# Electron in Atom

1. Dec 17, 2009

### dschmidt12

I know that electrons do not exist in orbitals, but instead in probability density clouds around the atom. I know that these electrons are quantized and can only have certain energies in the atom. I also know that it is due to the Heisenberg Uncertainty principle that these atoms do not crash into the nucleus and remain at the Bohr radius or farther away from the atom. However, I am having a lot of trouble piecing this all together. Is it that the electron is most likely to be found at the Bohr radius and its probability drops off as you get farther from that point, and does Schrodinger's equation help you find the probability of finding that electron in a specific section of the cloud? I suppose my question really is what the electron probability density says about where you can find the electron in the atom. Also, when the electron absorbs or releases a photon, does it actually "move" in the atom, does its probability density change? Any help would be greatly appreciated!

2. Dec 17, 2009

### rhenretta

I could be mistaken, there are plenty of people on here who are more conversant on the subject than I, but the Heisenberg uncertainty principle doesn't describe any forces. The electron is kept at distance from the nucleus by photon exchange.

As far as putting this all together, I personally feel we are at a point in physics where we have a very complicated system to describe nature. It was this very reason that Einstein hated QED, and said his famous quote, often paraphrased "God does not play dice". Actual quote is

"Quantum mechanics is certainly imposing. But an inner voice tells me that it is not yet the real thing. The theory says a lot, but does not really bring us any closer to the secret of the 'old one'. I, at any rate, am convinced that He does not throw dice.

I philosophically see it akin to The Ptolemaic Model where gears were used to describe the apparent wandering of planets, yet keeping the earth as the center. It allowed for accurate predictions, but was far more complicated than it needed to be with a simple perspective change. I think that is where we are at now.

3. Dec 17, 2009

### Bob S

See plots of electron radial wave function probability curves in Figs. 2 through 5 in
http://panda.unm.edu/Courses/Finley/P262/Hydrogen/WaveFcns.html [Broken]
As an electron moves to a higher state (quantum numbers), it moves further from the nucleus as seen in radial probability curves.
Bob S

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4. Dec 18, 2009

### alxm

Well, they don't exist in 'orbits' as in planets circiling the sun. "Orbitals" are these probability density clouds (although the idea of a 'probability density cloud' is not a full depiction either). Each possible orbital corresponds to a different state of the electron, i.e. different energy, angular momentum, etc.

Electrons are actually near the nucleus (and certainly within the Bohr radius) all the time. For an s-type orbital it's their most likely location, even. Also an s-orbital (e.g. in Hydrogen or Helium) has 0 angular momentum, so if anything the electron is actually moving in and out from the nucleus, rather than in a circular pattern around it.

They do not stay near the nucleus because the more you concentrate the probability density of a particle around the nucleus (or any point in space), the higher its momentum. (and thus, kinetic energy). So in QM an electron cannot lose energy simply by getting nearer the nucleus, unlike the classical-mechanical picture. You can rationalize this with the Uncertainty Principle.

The orbitals are the solutions to the Schrödinger equation, and they tell you not just the energy and this probability density, but all the information about the system; it's angular momentum, polarizability, etc.

Whether electrons 'move' comes down to semantics. I feel it's fine saying that they do, but I know a professor of quantum chemistry who feels they don't.

The orbitals don't change with time (they're by definition the time-independent solutions), but the overall wave function can do so. This doesn't correspond to spatial motion of the electron though, but rather the electron moving between different orbitals (an electron in an excited state will not stay there, so the ground-state is the only truly time-independent state)

So your electron in the ground state has a fixed probability density (or charge density) around the atom - which would indeed make it seem as if they're stationary. Nor do the electrons have trajectories in the classical sense. If detected at one position you can't say where it'll be the next time you measure it.

"And yet it moves!" as Galileo http://en.wikipedia.org/wiki/E_pur_si_muove!" [Broken]. The reasons why I think it's fine to say they're moving is because they're moving in the quantum-mechanical sense, and quantum mechanical motion does become classical mechanical motion as you move to macroscopic systems. So while you have to give up classical concepts like 'trajectory', I think it might do more harm than good to give the (incorrect) impression that the electrons are stationary.

They're not stationary, because 1) they still have kinetic energy and 2) the kinetic energy changes due to correlation of motion, a purely dynamical effect.

To explain what that is: If you have two electrons, then they won't 'move' independently of each other. Since they repel, they'll 'avoid' each other, and so follow different patterns of motion. The kinetic energies of the two particles is horribly intertwined - (http://digitalcommons.uconn.edu/chem_educ/8/" [Broken] a derivation). If you view the electron-electron interaction as simply one electron moving in the charge-density field (equalling the probability density) of the other electron, then you don't get the correct kinetic energy. The difference being what's called 'correlation energy'.

This is why I said the 'probability cloud' way of looking at things isn't entirely correct. It's basically 'wave-like behavior' (as opposed to the 'particle-like' idea of an electron following a planetary orbit). Neither picture is fully correct, although the density-cloud way of looking at things is certainly more correct.

When an electron absorbs or emits a photon, it means its electronic energy has changed by the corresponding amount, it's moved from one orbital to another. so yes, the probability density changes.

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5. Dec 18, 2009

### RedX

Is it even possible to measure the position of an electron in an atom to verify the probabilities?

The equations might say that the electron should be here or there most of the time, but isn't the only thing that can be measured are the energy of excited states?

Also, the electron ought to have zero average momentum, if it's in a definite energy state. If it's in a linear combination of states, then the electron can have an average momentum and bounce back and forth within the atom. What state is the electron usually in? The ground state, or a combination of the ground state and first excited state, etc?

6. Dec 18, 2009

### alxm

http://blogs.nature.com/news/thegreatbeyond/2009/09/electron_clouds_seeing_is_beli.html" [Broken], even more-or-less directly. But of greater practical significance is the indirect case - molecules have the geometry they have because of what the orbitals look like. Quantum mech has an excellent track record in reproducing that, and the structure of thousands of molecules are 'predicted' ab-initio every day.

All electronic orbitals/eigenstates (to the extent that the overall state is time-independent) have a zero expectation value for the momentum.

The ground state, by far. Or molecules wouldn't be very stable. E.g. the first excited state for H2 is the anti-bonding $$\sigma^*$$ orbital. Yet you won't find many of the molecules dissociating spontaneously at room temperature.

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7. Dec 18, 2009

### RedX

Isn't the geometry of a molecule determined experimentally by whether it creates a dipole moment? So for example, for CH4, we would be clueless (unless we can detect higher poles), but H20 would have dipole moment.

Do we observe geometry in the same way as we observe electron clouds?

How would we know if there were a bunch of disassociating $$\sigma^*$$? It would release some photons we could detect over background?

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8. Dec 20, 2009

### alxm

I'm not sure what you mean. Obviously the dipole moment can give a clue to structure, in fact the lack of dipole moment for methane (together with the stochimetric and thermodynamic evidence that it consists of CH4 with four identical bonds) is sufficient alone to determine that it must have a tetrahedral shape.

As a historic aside, the tetrahedral structure of sp3-carbon/methane was known prior to QM and QM's theoretical justification for it was an early triumph of quantum chemistry. Done by Pauling who flubbed and said he'd done the calculation to predict it (1928) when he hadn't yet done the full thing, which he eventually published in 1931. (It's also in his famous book)

Just about every method that could possibly be used to determine molecular geometries is used to some extent; IR, FTIR, Raman, CD, NMR, EPR, X-ray crystallography, neutron diffraction, and the plain ol' "chemical properties".

You could easily detect the presence of H* atoms spectroscopically. But just more generally: There's a reason why UV destroys so many chemicals and causes cancer, etc, as the UV range is where the next (electronic) excited state typically is. Molecules would be a lot less stable, and chemistry as we know it pretty different, if they stayed in excited states to a significant extent.
(Caveat: Transitions between states in the same orbital, e.g. different d-orbitals, is often down in the visual range, which is why transition-metal complexes are so often strongly colored)