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Forgive me if the answer to this is obvious. I have no formal physics training, and I remember that when I asked my physics teacher this, she just frowned and said "Good question."

An electron is negatively charged. A proton is positively charged. Based on basic principles, it seems like it would be logical for the electron cloud of an atom to "collapse" into the nucleus and become a part of it (especially since the electrons are so much lower mass). Why doesn't this happen? How do electrons maintain separation from the protons in the nucleus, when the opposite chargers ought to draw them together?

I considered that perhaps the charge in the electron was too minuscule in comparison to the proton (like having a negatively charged magnet on Saturn while all of Earth was positively charged; obviously the magnet wouldn't just be drawn to the Earth because the forces weren't strong enough to act over that distance). But if that was or is the case, I would expect some other chemical behaviors to not exist. For example, the whole phenomenon of water being a "dipole". If the charge of the electron is too weak to interact with the proton, how could the oxygen in water more strongly attract them than hydrogen? I get that the oxygen has more protons, and thus more positive charge in the nucleus, but that still would seem to support that the oxygen atom's own electrons should be attracted to it...

Can anyone explain the phenomena occurring here or simply point out the flaw in my thinking?

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marked as duplicate by DarenW, Abhimanyu Pallavi Sudhir, Qmechanic Nov 30 '13 at 11:43

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ I'd say that is because the energy is "quantized". The electron's energy can't have any value, it can only have a multiple of the elementary energy $h \nu$. $\endgroup$ – Ana S. H. Nov 30 '13 at 6:52
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    $\begingroup$ The uncertainty principle prevents the innermost electron from simply becoming completely localized on top of the proton (because, if the electron is at rest, there will be a corresponding spread in its momentum values, translating to a spatial spread of its wavefunction), and then the other electrons are kept even further out by Pauli exclusion. $\endgroup$ – Mitchell Porter Nov 30 '13 at 6:52
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    $\begingroup$ Possible duplicates: physics.stackexchange.com/q/20003/2451 , physics.stackexchange.com/q/9415/2451 and links therein. $\endgroup$ – Qmechanic Nov 30 '13 at 8:01
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As Mitchell says in his comment, this is related to the uncertainty principle.

The uncertainty principle states that if you have some system with a position $x$ and a momentum $p$ then there is an uncertainty in the position, $\Delta x$, and an uncertainty in the momentum, $\Delta p$, related by Heisenberg's uncertainty principle:

$$ \Delta x \Delta p \approx \hbar $$

In the case of the hydrogen atom the uncertainty in the position of the electron is roughly the size of the atom i.e. we know the electron is in the atom somewhere, but we don't know exactly where. This means we have an uncertainty in the momentum given by:

$$ \Delta p \approx \frac{\hbar}{\Delta x} $$

If you try and force the electron closer to the nucleus you make $\Delta x$ smaller because you know more precisely where the electron might be. But momentum is proportional to speed, and increased speed means increased energy. So by trying to confine the electron you increase its energy. The size of the hydrogen atom is a balance of the electrostatic attraction and the uncertainty principle.

If, as you say, you're a physics non-nerd then what follows might be a bit over the top, but I'll post it anyway because it's a nice illustration of what happens. Suppose the radius of the hydrogen atom is $r$ then it seems reasonable to say the uncertainty in position is $r$, in which case Heisenberg's equation tells us:

$$ \Delta p \approx \frac{\hbar}{r} $$

Now momentum is related to energy by:

$$ E_1 = \frac{p^2}{2m} $$

and the electrostatic energy of the electron is:

$$ E_2 = - k_e \frac{e^2}{r} $$

So if we say the momentum of the confined electron is about $\Delta p$ then it's total energy is:

$$ E = \frac{\hbar^2}{2mr^2} - k_e \frac{e^2}{r} $$

This gives us an equation that tells us how energy changes with the size of the atom, and this graph shows the energy as a function of the radius $r$:

Hydrogen

The minimum is at $r = 0.53$ Angstroms and the minimum energy is 13.6eV. Amazingly these are the correct values for the hydrogen atom. The ionisation energy of hydrogen is 13.6eV, and 0.53 Angstroms is the Bohr radius.

Now this is a pretty rough calculation, and to be honest I carefully chose the form of the uncertainty principle that gives the right answer. Still, I think this nicely shows how the uncertainty principle is linked to the size of the hydrogen atom.

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  • $\begingroup$ The answer makes sense to me. However another question has arised in my mind. The answer assumed the uncertainity doesn't apply to the protons right? Is it because the proton is too heavy so that it has very low wave property? $\endgroup$ – theGD Mar 9 '14 at 22:57
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    $\begingroup$ @GunDeniz John has left it out, but you can perform a standard transformation to the center of mass coordinates and get an effective problem with a particle of "reduced mass" $\mu = (m_em_p)/(m_e + m_p)$ moving in a fixed field. In the case of hydrogen $\mu$ is very nearly equal to $m_e$, though careful spectroscopy of protium (normal hydrogen) and deuterium (heavy hydrogen) can see the difference in the reduced masses. $\endgroup$ – dmckee Jan 24 '16 at 5:38

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