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according to bohrs atomic model, the orbits of electrons are quantized and cant have an arbitrary radius, the electron revolving around the nucleus in this orbit, but when I was taught the quantum mechanical model of the atom we were introduced to the concept of orbitals $(s,p,d,f)$ which had different shapes like a sphere, dumbell, double dumbell, and triple dumbell I am unable to comprehend how an electron revolves around the nuclei in these orbits(if it even does because many a time I have been conveniently told that it's merely a cloud of probability of finding an electron and if it is so how does it satisfy $mvr = n\frac{h}{2\pi}$.)

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sandy
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3 Answers3

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For your question, we need to pick up de Broglies concept of electrons as matter waves. Cosider an electron moving (note: moving, NOT revolving) around a nucleus specifically as a wave.

He proposed that when an electron wave moves about, both constructive and destructive in terference must happen. But if destructive interference were to happen inside an atom, the atom would not likely be stable.

For constructive interference, he proposed that

\begin{gather*} 2\pi r\ =\ n\lambda \\ \end{gather*}

Also, by his own hypothesis,

\begin{gather*} \lambda \ =\ \frac{h}{p} =\frac{h}{mv}\\ \end{gather*}

So, in the end, by simplifying the above equation,

\begin{gather*} mvr\ =\ \frac{nh}{2\pi }\\ \end{gather*}

  • my doubt wasn't exactly the derivation of mvr = n(h/2pi ) if the electron is behaving like a wave does it still have an angular momentum moreover can it even be said to be revolving the nucleus? – sandy Apr 24 '20 at 11:31
  • for that question please visit this site. It should provide some clarification.https://wtamu.edu/~cbaird/sq/2014/12/01/does-an-electron-in-an-atom-move-at-all/ – Ishraaq Parvez Apr 24 '20 at 11:51
  • It's time for you to begin to abandon the idea of an electron as a particle or a wave. It's neither; it's something else, something for which we have no example in the macroscopic world. We have no familiar concepts that can be used as a metaphor. – garyp Dec 22 '20 at 13:30
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First of all there are multiple ways to interpret what the wavefunction means (Copenhagen, many worlds, Bohmian mechanics etc.). The most 'textbook' interpretation is the Copenhagen interpretation. It says that the electron is the wavefunction. If you perform a measurement it collapses the wavefunction to one of the possible outcomes of the measurement. For example you could measure the spin which is a quantum property that is similar to spin angular momentum. Spin for an electron can have two values: 1/2 (called spin up) and -1/2 (called spin down). Before measurement the electron can be in a superposition of states. For example the electron could be 70% spin up and 30% spin down. It is both at the same time and this is what makes quantum mechanics weird. If you measure the spin there's a 70% chance of finding spin up after which the electron is 100% spin up and 0% spin down. It has 'collapsed' to spin up.

Similarly the electron is in a superposition of locations. It doesn't have a single position similar to how a water wave doesn't have one position. It is spread out in space and the wave function tells you how likely it is to find it at each location.

Now what do these orbitals have to do with angular momentum? Your picture shows what's called the real representation of the atomic orbitals$^\dagger$. The real representation is the easiest to sketch so this is why it's the most common. In the complex representation it's easier to see how the quantum numbers are related to angular momentum. I highly recommend this applet to see the orbitals simulated. If you pick the complex representation and a state with $m>0$ you get a picture like this.

enter image description here

The rainbow colors rotate clockwise. The colors represent the phase$^\dagger$ of the wavefunction. A phase that rotates in space roughly corresponds to momentum. For example a wave that moves to the right would look like this:

enter image description here

So you can kind of see that orbital with $m>0$ kind of looks like something with momentum that rotates in a ring. But to prevent oversimplification I have to stress that the probability density is constant in time for all these orbitals. They aren't moving. They are static distributions of the electron.

$\dagger$ If you haven't encountered complex numbers yet this paragraph may be a bit confusing. I can try to explain this without mentioning complex numbers if you want.

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    Thanks for the link to the app! I'm sure I'll get good use of this in my QM and Nuclear classes. There is a link to a collection of others at the web page. – Bill N Jan 21 '21 at 20:33
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Your question should have been answered within the very first week of a (good) general chemistry course, but unfortunately that doesn't always happen; the electron is not an object rattling around within these weirdly shaped shells (which aren't even correct descriptions as shown in the picture!), but rather, the electron IS a smeared out wave propagating about space in three dimensions which can be thought to sort of look a bit like the drawings you have given depending on what state it is in. To understand why this is requires quite a bit more technical background and math that ultimately isn't too hideous. But the key takeaway is that the discovery of wave-particle duality--the manifest behavior of material particles as both particulate and wavelike in different contexts--radically changed the way we even conceive of the structure of matter. This led to the development of quantum mechanics, which gives the Heisenberg uncertainty principle (one way of explaining why electrons aren't simply "attracted" all the way into the nucleus), as well as a proper quantum mechanical description of atoms such as the hydrogen atom. The hydrogen atom's exact quantum mechanical description gives rise to the four quantum numbers you have mentioned, along with all the rules for how to order and interpret states. I hope this helps!

Matt Hanson
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