atomic-physics
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 🙂
Things are not empty space. Our classical intuition fails at the quantum level.
Matter does not pass through other matter mainly due to the Pauli exclusion principle and due to the electromagnetic repulsion of the electrons. The closer you bring two atoms, i.e. the more the areas of non-zero expectation for their electrons overlap, the stronger will the repulsion due to the Pauli principle be, since it can never happen that two electrons possess exactly the same spin and the same probability to be found in an extent of space.
The idea that atoms are mostly "empty space" is, from a quantum viewpoint, nonsense. The volume of an atom is filled by the wavefunctions of its electrons, or, from a QFT viewpoint, there is a localized excitation of the electron field in that region of space, which are both very different from the "empty" vacuum state.
The concept of empty space is actually quite tricky, since our intuition "Space is empty when there is no particle in it" differs from the formal "Empty space is the unexcited vacuum state of the theory" quite a lot. The space around the atom is definitely not in the vacuum state, it is filled with electron states. But if you go and look, chances are, you will find at least some "empty" space in the sense of "no particles during measurement". Yet you are not justified in saying that there is "mostly empty space" around the atom, since the electrons are not that sharply localized unless some interaction (like measurements) takes place that actually forces them to. When not interacting, their states are "smeared out" over the atom in something sometimes called the electron cloud, where the cloud or orbital represents the probability of finding a particle in any given spot.
This weirdness is one of the reasons why quantum mechanics is so fundamentally different from classical mechanics – suddenly, a lot of the world becomes wholly different from what we are used to at our macroscopic level, and especially our intuitions about "empty space" and such fail us completely at microscopic levels.
Since it has been asked in the comments, I should probably say a few more words about the role of the exclusion principle:
First, as has been said, without the exclusion principle, the whole idea of chemistry collapses: All electrons fall to the lowest 1s orbital and stay there, there are no "outer" electrons, and the world as we know it would not work.
Second, consider the situation of two equally charged classical particles: If you only invest enough energy/work, you can bring them arbitrarily close. The Pauli exclusion principle prohibits this for the atoms – you might be able to push them a little bit into each other, but at some point, when the states of the electrons become too similar, it just won't go any further. When you hit that point, you have degenerate matter, a state of matter which is extremely difficult to compress, and where the exclusion principle is the sole reason for its incompressibility. This is not due to Coulomb repulsion, it is that that we also need to invest the energy to catapult the electrons into higher energy levels since the number of electrons in a volume of space increases under compression, while the number of available energy levels does not. (If you read the article, you will find that the electrons at some point will indeed prefer to combine with the protons and form neutrons, which then exhibit the same kind of behaviour. Then, again, you have something almost incompressible, until the pressure is high enough to break the neutrons down into quarks (that is merely theoretical). No one knows what happens when you increase the pressure on these quarks indefinitely, but we probably cannot know that anyway, since a black hole will form sooner or later)
Third, the kind of force you need to create such degenerate matter is extraordinarily high. Even metallic hydrogen, the probably simplest kind of such matter, has not been reliably produced in experiments. However, as Mark A has pointed out in the comments (and as is very briefly mentioned in the Wikipedia article, too), a very good model for the free electrons in a metal is that of a degenerate gas, so one could take metal as a room-temperature example of the importance of the Pauli principle.
So, in conclusion, one might say that at the levels of our everyday experience, it would probably enough to know about the Coulomb repulsion of the electrons (if you don't look at metals too closely). But without quantum mechanics, you would still wonder why these electrons do not simply go closer to their nuclei, i.e. reduce their orbital radius/drop to a lower energy state, and thus reduce the effective radius of the atom. Therefore, Coulomb repulsion already falls short at this scale to explain why matter seems "solid" at all – only the exclusion principle can explain why the electrons behave the way they do.
The electrons in an atom do not have a position. They are delocalised over the whole atom. So the question of how the positions of the electrons are arranged is meaningless.
In the macroscopic world we take it for granted that an object, e.g. a tennis ball, has a well defined position. However when we get down to the size range where quantum effects become important we find that quantum objects like electrons are described by wavefunctions that are spread out over some region of space. In particular the electrons in atoms are not like little balls but are more like fuzzy clouds, and these fuzzy clouds all overlap with each other.
We can get an approximate description of the electron distribution in atoms by using the self consistent field approximation, and this gives us the well known $s$, $p$, $d$, $f$, etc atomic orbitals:
The total electron density is then obtained by summing up the orbitals that are populated by electrons. As it happens I did a final year project on calculating the electron densities in atoms and it is surprisingly boring. In the absence of any external perturbation the electron density is spherically symmetric and falls off in a roughly exponential fashion with distance from the nucleus.
Best Answer
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 $r$:
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.