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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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    $\begingroup$ 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? $\endgroup$ – anna v Jan 29 '12 at 17:18
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    $\begingroup$ @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. $\endgroup$ – HelloGoodbye Mar 4 '14 at 12:39
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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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  • $\begingroup$ You mean the stationary state of electron? $\endgroup$ – Revo Jan 30 '12 at 22:11
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    $\begingroup$ 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. $\endgroup$ – Luboš Motl Jan 20 '13 at 8:11
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    $\begingroup$ The scaling with Z is approximately a proportionality to Z for the hydrogenlike system, so essentially $v\sim Z\alpha c$. $\endgroup$ – Ben Crowell Apr 16 '13 at 3:16
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    $\begingroup$ Dear @Sidd, every Hermitian linear operator is an observable and may be measured in principle. The required apparatus may be simple or contrived but it may always be in principle constructed. The nonzero value of the velocity doesn't contradict the state's being stationary. Stationary states allow things like velocities to be nonzero as long as their statistical distributions etc. don't depend with time. De Broglie-Bohm theory emulates the right theory, quantum mechanics, to some extent but not the full extent including many bodies, spin, and the measurement event and events afterwards. $\endgroup$ – Luboš Motl Oct 27 '15 at 5:51
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    $\begingroup$ Dear @Jung, right but the radius of the atom is really $\Delta r$, the obstacle preventing you from setting $r=0$. And when the uncertainty principle implies that $m\cdot \Delta v \geq \hbar / \Delta r$, it again means that the average absolute value of the velocity must also be at least $\Delta v$, otherwise $v$ would be too accurate. So you're right but effectively wrong. When we want $r=0$ and $v=0$ but the uncertainty principle prevents us from setting these things, then effectively $\bar r=\Delta r$ and $\bar v=\Delta v$. Does it make sense? $\endgroup$ – Luboš Motl Jan 30 at 14:47
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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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    $\begingroup$ And what about kinetic energy of an orbital electron? $\endgroup$ – voix Jan 29 '12 at 18:02
  • $\begingroup$ The kinetic energy is due to what I am calculating as the momentum uncertainty. $\endgroup$ – FrankH Jan 29 '12 at 18:46
  • $\begingroup$ 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. $\endgroup$ – Anixx Apr 16 '14 at 22:10
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    $\begingroup$ It is nonsense to claim that for an electron in an orbital there isn't an average speed, or a minimum or a maximum speed. The squared absolute value of the wave-function in $p$-representation is a perfectly sound probability distribution for the kinetic energy. From this one can obtain all the statistical properties for the speed of the electron. $\endgroup$ – M. Wind May 6 '15 at 2:02
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This sounds absurdly simple, but for order of magnitude results, it matches well with online results and other results here. People tend to think that good old Newtonian dynamics are useless on the atomic scale, but they still apply if the velocity is well below light speed.

Simply take the electric force between the electron and nucleus, convert it to an acceleration by a = F/m, and then balance that acceleration with centripetal acceleration of velocity squared over r. That is, v = SQRT(rF/m).

I know, you might say that doesn't work, an electron is not like a satellite at a point location, but is rather smeared out in an electron cloud about an orbital. But remember - the definition of the orbital is a path at which each point is a balance between kinetic and potential energies. So each location on a spherical orbital has to be a point of balance between centripetal acceleration and electrostatic attraction.

enter image description here where epsilon-zero is 0.00000000000885 and the charge q of an electron is 1.606e-19 coulomb.

Hydrogen example:
enter image description here

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  • $\begingroup$ This isn't really a Newtonian calculation. You put in a value for $r$, which has to be taken either from experiment or from a quantum-mechanical calculation. (By the way, please mark up your math using mathjax.) $\endgroup$ – Ben Crowell Apr 15 at 22:30

protected by Qmechanic Oct 29 '13 at 15:19

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