table have electrons in spherically symmetric, no-rotation states, with no orbital angular momentum. The fifth, boron, has one unpaired electron in a "p-state", with orbital angular momentum of one quantum unit. In the lowest energy state that' turns out to be anti-aligned with the electron's spin angular momentum. So in that state the magnetism contains contributions from both orbit and spin. However, since in this particular type of atom they're oppositely aligned the net boron magnetism is very small, because there's only half as much spin angular momentum but the magnetic moment to angular momentum ratio is very close to twice as large for spins as for orbits.

Working your way up the periodic table, there are some quantum rules (e.g. Hund's rule) determining what the total spin angular momentum is and the total orbital angular momentum, and whether they're aligned or anti-aligned. For the noble gases, these always come out to zero- the spins are paired up with opposites and so are the orbits. For atoms with odd numbers of electrons there's always some of each type of moment left.

This a rather complex topic, so I hope these few remarks can just help get you started.

Mike W.

The orbital magnetic contribution depends very strongly on whether the atoms exist separately in a gas or interacting in a crystal. I'm guessing you're interested in the separate-atom case. The first four elements in the periodic table have electrons in spherically symmetric, no-rotation states, with no orbital angular momentum. The fifth, boron, has one unpaired electron in a "p-state", with orbital angular momentum of one quantum unit. In the lowest energy state that' turns out to be anti-aligned with the electron's spin angular momentum. So in that state the magnetism contains contributions from both orbit and spin. However, since in this particular type of atom they're oppositely aligned the net boron magnetism is very small, because there's only half as much spin angular momentum but the magnetic moment to angular momentum ratio is very close to twice as large for spins as for orbits.

Working your way up the periodic table, there are some quantum rules (e.g. Hund's rule) determining what the total spin angular momentum is and the total orbital angular momentum, and whether they're aligned or anti-aligned. For the noble gases, these always come out to zero- the spins are paired up with opposites and so are the orbits. For atoms with odd numbers of electrons there's always some of each type of moment left.

This a rather complex topic, so I hope these few remarks can just help get you started.

Mike W.

*(published on 10/25/2011)*

Q:

it sounds like your saying the whole probability density of the p moves? Also my friends don't believe me and we had a pretty long argument and they say the only way the orbital can move is if a photon is being released. is there a name for this phenomena or some website or research paper posted on this?

- Sera Shane (age 19)

St. Paul, MN USA`

- Sera Shane (age 19)

St. Paul, MN USA`

A:

Your friends are mixed up, or maybe there's been a miscommunication about what the question was. I think what they have in mind is that the electron cannot go from one orbital state to another, with lower energy, without somehow getting rid of the energy. In nearly every case, the way the energy leaves is a one or more photons.

As for whether there can be some orbital motion is a state with fixed energy, the answer is just plain yes. Any beginning book on quantum mechanics describes a set of possible states for the electrons. It's almost universal to list states of definite energy and usually of definite angular momentum as the basis on which to describe all the possible states. A state of definite angular momentum has the sort of motion your talking about. Quite unlike a classical picture, however, the position of this rotating cloud is not changing at all in time. in order to get positions to change, you need a state with more than one value of energy.

Mike W.

*(published on 11/01/2011)*

Q:

You say the electron wave "just plain means distributed, like the way a water wave is just plain spread out". A water wave is caused by moving water molecules so what is the electron wave caused by?

- John (age 59)

St Julians, Malta

- John (age 59)

St Julians, Malta

A:

It's true that a water wave is a particular pattern of moving water molecules. That doesn't mean that every other wave has to also be made up of some smaller parts. A wave might be one of the basic ingredients of the universe.

With that said, however, it's also possible that each of the known particle waves is made up of something deeper. For example, it's possible that the strings of string theory are the things that are waving to make up an electron wave. Of course we don't yet know if string theory is a correct description. Let's imagine that it is. Then what are the strings themselves made of? Maybe there's a deeper layer. How far would that sort of thing go? Would it ever stop, or would there be an infinite number of levels? If there's a deepest level, is it also some sort of continuous differential equation, like a wave equation, or is it some sort of discrete math, like a cellular automaton?

Some people think they know the answers to those questions, but they don't agree. I think we just don't know.

Mike W.

*(published on 03/22/2014)*

Q:

Expert response Mike. The web needs more websites like yours. I think you covered nearly all imaginable answers. But none seem satisfactory. Basic or discrete ingredients cannot just exist automatically and infinity is undefinable. Do you think the real answer is beyond human imagination or theory?

- John (age 59)

St Julians, Malta

- John (age 59)

St Julians, Malta

A:

Thanks for the kind words.

As for your question, beats me.

Mike W.

*(published on 03/24/2014)*

Q:

Is really electron moves with the speed of light around the proton and how it moves in our body also? Why we don't feel it?

- AKM Jahurul Islam (age 56)

Dhaka Bangladesh

- AKM Jahurul Islam (age 56)

Dhaka Bangladesh

A:

They don't travel at the speed of light, ever. In ordinary light atoms and in your body the speeds (relative to the nucleus or to your body) are actually very low compared to the speed of light.

We consist of those electrons, etc. When they're doing nothing out of the ordinary, they don't trigger nerve signals so we don't feel them. What would be the evolutionary point of being bombarded by sensations that nothing special is happening? If their motions become very unusual, say when you get an electric shock, they do trigger nerve signals and you do feel them.

Mike W.

*(published on 08/18/2015)*