34
$\begingroup$

A nucleus consists of protons and neutrons. Both are extremely heavy compared to electrons. Then how come they are contained within an extremely tiny space? And why does the atom consist of 99.999% empty space?

I do not understand the mathematics regarding this one bit. Please give me an explanation of this phenomenon in words as far as possible :)

$\endgroup$
37
$\begingroup$

Suppose we have two particles that attract each other. This could be an electron attracted to the nucleus by the electromagnetic force, or two nucleons attracted by the strong force, but let's keep it general for now. Suppose these two particles are separated by a distance $r$:

Particles

If the particles were for example an electron and a proton there would be an attraction between them and as we decrease $r$ the energy $E$ will decrease according to:

$$ E = -\frac{A}{r} $$

for some constant $A$ that tells us how strong the attractive force between the particles is. The particles want to reduce their energy, so they will try and make the distance $r$ between them as small as possible i.e. they will try to merge together.

But Heisenberg's uncertainty principle tells us that:

$$ \Delta x \Delta p \ge \frac{\hbar}{2} $$

that is, if we localise a particle to within a distance $\Delta x$ its momentum becomes uncertain by an amount $\Delta p$. In our system of two particles we can, in an arm waving way, say that the uncertainty in position is around the distance between the particles. A quick rearrangement of the equation above tells us that the momentum uncertainty is related to $r$ by:

$$ \Delta p \ge \frac{\hbar}{2r} $$

The reason this matters is that the energy of a system is related to its momentum by:

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

and if we take the $\Delta p$ we calculated above and put it in this equation we get:

$$ E = \frac{\hbar^2}{8r^2m} $$

So as we move the particles together the uncertainty principle means their energy increases, and this opposes the attractive force between the particles. The particles are going to end up at a distance where these two effects balance out, i.e.

$$ \frac{A}{r} = \frac{\hbar^2}{8r^2m} $$

and rearranging this for $r$ gives:

$$ r = \frac{\hbar^2}{8Am} $$

Since this is a very approximate argument let's ignore the constants and just write:

$$ r \propto \frac{1}{Am} $$

And this immediately tells us why nuclei are smaller than atoms. The mass of nucleons is about 2,000 times greater than the mass of electrons, and our equation tells us that size reduces as mass increases. Also nuclear forces are stronger than electromagnetic forces, i.e. the constant $A$ is greater for nuclear forces, and the equation tells us that as $A$ increases $r$ decreases. Both effects mean the size of nuclei is going to be smaller than the size of atoms.

$\endgroup$
  • $\begingroup$ exotic atoms demonstrate this, en.wikipedia.org/wiki/Exotic_atom $\endgroup$ – anna v Feb 18 '15 at 12:06
  • 7
    $\begingroup$ this is a good answer, but I'd really like to see the other part of the question (why atoms are 99.999% empty) answered, preferably early on before readers give in to "oh man too much math, science is hard!" $\endgroup$ – Shep Feb 18 '15 at 18:22
  • 1
    $\begingroup$ @Shep if science were easy, everyone would do it! $\endgroup$ – Michael McGriff Feb 18 '15 at 18:54
  • 4
    $\begingroup$ I really don't like this argument of setting the uncertainty in position equal to the distance, or uncertainty in momentum equal to the momentum, at least not without some better justification. I've seen it lead to too many erroneous conclusions. $\endgroup$ – David Z Feb 18 '15 at 23:08
  • 1
    $\begingroup$ You've used the same symbol $E$ for the potential energy of the interaction and the kinetic energy of the particles. That might confuse some readers (including perhaps the OP), especially since you then equate the negative of the potential energy to the kinetic energy. $\endgroup$ – Andreas Blass Feb 19 '15 at 16:43
22
$\begingroup$

The first think you need to understand is that an atom is not 99.99% empty space. It's just a misconception which refuses to die, for several reasons:

  • The "gee-wiz" or "far-out" factor (depending on your decade) is strong with this one. It's a convent sound-byte that is simultaneously awe-inspiring and simple to understand.
  • We continue to depict atoms as very small solar-systems with the electrons orbiting the nucleus. Some may argue that this is more "intuitive" than the more accurate picture involving orbitals, but it really has its roots in the Bohr model of the atom, which is interesting for historical reasons but otherwise not particularly valuable.

The last point generally gives rise to a picture of a particle zipping in from outside and slamming into an electron or a nucleus, like an asteroid would come and hit the earth. Clearly to have such a collision, the particles must have some spacial extent, and if they have some spacial extent there's a notion of "occupied" and "empty" space.

This image is wrong. The spatial extent of the electron is that of the atomic orbital, which is roughly the size of the atom. The spatial extent of the nucleus is much smaller, as explained in another answer. Any notion of "empty space" is just misleading.

So where does anyone get these "mostly empty" ideas? Well, it turns out that looking at the nucleus isn't very easy with visible light (basically because visible photons are too spatially extended to be a good probe), so the only way to resolve the nucleus is by hitting atoms with high energy particles. At the energies where you can resolve a nucleus, the probability of a particle interacting with an electron is quite low. Physicists translate this lower probability into geometrically-loaded language by talking about the cross section of the electron, so in this experiment we see a nucleus and a "small" electron, i.e. mostly empty space.

The important thing is that the electron was never actually "small", so space was never actually empty. The whole viewpoint is an oversimplified framing of the first experiments where the atomic structure was resolvable.

$\endgroup$
  • $\begingroup$ Is JLAB wrong then? $\endgroup$ – Dan Dascalescu Feb 18 '15 at 23:25
  • 2
    $\begingroup$ @DanDascalescu, I guess it depends on which science writer you ask: "The bottom line is that the old picture of the electron spinning around in an orbit (like a tiny solar system) is simply not right." $\endgroup$ – Shep Feb 18 '15 at 23:41
  • 1
    $\begingroup$ @DanDascalescu, but yes, I'd say they are being very confusing with that question and the several that follow it: by the logic that they call the electron cloud "empty" they could also call the proton "empty" since it's actually made up of quarks. If you're going to say an atom is 99.999999% empty, you should just say "all of space is empty". It's a fun metaphysical point, just not a very useful definition. $\endgroup$ – Shep Feb 18 '15 at 23:47
8
$\begingroup$

The protons and Neutrons in the Nucleus feel the strong nuclear force (or rather the strong force's equivalent to the van-der-Waals force). Since this force is much stronger than the electrostatic repulsion between the protons, atomic nuclei are so tiny.

The electrons on the other hand do not feel the strong force at all - that's just one of their basic properties. Ergo they are only bound through the electromagnetic force and since this is much weaker, it leads to much larger orbitals.

$\endgroup$
6
$\begingroup$

This may not be a direct answer to your question - but you're wrong in your assumption. What you described is not the case every time.

Matter of fact, there do exist exotic forms of matter - where there is basically no space between atomic nuclei at all; all the atoms can be squished together into a soup of protons and neutrons under the effects of extreme gravity.

This is in fact what happens in a neutron star. A neutron star is theorized to be composed of a number of layers, similar to the way our planet is composed of layers, with the less dense layers at the top and the denser stuff closer to the centre, with the much greater pressures keeping them that way.

The outer-most layer of the neutron star, the crust, will still have some space between nuclei of protons/neutrons and will thus still have an identifiable concept of atoms, albeit very much clustered up. But as we move towards the centre, all that empty space will disappear, gravity will overcome electron repulsion completely and the atoms will essentially lose all their empty space and consist of just their nuclei, as their orbiting electrons are 'captured' by the protons of their nuclei (and subsequently transformed into neutrons - hence neutron star). Go further still, and all distinctions of individual atoms and nuclei will disappear under the weight of more and more pressure; all these neutrons will basically just form an ocean.

Why am I telling you all this? Well actually there is a clue here, for your answer. Namely, that in order for an atom to NOT be 99.999% empty space, you would need to overcome its electrostatic repulsive forces with a force so great, that you would only be able to achieve it under the gravity of a star that's as dense as our sun would be - if you took it and shrunk it to the size of a small city. That's how strong the force ensuring that 99.999% empty space is, and what sort of pressures it would take to overcome it. In the absence of any such overwhelming force, the electromagnetic repulsive forces within the atom would ensure the sizes that atoms hold currently.

$\endgroup$
2
$\begingroup$

I think your question has more to do with the electromagnetic force than the nucleus. It's not the nucleus that defines empty space in the atom so much as the electron orbitals, and a bit like the earth orbiting the sun, or, Pluto orbiting the sun - Orbits can be much much larger than what they orbit.

It's electron orbits that determine both the size and, one could say, the chemical nature of atoms (though the chemical nature is often associated with the number of protons in the Nucleus), but the number of protons in the nucleus determines the electron orbits and it's those electron orbits that make the chemical bonds and give the atom it's characteristics.

In a sense, it's not really empty space at all cause it's full of the electromagnetic force and while the electron itself is itsy bitsy small compared to the atom, the electron field is about the size of the atom.

But, if the question is, why electrons orbit as far from the Nucleus as they do. That's above my pay-grade.

$\endgroup$

protected by Qmechanic Feb 18 '15 at 21:21

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.