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thank you for your time. I was wondering about the correlation between volume and probability of finding an electron.

we know that if we move away from the nucleus, we are going to get less probability of finding an electron. this happens with a 1s sub-level.

But when i want to know ( graphically ) the probability of finding an electrons in a 2s sub-level, then the volume comes up.

my professor during the lessons said "The PROBABILITY increases because the PROBABILITY DENSITY decreases, but the VOLUME increases."

he also said that this is true to a certain point, since after that it (volume) starts to decrease.

he then said that the DENSITY OF PROBABILITY is 0 in then nucleus.

what i wrote above it's probably messy. i'm just looking for a better explanation of this concept since i didn't find anything online that talks about density of probability and volume.

A.K.
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Ale Chero
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  • Outside of context, what your professor said makes zero sense, or maybe even less (that is, it destroys whatever little sense we could have had previously). The probability increases, he says. *Probability of what?* – Ivan Neretin Feb 22 '19 at 11:07
  • there are lots of answers on this site addressing this eg: https://chemistry.stackexchange.com/questions/92244/what-is-the-difference-between-%cf%88-%cf%88%c2%b2-radial-probability-and-radial-distribut?rq=1 – Buck Thorn Feb 22 '19 at 12:06
  • also: https://chemistry.stackexchange.com/questions/84726/most-probable-point-for-finding-an-electron-in-the-1s-orbital-of-a-hydrogen-atom – Buck Thorn Feb 22 '19 at 12:07
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    This part is wrong: "he then said that the DENSITY OF PROBABILITY is 0 in then nucleus." – Buck Thorn Feb 22 '19 at 12:07

1 Answers1

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There is a straightforward relationship between g(r) - the radial distribution function of the electron (probability per unit radius) - and $\rho(r)$ - the electron density (probability per unit volume).

You can write that

$$\mathrm{g(r) = 4\pi r^2\rho(r)}$$

Here all that's been done is to integrate $\rho(r)$ over a concentric area $4\pi r^2$ surrounding the origin, removing the angular dependence of the density. This works for an s-orbital function with no angular variations in the density.

The $r^2$ factor means that even when the density becomes small away from a nucleus, the total radial probability can be substantial (and reaches a maximum). Inversely, you can have a very high volume density but very low radial probability, for instance at the origin of an s-orbital, where $r^2$ goes to 0.

This is shown schematically (not intended to depict a real H-atom) in the following:

enter image description here

where the red curve illustrates the behavior of g(r), the blue curve of the density.

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Buck Thorn
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