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I was wondering in orbitals that have nodes such as the p orbital, how does the electron move from one lobe to the other? I know that people say it is because of the wave/particle nature of electrons but they don't actually explain how they travel across the node without actually passing it.

My teacher says it is because that electrons are able to act like waves and hence like how EM waves can pass through a wall, electrons are able to pass through the node. However when he says that electrons are like a wave, does he mean that it is literally a wave so does an electron have a wavelength, amplitude, frequency, etc? Because I remember reading somewhere which stated that electrons aren't literally like waves. However they can be mathematically described using concepts that apply to waves. Is this true?

From the comments, they state that electrons can be thought as standing waves. After looking at the internet I came across the following image which explained how electrons act as a standing wave:

electron as a standing wave

However this doesn't make sense to me. Don't electrons exist in orbitals not in orbits around the nucleus? If so, isn't this wrong? Also, since electrons move in orbitals, how can there be any end points as it can move anywhere and its movement isn't restricted to a close loop. Hence how can it act as a standing wave?

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  • $\begingroup$ I think I've seen a very similar question before. In short, yes, it's a wave. Think of it as a standing wave of peculiar configuration, so the concepts of wavelength and frequency do not apply. (There is a so-called de Broglie wavelength, but that's defined for a free electron). $\endgroup$ – Ivan Neretin Aug 30 '15 at 11:12
  • $\begingroup$ You say, 'in short, yes, it's a wave'. I am curious to know what it is actually. $\endgroup$ – Nanoputian Aug 30 '15 at 11:15
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    $\begingroup$ An orbital is not a physical observable. It is indeed just a mathematical construct to solve Schrödinger equation, and not unique. E.g. the p-orbitals could be used the complex set p(0), p(-), and p(+) or as p(z), p(x), and p(y), which is more familiar. The properties derived from this wavefunction are, however, real. Integrate it's square over all space and you get the probability distribution. It's 0 at the nodes. The electron IS on both sides of the nodes but never exactly at the node. Hence it does not travel across it, it just exists at both sides of it... $\endgroup$ – Fedor Aug 30 '15 at 20:59
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    $\begingroup$ Correct if I am wrong, I thought that when you square the wavefunction you get the probability distribution of the electron being at a particular place. Doesn't this just mean that it is possible for the electron to be at different places (i.e different lobes) but in reality the electron is actually only in one position, we just don't know for certain where it is. Because if we were to observe it, we would find that the electron would only be at one position, not at two places at the same time. $\endgroup$ – Nanoputian Aug 30 '15 at 21:08
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    $\begingroup$ I've uploaded 3 pages from Griffiths, Introduction to Quantum Mechanics 2nd ed here. You might find it relevant (it's a very good book but much more oriented towards physics). imgur.com/MOWuMA6,p37pwpQ,6Aopv68 $\endgroup$ – orthocresol Aug 31 '15 at 6:22
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Electrons (indeed all matter and energy) behave both as wave and as particle, depending how the observation is made. The most obvious example is the single-electron interference pattern.

"In 1974 the Italian physicists Pier Giorgio Merli, Gian Franco Missiroli, and Giulio Pozzi repeated the [interference] experiment using single electrons, showing that each electron interferes with itself as predicted by quantum theory. In 2002, the single-electron version of the experiment was voted "the most beautiful experiment" by readers of Physics World." Wikipedia.

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  • $\begingroup$ could you also explain the image that I have uploaded on my question? Don't electrons exist in orbitals not in orbits around the nucleus? Therefore since there is no endpoints to its movements, how can an electron act as a standing wave? $\endgroup$ – Nanoputian Aug 31 '15 at 7:07
  • $\begingroup$ If electrons actually moved in circles or orbitals (or any closed path), they would radiate energy until they fall into the nucleus. The diagram is not a physical path but a graph of the probability of finding the electron in a particular experiment. $\endgroup$ – DrMoishe Pippik Aug 31 '15 at 23:19
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Electron properties are well described by treating them as particles or waves, and the de Broglie equation is useful because it relates wavelength and momentum which are wave and particle properties, respectively.

Remember that there are limits to any analogy!

Electron as standing wave: You can make standing waves with a slinky toy spring, and it takes more energy to get more nodes. Your diagram with n=8 is quite high in energy, but you can probably get n=2. You can say where the spring is going to move for each standing wave and make the analogy that the spring is the electric field and the movement is the electron.

Your diagram tries to show how a standing wave can be in an orbit by bending the 1D slinky toy analogy into a circle. This might be useful in understanding that only standing waves make orbits through a constructive interference analogy.

Electron as particle: If you want to look for the electron particle that has the energy of a standing wave then the probability curve of finding it in 1D is going to look like the slinky toy.

Electron orbitals: These are representations of the probability curve in a 3D for electrons in orbit.

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