In most images of $1s$ subshell I see that there's no node shown at the center, and even the formula $n-\ell-1$ gives 0 as the answer.

But, isn't the nucleus experimentally proven to be at the center? And as the mass should be much more than individual electrons, I'm assuming the nucleus doesn't have large probability clouds like individual electrons do - i.e. the nucleus is mostly at the center.

If my assumption is correct, then shouldn't it be impossible for the electron cloud to occupy any space at the center as it's already taken up? Even for the other orbitals there seems to be empty space in the center, so why not for the $s$ orbital?

I did see one image of the $s$ orbitals where it appeared like they showed a node but that was just one image, every text or any other image says otherwise. So, could someone verify and explain whether there is a node or not?

  • $\begingroup$ s orbitals actually have an antinode at the centre, see the drum membrane anims here: en.wikipedia.org/wiki/… However, although the probability density is high in the nucleus, you need to integrate that probability density to get the actual probability, so the probability of finding an s electron in the nucleus is quite small, since the nucleus is tiny, as explained here: physics.stackexchange.com/a/230103/123208 $\endgroup$
    – PM 2Ring
    May 16 at 15:05
  • $\begingroup$ @PM2Ring I'm a little confused - there's an antinode but at the same time the probability of finding an electron is...small? Wouldn't that be contradictory? Unless I'm confusing the meaning of antinode somehow... $\endgroup$
    – Nirabhra
    May 16 at 15:36
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    $\begingroup$ @Nirabhra the electron is not prevented from being in the nucleus. In fact, for electron capture it both happens and is important for the decay $\endgroup$
    – Dale
    May 16 at 16:40
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    $\begingroup$ @Dale yeah I think I understand better seeing all these answers - knowing that it's not actually being prevented makes a lot more sense since my whole question was based on 'how can it have a high chance of being in a place where it SHOULDN'T be able to' $\endgroup$
    – Nirabhra
    May 16 at 16:42
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    $\begingroup$ Possibly related. Note that the volume of a nucleus is about $10^{-15}$ the volume of a sphere with the Bohr radius, so the overlap between an electron and its nucleus can be small even if the electron’s probability density is larger inside the nucleus than in any other comparable volume. $\endgroup$
    – rob
    May 16 at 22:11

2 Answers 2


These subshells are just the Spherical Harmonics. As you can see from that Wikipedia article, the S orbitals do not have a node in the center.

There is no reason why an electron couldn't be "inside" a nucleus. As long as particles differ in some "quantum number", there is no physical law that would forbid that. For example, an electron and a proton are different. Or two electrons with different spin.

It is that for that very reason, namely that S-orbital electrons do have a finite probability of being within the nucleus, that some isotopes decay via electron capture.

  • $\begingroup$ @PM2Ring thx fixed $\endgroup$
    – rfl
    May 16 at 16:12
  • $\begingroup$ took a while having to read other answers and trying to understand, but I think I get it now, thank you for the help $\endgroup$
    – Nirabhra
    May 16 at 16:45

The idea that 'space is already taken up' is a classical concept which one has to learn to drop. The nucleus and the electron can be at the same point - or, equivalently, if you consider a small volume $dV$ at the origin, the probability that both the nucleus and the $s$ electron are in that volume tends to some finite number times $dV$ as $dV$ tends to zero.

Classical objects can't exist in the same space because of the overlapping electron wavefunctions and the Pauli principle. At the atomic level objects can occupy the same space - and they do.

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    $\begingroup$ I see what you want to say, but, as it is, your statement is wrong. The probability that the 𝑠 electron is in a volume $dV$ around the nucleus is $\left| \psi_{1s} \right|^2 dV$ and goes to zero with $dV$. It is the probability density $\left| \psi_{1s} \right|^2$ that tends to some finite number. $\endgroup$
    – GiorgioP
    May 16 at 17:56
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    $\begingroup$ @GiorgioP - "the probability that both the nucleus and the s electron are in that volume tends to some finite number times dV as dV tends to zero" seems as a good way to say "probability density goes to zero" to a non technical audience. Please notice that it says tends to some finite number times dV, not just tends to some finite number. $\endgroup$
    – Pere
    May 16 at 18:23
  • $\begingroup$ Thanks Pere! I was trying to avoid introducing the difference between probability and probability density. $\endgroup$ May 17 at 12:32

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