The Heisenberg uncertainty principle implies that it is not possible to calculate the position and velocity of a body at the same time accurately. Electrons follow this principle since their orbitals are just probabilities where they may occur. But in the case of protons, we are kind of certain about their position in the atom. So shouldn't they have a very, very, very high velocity to balance the certainty in position?
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2 Answers
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You are using the Heisenberg uncertainty principle to relate the uncertainty in position $x$ to the uncertainty in velocity $v$.
However, the quantitative version of this principle actually is
$$\Delta x\cdot\Delta p\geqslant\tfrac12\hbar $$
where $\Delta x$ is the uncertainty in position $x$ and $\Delta p$ is the uncertainty in momentum $p$.
Certainly, since $$p=m\cdot v$$ or $$v=\frac pm$$ where $m$ is mass and $v$ is velocity, you may also use the uncertainty principle to compare the uncertainty in position $x$ to the uncertainty in velocity $v$.
However, the mass of a proton is $m_\mathrm p=1.672\,621\,923\,69(51)\times10^{-27}\ \mathrm{kg}$[CODATA2018], whereas the mass of an electron is only $m_\mathrm e=9.109\,383\,7015(28)\times10^{-31}\ \mathrm{kg}$[CODATA2018]; i.e. a proton is over a thousand times heavier than an electron.
Therefore, a given uncertainty in position $x$ and the corresponding uncertainty in momentum $p$ result in a smaller uncertainty in velocity $v$ for the proton.
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But in the case of protons, we are kind of certain about their position in the atom.
Well, yeah, kind of certain. The very notion of molecular geometry arises in the Born-Oppenheimer approximation. Nuclei are much heavier than electrons so that when solving the electronic Schrödinger equation they can be assumed to be stationary. This clearly violates the uncertainty principle since we precisely know both position (whatever it is) and momentum (zero by assumption) of each and every nucleus simultaneously. But the thing is that when we solve the electronic Schrödinger equation nuclei are treated as classical particles, so everything is fine. Kind of fine, I mean.
So shouldn't they have a very, very, very high velocity to balance the certainty in position?
No, on the contrary, nuclei are assumed to be perfectly stationary classical particles. That's the trick. And this is, of course, an approximate description of the reality which usually works fine, though.
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2$\begingroup$ "so that when solving the electronic Schrödinger equation they can be assumed to be stationary" There is no need for that, and the usual (at least in physics texts) solution involves the canonical transformation into CoM coordinates using the reduced mass. What the OP described as the motion of the electron is actually the motion of the difference in position. Motion of the proton only makes up about .06% of that, but it is there. This has the effect of allowing you to predict the spectral differences between protium, deuterium and tritium from the changes in the reduced mass. $\endgroup$ Oct 24, 2015 at 16:10
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2$\begingroup$ @dmckee, in chemistry, there is a need to assume at first that nuclei are stationary, since otherwise the very notion of molecular geometry is undefined. So, yeah, we do CoM separation as usual, but that is not yet chemistry, it still physics. To get into chemistry you have to introduce the BO (or adiabatic) approximation in which at the first step (electronic SE) nuclei are treated as stationary classical particles. And this is the step where orbitals arise, so electrons are still quantum particles, but not nuclei. $\endgroup$– WildcatOct 24, 2015 at 16:43
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$\begingroup$ Hmmm ... that's what I get for stepping out of my area of expertise. But it seems that you could still define the molecular geometry in terms of the CoM's of the atomic systems. Or are there surprises lurking in the details? $\endgroup$ Oct 24, 2015 at 17:00
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1$\begingroup$ @dmckee, well, true, for an atomic system you don't need to invoke the BO approximation, but still even in atoms the nucleus is treated classically in chemistry. Except for some very rear cases when the quantum description is unavoidable, such as, for instance, the accurate studies involving the proton transfer: it is known that quantum tunneling is important there. But usually we treat nuclei as classical particles. $\endgroup$– WildcatOct 24, 2015 at 17:14
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2$\begingroup$ Physically, of course, nuclei are not classical particles. But what I wanted to say, is that orbitals (mentioned by OP) arise after it is already assumed that nuclei are classical particles, so at this stage there is no sense anymore to apply the uncertainty principle to them. Thus, there is no paradox out there, just an approximation. $\endgroup$– WildcatOct 24, 2015 at 17:17