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So the way I understand the question is that it asks us at what r (distance from nucleus) and at what angles is the probability going to be maximum. The way I see probability of finding at a point in a meaningful sense is, just like in 1D probability of finding particle at a point is taken as probability it is in dx region beside it, we find probability of finding it in a small volume near the point of interest. Now the probability density of 1s orbital times dV in polar coordinates will end up containing a sine term (since small volume in polar cooditnates contains sine of angle with z axis). Thus the probability is dependent on not just r but also angle with z axis, which isn't expected for such a spherically symmetric orbital. I am confused here.

Of course one could say that since this is spherically symmetric, probability depends only on r and thus use only radial probability distribution and maximise it. But why is the above way which seems more general giving an apparently different answer.

Sorry if the above part isn't clear or is messy. Just to clarify things, I am not asking a "how to find" question. One way to look at my question is that if instead of maximising radial probability distribution, if I evaluate probabilities in small volume around points and maximise it, my answer should match with the one I get from maximising radial probability distribution. Consequently, I should get all points on sphere with Bohr radius but on maximising probability in small volume around a point, I get points at Bohr radius and on XY plane only (since for maxima, sine must be 1 meaning perpendicular to z axis).

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marked as duplicate by Mithoron, Todd Minehardt, airhuff, ron, Jon Custer Nov 20 '17 at 18:07

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  • $\begingroup$ You don't need $dV$. It's perfectly reasonably to ask what the probability of finding an electron is at a distance between $r$ and $r+dr$ from the nucleus. That's not any different from your 1D case. $\endgroup$ – Zhe Nov 14 '17 at 3:30
  • $\begingroup$ Agree, but taking dV also should give me the same answer, but instead we get an angle dependence also. This is what I am not able to understand. $\endgroup$ – Niket Nov 14 '17 at 3:36
  • $\begingroup$ @Tyberius I did read the answers for the above question, but my question is why is there angle dependence of probability when there shouldn't be? Maybe I am comprehending things in the wrong way, but I am stuck here. $\endgroup$ – Niket Nov 14 '17 at 3:40
  • $\begingroup$ @Niket at least in the answers I saw on that post, there was no angle dependence to the probability or probability density. $\endgroup$ – Tyberius Nov 14 '17 at 3:51
  • $\begingroup$ @Tyberius But it does seem to have as I have written in my question. What is wrong in my interpretation? Shouldn't density at a point times small volume around that point give us probability of finding electron there? Consequently this should turn out to be same on all points on a sphere, but it doesn't. $\endgroup$ – Niket Nov 14 '17 at 3:56
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You are right in saying that an s orbital is spherically symmetric, hence the probability density must not depend on the angles. Also, you are right in saying that $dV$ in polar coordinates contains $\sin\theta$. So what? You don't have to multiply or divide anything by that $dV$ (nor do you need to even know what it is). Old man Schrödinger took care of that for you. $|\psi^2|$ already gives you the probability density. Just as expected, it is a function of $r$ and not of the angles.

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  • $\begingroup$ We have probability density right there, but since we are interested in actual probability at a point (i.e. around it) and not just probability density, shouldn't we multiply density by volume to get the actual probability? $\endgroup$ – Niket Nov 14 '17 at 4:52
  • $\begingroup$ There is no such thing as actual probability at a point. If you try to define it in a natural way, it will turn out zero at any point. To have a non-zero probability, you have to integrate over some finite volume. To compare points, you have to use probability density; that's what it is designed for. $\endgroup$ – Ivan Neretin Nov 14 '17 at 5:03
  • $\begingroup$ But we can use simply probability density to compare points only when all of them have same dV, which they don't have. Otherwise I get the point. $\endgroup$ – Niket Nov 14 '17 at 5:12
  • $\begingroup$ Points don't have dV. They are points. They have zero size and zero volume. dV is a feature of a particular coordinate system. $\endgroup$ – Ivan Neretin Nov 14 '17 at 5:15
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    $\begingroup$ You have it all backwards. These are different kinds of maximum. You may ask about the point with maximum probability density, or you may ask about the spherical shell between r and r+dr having maximum probability. These questions are different and hardly related, and so are their answers. If you need a point, just use density. If it has a maximum at 0, then so be it. Who cares there's no volume there? There's no volume at any single point anywhere, and we are comparing points. Otherwise, if you are comparing spherical shells, then integrate over a spherical shell and go with that. $\endgroup$ – Ivan Neretin Nov 14 '17 at 6:15

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