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E.g. in exploding nuclear bomb or some other big explosions.

I mean if the speed of electrons as waves/particles is a constant or changes according to other "forces" involved?

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closed as not a real question by Waffle's Crazy Peanut, user1504, Emilio Pisanty, Manishearth Apr 27 '13 at 8:45

It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

The velocity of the electrons depends on their energy (kinetic) energy. The speed of an electron absolutely isn't constant. Because the mass of an electron is so low they're often traveling at relativistic speeds so you the classical $E = \frac{1}{2} m v^2$ often isn't a very good approximation. – Brandon Enright Apr 19 '13 at 7:13
Can you expand a bit on what you mean? Are you thinking about the speed of electrons bound to atoms, or electrons emitted due to ionisation in the explosion? – John Rennie Apr 19 '13 at 8:42
After an explosion, of course all kinds of particles will be flying away from the centre of the explosion, including electrons. That's obviously a change in their speeds. But your question probably hinges on the speed vector versus the absolute speeds. – MSalters Apr 19 '13 at 11:37
up vote 1 down vote accepted

@Brandon is correct. You can compute the average kinetic for any free particle using the equipartition theorem, which gives $\langle E \rangle = \frac{1}{2}k_B T$ per quadratic degree of freedom, where $T$ is the temperature and $k_B$ is Boltzmann's constant. For free particles in 3D this gives $\langle E \rangle= \frac{3}{2} k_B T$; equating $\frac{1}{2} m_e v^2 = \frac{3}{2} k_B T$ (in a classical approximation) shows that the root-mean-square velocity $v$ of free electrons is temperature dependent.$^1$

In the above, the velocities are measured with respect to the environment at temperature $T$ -- we don't even need to consider the fact that the electrons you're talking about are presumably on Earth, being accelerated around the Sun in centripetal motion.

Even light will change its speed $c=c_0/n$ in an explosion, because the index of refraction $n$ of the surrounding medium will become inhomogeneous and fluctuate: it's only the speed of light $c_0$ in vacuum that's constant.

$^1$Note, as also pointed out by Brandon, electrons often move at relativistic speeds, so $\frac{1}{2}m_e v^2$ is a poor approximation. $v$ in this formula can exceed the speed of light $c$, for example. To quantitatively calculate the velocity you need the correction from special relativity. This doesn't qualitatively change the answer though.

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@Brandon maybe this should have been your answer... – Douglas B. Staple Apr 19 '13 at 13:38
Btw, in case you didn't know, Brandon won't be pinged, since he technically hasn't participated in this post. You can ping him in the comments to the question, however. See… – Chris White Apr 19 '13 at 17:07

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