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I am wondering how fast electrons travel inside of atomic electron orbitals. Surely there is a range of speeds? Is there a minimum speed? I am not asking about electron movement through a conductor.

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from the way you phrase your question you give the impression you do not know much about quantum dynamics. Have a look at en.wikipedia.org/wiki/Atomic_orbital . Can you understand the article? –  anna v Jan 29 '12 at 17:18
@anna v: For all particles, you can calculate an expectation value for the kinetic energy. This kinetic energy corresponds to a speed. This is true even for trapped, "stationary", particles. –  HelloGoodbye Mar 4 '14 at 12:39

4 Answers 4

There are actually two orbital velocities involved by the electron, one around the nucleus and one around the common center of mass of the electron and the nucleus. The same applies to the Moon-Earth system where the Moon's velocity around the Earth is slightly faster than the "sidereal" velocity of the Moon around their common center of rotation, the center of mass, which is situated a short distance inside the Earth's surface. The velocity of an electron in its innermost orbit around the nucleus of an atom equals the product of the velocity of light and the fine structure constant multiplied by Z, the atomic number. The velocity of the electron around the common center of mass is less by M/(M + m), where M and m are the masses of the nucleus and electron respectively. The Schrödinger equation is based on wave functions and does not deal with electrons as an orbiting particles. However the weakness of the Schrödinger equation is that it is a non-relativistic equation and does not provide exact solutions.

Please look at http://www.colutron.com/download_files/Quantum.pdf It might be controversial but contains exact solutions and the numbers come out right.

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Dear Lars: For your information, Physics.SE has a policy that it is OK to cite oneself, but it should be stated clearly and explicitly in the answer itself, not in attached links. –  Qmechanic Feb 17 '13 at 20:11

The state of an electron (or electrons) in the atoms isn't an eigenstate of the velocity (or speed) operator, so the speed isn't sharply determined. However, it's very interesting to make an order-of-magnitude estimate of the speed of electrons in the Hydrogen atom (and it's similar for other atoms).

The speed $v$ satisfies $$ \frac{mv^2}2\sim \frac{e^2}{4\pi\epsilon_0 r}, \qquad mv\sim \frac{\hbar}{r} $$ The first condition is a virial theorem – the kinetic and potential energies are comparable - while the second is the uncertainty principle. The second one tells you $r\sim \hbar / mv$ which can be substituted to the first one (elimination of $r$) to get (let's ignore $1/2$) $$ mv^2 \sim \frac{e^2 \cdot mv}{4\pi\epsilon_0\hbar},\qquad v \sim \frac{e^2}{4\pi\epsilon_0\hbar c} c = \alpha c $$ so $v/c$, the speed in the units of the speed of light, is equal to the fine-structure constant $\alpha$, approximately $1/137.036$. The smallness of this speed is why the non-relativistic approximation to the Hydrogen atom is so good (although a non relativistic kinetic energy was assumed from the start): the relativistic corrections are suppressed by higher powers of the fine-structure constant!

One could discuss how the speed of inner-shell electrons and valence electrons is scaling with $Z$ etc. But the speed $v\sim \alpha c$ would still be the key factor in the formula for the speed.

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You mean the stationary state of electron? –  Revo Jan 30 '12 at 22:11
It's an order-of-magnitude estimate, so I mean any bound state - any state for which the probability is nearly 100% at all times for the electron to be less than $r<r_0$ for some large but fixed $r_0$. Mathematically, they're arbitrary linear superpositions of the bound (discrete spectrum) energy eigenstates: the estimate above works for them. –  Luboš Motl Jan 31 '12 at 7:23
I'd think that any electron being accelerated would radiate. Perhaps in a stationary state the electron is best described as a motionless cloud? –  Paul J. Gans Nov 4 '12 at 2:04
Dear Paul, the electron in the ground state can't radiate because there isn't any state with lower energy. The existence of a lower-energy state would be needed by energy conservation (photon carries a positive energy away) but it just doesn't exist, so the probability of transition to this non-existent state is clearly zero. Your comment shows that you are trying to find a classical model - a cloud? - but it's just wrong. You must start to think quantum. –  Luboš Motl Jan 20 '13 at 8:11
The scaling with Z is approximately a proportionality to Z for the hydrogenlike system, so essentially $v\sim Z\alpha c$. –  Ben Crowell Apr 16 '13 at 3:16

So, what about this?

"It takes about 150 attoseconds for an electron to circle the nucleus of an atom. An attosecond is $10^{-18}$ seconds long, or, expressed in another way: an attosecond is related to a second as a second is related to the age of the universe," says Johan Mauritsson, an assistant professor in atomic physics at the Faculty of Engineering, Lund University. He is one of seven researchers behind the study, which was directed by him and Professor Anne L'Huillier."

Doesn't that give us a defined speed? Or does that make the 'orbit' totally unreliably uncertain?

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What speed would that imply for the electron? –  Colin McFaul Apr 15 '13 at 22:36
The researcher's site is atomic.physics.lu.se/research/attosecond_physics and their paper is arxiv.org/abs/1012.3863 –  Brandon Enright Apr 16 '13 at 2:31

This is the realm of quantum mechanics and classical notions about point like electrons travelling at certain speeds don't really apply in this domain. So there isn't an average speed or a minimum speed or even a maximum speed (except for the speed of light which is the maximum speed for any particle with mass).

The closest you can come to having any concept of speed for an electron in an orbital would be to apply the Heisenberg uncertainty relation which states that $$\Delta x \Delta p \geqslant \hbar$$ So if you plug the size of the orbital in for $\Delta x $ and solve for $ \Delta p $ you would have an estimate for the uncertainty in the momentum which you could then relate to the uncertainty in speed.

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And what about kinetic energy of an orbital electron? –  voix Jan 29 '12 at 18:02
The kinetic energy is due to what I am calculating as the momentum uncertainty. –  FrankH Jan 29 '12 at 18:46
speed of light is not the maximum speed for an electron in atom because even faster movemen would not violate causality. Particularly, you cannot distinguish a virtual electron in the atom from a real one. So there is no higher limit. –  Anixx Apr 16 '14 at 22:10

protected by Qmechanic Oct 29 '13 at 15:19

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