When an electron revolves around the nucleus in P or d-orbitals why does not it collide with the nucleus.
I mean to say that the shape of the orbital narrows near the nucleus , so shouldn't it collide with the nucleus?

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  • $\begingroup$ I'm not too certain those will help the OP. The misconception seems to be around the fact that those diagrams of orbitals show 90% confidence interval surfaces, not surfaces of constrained motion. $\endgroup$
    – user10851
    Jul 4, 2013 at 6:12
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    $\begingroup$ @M.Tarun In any event, only s-orbitals have nonzero probability at the very center. The 2p orbital, for instance, has a probability distribution $(1/96\pi a_0^2) r^2\mathrm{e}^{-r/a_0} (1-\sin^2(\theta)\cos(2\phi))$ in some choice of spherical coordinates, which does indeed vanish for $r = 0$. $\endgroup$
    – user10851
    Jul 4, 2013 at 6:23
  • $\begingroup$ On @ChrisWhite's thoughts I'll re-open this. It brings up a different set of issues. Basically the meaning of "touch" or "collide" is not the same in the world of the very small. $\endgroup$ Jul 4, 2013 at 14:25
  • $\begingroup$ consider the orbital as a twisted loop.It has a point of intersection. In this case I think the electron should go through that point and collide with nucleus. $\endgroup$
    – M.Tarun
    Jul 5, 2013 at 17:38

2 Answers 2


First let's take a step back and try to explain what exactly is being shown in "pictures" of orbitals. When you solve the Schrödinger equation for an electron around a nucleus, you can describe the resulting wavefunction any number of ways. One of the most common is in the position basis, so that the (square magnitude of) the wavefunction gives the probability density for the electron to be "found" at that location if you were to instantly measure its position (i.e. force it into an eigenfunction of the position operator).1

So all we have is this "cloud" - a nonnegative function of $\mathbb{R}^3$ describing where the electron might be. Since it is hard to draw functions of 3D space, what people often do is draw a single surface, usually a surface of constant probability density that contains, say, 90% of the total probability inside of it.2 If you think of the cloud as having varying mass density, we want to depict a natural region wherein we can find 90% of the total "mass."

Just because this surface comes to a point doesn't mean the electron is forced to that point. In fact, such simple diagrams don't really say anything about how the probability density is distributed inside the orbital. Moreover, if you believe the probability density doesn't do anything crazy (e.g. go off to infinity), the shrinking of the surface to a point tells us the electron is actually unlikely to be found near the center.3

As I finish writing this I realize it's something of a loose, mathy explanation for what those shapes are supposed to be. On the other hand, Jim's answer gives a far better explanation of physically what's going on.

1 For more explicit formulas, see an answer I wrote here.

2 The reason I specify a surface of constant density is because there are infinitely many surfaces that contain 90% of the probability. Start with one, expand it a little over here, shrink it a little over there, and you're left with a different 90% surface.

3 In fact, the process whereby a proton captures an electron really only happens with s-orbital electrons. All other orbitals have a node at the very center, so the electrons almost never "come close" to the nucleus.


The shape you see associated with the d-orbital is not the actual shape of the d-orbital. That is a probability cloud of the location of the electron. Outside of the shape, or at least in the not shaded-in regions, the probability of finding the electron drops significantly. This means that the narrowing in the shape of the d-orbital indicates that the region where the electron could be found is shrinking. This can be extended to mean that at the nucleus, the probability cloud effectively doesn't exist. This means not that the electron should collide with it, but that the electron cannot be found there ever.

If you are now wondering how the electron gets from one region of the point cloud to another if they it cannot travel through the nucleus, it is simply a matter of treating the electron as a wave. The waveform exists at one location or another. Like a sound wave, it is possible to have regions of destructive interference where the wave doesn't exist but still have the wave exist on either side. Similarly, the probability cloud shows where you can find the electron. It does not say that the electron is never outside of this region nor does it comment on the velocity or motion of the electron. It can move from one region to another non-connected one; the cloud only implies that you will never (never meaning high statistical improbability) measure the electrons position outside of the region.

Having now read Chris' answer, it excellently fills in my loose description of the important mathematics behind the concepts I was attempting to convey. Anyone reading this answer should definitely read that one as well; they are complimentary.

  • $\begingroup$ I see no logic in your inference. At first, "significant dropping" does not mean elimination. Secondly, the fact that the cloud of probability narrows down to a single point does not mean that electron does not pass through the point. It actually means the opposite: electron has no way to bypass the point. Making things logical also removes the paradox with traveling from one part of the cloud to the other. Try to find another explanation. I would say that as electron comes closer to the nucleus, it accelerates to ∞, which means ∞ speed and thus, 0-probability. But, this is a paradox. $\endgroup$
    – Val
    Jul 4, 2013 at 16:24
  • $\begingroup$ @Val Like it or not, the electron does not travel across the nucleus. Many sources claim it uses quantum tunneling to move to other areas. Also, I can say with 100% certainty that it does not accelerate to $\infty$ since that is slightly above c. There is no paradox, I simply did not want to provide all of the extensive mathematical detail required to fully explain it. I recommend conducting more research into QM; the essence of my explanation rings true with the material from that subject $\endgroup$
    – Jim
    Jul 4, 2013 at 17:39

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