# Probability and Areas Around Atomic Nuclei

1. Jun 24, 2011

### Infrasound

As a layperson...(Who is trying to understand the atom in a bit more detail).

In the past, I have most often seen the atoms drawn with electron clouds that have bands of varying density. I understand the density of the bands represent areas of higher probability.

I have often seen models of atoms drawn with clouds which are most dense near the nucleus with decreasing density as the radius increases.

Question:

Is there actually an intuitive way of understanding WHY atoms have these bands of higher probability? Is there an intuitive reason why the cloud is "thicker" in some areas/bands/distances from the nucleus?

Is there a mathematical reason why the cloud is "thicker" in some areas/bands/distance from the nucleus?

Or, are the bands just a fact of nature (sort of a "that's just the way it is as far as we can tell), which we have measured and accepted without and deeper understanding of cause?

Last edited: Jun 24, 2011
2. Jun 24, 2011

### danR

I would have to see a picture. This will have to do until a more expert answer. As another layman, the bands may be graphic artifacts, but I suspect they are the 'shells' that electrons are constrained to inhabit according to their energy states. The various densities (shades of gray?) do indeed represent probabilities, but for many textbooks they may be informally rendered without any precise numerical representation.

Because of Pauli's exclusion principle, no two electrons may occupy exactly the same position, state and spin; and discrete bands would represent the habitation of exactly one electron's probability field.

3. Jun 24, 2011

### SpectraCat

It's tough to answer those questions in a satisfactory way without getting into the math, which is often where the layperson gets lost. You may already be familiar with some of the following points, but bear with me, and I hope it will be clearer to you in the end how it all works.

Basically, quantum mechanics tells us that the electrons around an atom are distributed as waves in 3-D space .. we call these "wavefunctions". Furthermore, QM tells us that they are standing waves, analogous to harmonics of a violin string:

but in 3-D. As you can see, this means that the places in space where they have zero amplitude (called "nodes" stay fixed, as do the places where the amplitude is largest, although the latter points change phase (i.e. oscillate up and down) with time. Notice in the image that each harmonic has one more node than the next, and that each harmonic goes to zero at the end of the string. This is also true of the electron wavefunctions around atoms .. only certain solutions with nodes in well-defined places are allowed, and each wavefunction has a characteristic energy associated with it, which increases with the number of nodes (the lowest energy state, or "ground state", doesn't have any nodes, although it still has a non-zero energy). We call this "quantization of energy states", since only certain values of energy are allowed.

One of the fundamental postulates of QM is called the "Born interpretation". What that tells us is that if we take the square of the wavefunction representing one of the electronic energy states, then that tells us about the probability of finding an electron at a given position around the atom. Now, remember that the amplitude of the wavefunction at a node is guaranteed to be zero .. so that means that the probability of finding the electron at the position corresponding to a node is also zero. On the other hand, the probability of finding the electron at the peak or valley of a wave is rather high.

So, when you see images like the ones you are describing, the more intense bands represent the peaks of the electronic wavefunction (really the square of the wavefunction), while the regions with lower probability are concentrated around nodes of the wavefunction.

Now, you may now want to know *why* the quantum solutions to the electronic structure of atoms are waves in the first place. For that I suggest that you read the wikipedia pages on the de Broglie wavelength and the Heisenberg Uncertainty Principle to start, and then come back and ask more questions as they arise.

[NOTE: One technicality point I neglected in the above description is that fact that the solutions I describe are really for one-electron atoms only ... that is because electrons are charged particles that repel each other, which makes the solutions of multi-electron atoms much more complicated. However, most of the time the sorts of pictures you describe are for the H-atom solutions, like these:

In any case, the explanation I gave forms the basis for understanding electronic distribution around all atoms.

Last edited: Jun 24, 2011
4. Jun 25, 2011

### danR

5. Jun 25, 2011

### Infrasound

Thank you very much for the diagrams. I think I understand the principles of harmonics depicted above. My new question, however, would be: What is waving? Is it the probability of an electron at the various locations along the "string" which is waving?

Also (and hopefully I'm not changing the topic too much), what do physicists mean when they talk about the "momentum" of a particle, like a photon. I thought all photons were the same, and move at the same speed?

6. Jun 25, 2011

### danR

they will have differing momenta with the equation

p=h$\nu$/c

where p is the magnitude of the momentum p, and $\nu$ is the frequency of the photon, which is variable, of course.

If I have that more than half-way right.

7. Jun 25, 2011

### alxm

In case you want an elaboration, it may also help to look at http://wiki.chemprime.chemeddl.org/index.php/Tones_on_a_Drum_and_Orbital_Wavefunctions" [Broken] - that's basically the 2-dimensional version of the 1-dimensional standing waves on a string, so it's the intermediate between that and the 3-dimensional spherical harmonics which give the electron orbitals their overall shape.
Yes; the amplitude of the wave function is the probability that the electron is at that location.
Classically, momentum is velocity times mass: $$p = mv$$ But in quantum mechanics, 'velocity' isn't a well-defined concept, because particles don't have a definite location. So just as with position, you have a distribution of possible particle momenta, and the two are related. But if you use the classical definition of kinetic energy $$E = \frac{mv^2}{2}$$ you have the alternate definition $$E = \frac{p^2}{2m}$$ Which also holds in non-relativistic QM - only there, p is no longer a single value, but a probability distribution over the possible momenta, just as the spatial wave function you see above is a probability distribution over locations in space. When special relativity is taken into account, the equation is: $$E^2 = m^2 c^4 + p^2 c^2$$ which for a particle with zero momentum, reduces to the famous $$E = mc^2$$ This only applies to massive particles though, so not photons. Photons still have momentum and kinetic energy even though they have zero rest mass. They always move at the same speed, but their momentum instead determines their frequency/wavelength. (as per the equation danR posted)

Last edited by a moderator: May 5, 2017
8. Jun 25, 2011

### Infrasound

So, the wave is in a spheres around the nucleus, and in the places where the radius allows for constructive interference/phase matching as the wave makes its way around will be where the electron orbitals are found?

And the radii which cause destructive interference in the wavefunction would be areas where there is no orbital?

9. Jun 25, 2011

### Infrasound

-So because we can't really know where the electron currently is, or where it was in the past, we can't really talk about its "speed".

-But we know that electrons have momentum or energy because of what they can do when they jump around or move through a circuit.

Is that more or less correct?

Also, I know that physicists often talk about voltage (in a circuit for example). I know that this is going to sound incredibly dumb, but, what does voltage actually mean?

I used to think it was something like how hard electrons are being pushed through a wire.

Once again, thanks for your patience guys.

10. Jun 25, 2011

### danR

11. Jun 25, 2011

### alxm

No; there's no phase-matching or such. That's the de Broglie rationalization for the Bohr model of the atom, you have circular or elliptical 'orbits' around the atom which correspond to integer multiples of a wavelength. The Bohr model is wrong though. Electrons don't follow any specific path around the atom - things in QM don't move in nice trajectories, that would imply a well-defined position. They don't even necessarily move 'around' the atom - l signifies angular momentum, so the states where l=0 are states where the electron has no momentum perpendicular to the electron-nucleus line. In other words, to the extent that it can be described in moving in the classical sense, the motion is entirely in the inwards-outwards directions.

The orbitals are the solutions to the Schrödinger equation, and they're akin to standing waves in three dimensions. (so, 'wave-like' behavior). But the nodes (areas where the wave function/electron probability is zero) are part of the orbital; the whole orbital is 'where' the electron is. It's not a time average or any such thing, but the actual 'position' of the electron, as far as it's meaningful to talk about it.