I've been trying to find an answer to this question, but have come across contradictory answers, and have limited knowledge of quantum mechanics myself.

Almost all the threads (here and on Quora) state that the uncertainty principle prohibits the electron from entering the nucleus, for it would be required to travel with speeds greater than that of light, even though it has mass. However it is quite confusing, given that many support the probability of it being in the nucleus is not zero. Even if that is not true, wouldn't tunneling enable it to do it anyway?

Also when it enters, does it get captured by a proton to form a neutron as in a neutron star? If not what are the conditions required for this to happen?

I'd like to point out that this isn't a duplicate of Why don't electrons crash into the nuclei they "orbit"? as I am already aware of the quantum model, orbitals and principles like uncertainty and Pauli exclusion, and haven't been able to find a conclusive answer no matter where I've looked.

Its also different than What happens to Protons and Electrons when a Neutron star forms? although it's been a bit helpful (I think), but I am more interested in whether electrons and protons fuse into neutrons as soon as they get really close or there are more universal conditions than need to be satisfied before that happens (which obviously are in the specific case of neutron stars).

As Fermi brilliantly put it "Before I came here I was confused about this subject. Having listened to your lecture I am still confused. But on a higher level."

Edit: I wanted to get a bit more specific about the inconsistency I found. "In the quantum mechanical model of the electron, there is a finite probability of finding the electron within the nucleus. During the internal conversion process, the wavefunction of an inner shell electron (usually an s electron) is said to penetrate the volume of the atomic nucleus." Taken from the wikipedia page on internal conversion. Doesn't that come in conflict with the uncertainty principle based argument that the electron would have to travel faster than light?

Edit 2: To better explain my confusion I found two threads that summarize the arguments I find contradictory (although that could possibly be my flawed perception).

https://www.quora.com/Why-do-electrons-not-fall-into-the-nucleus-of-an-atom This is a quora thread about basically the same issue. The second answer by Rod Sid is a common explanation for why this phenomenon doesn't occur.

Yet according to Why do electrons occupy the space around nuclei, and not collide with them? it does. I hope you can see how these two can confuse people. Thanks for anyone that answers it in advance.

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    $\begingroup$ Yes it can: think i.e. to the reaction Potassium-40 to Argon-40. It happens via electron capture (K-capture). $\endgroup$ Jan 15, 2019 at 21:43
  • $\begingroup$ @PietroOliva Thank you for your answer, it gave me something to base my searches on. $\endgroup$ Jan 15, 2019 at 22:05
  • $\begingroup$ Contemplate the wavelength of a multi-MeV electron. $\endgroup$
    – Jon Custer
    Jan 16, 2019 at 0:18
  • $\begingroup$ Hmmm ... from my answer to your first link: 'An example of physics arising because they spend some time in the nucleus is so called "beta capture" radioactive decay in which $e + p \to n + \nu$ occurs within the nucleus.' $\endgroup$ Jan 16, 2019 at 3:12

4 Answers 4


You may want to take a closer look to electron capture phenomenon, also in this thesis.

  • $\begingroup$ Thank you very much for your reply, i was just familiarized with this phenomenon and I think you pretty much covered the second part of my question. I will take my time to review these papers. $\endgroup$ Jan 15, 2019 at 22:08

Actually, a similar question was asked by Feynmans dad to Feynman after he graduated from MIT and Feynman found it difficult to answer - which shows by the way, that it is a good quetion.

As Feynman pointed out, the problem lies with the premises of the question which presupposes things like entering or leaving. These are notions common to a purely material atomic theory, where atoms enter a new configuration or leave one.

In fact, what we have is identity change.

Whilst a neutron may decay into a proton and an electron, in no sense is a neutron ontologically speaking the same as a proton and an electron. The decay describes a process of ontological change. The neutron changes into a proton and an electron.

You might find it interesting that this one of the reasons that Aristotle dismissed the atomic theory, as he regarded such ontological changes as possible, and in fact neccessary, but the atomists merely saw change as a kind of permutation of matter; and this over two millenia ago!

  • $\begingroup$ Thanks for your response. Speaking of Feynman, it is stated in 6 easy pieces that the uncertainty principle keeps the electron from falling in, yet left me with more questions that it answered. Other than the transformation as you may call it of an electron and a proton to a neutron, is there more on its entering the nucleus? $\endgroup$ Jan 16, 2019 at 19:16

In general electron capture does not occur because the final state has much higher energy than the energy that the electron contributes. The simplest case is $^1H$ or the proton. The neutron rest energy is $1.3~MeV/c^2$ larger than that of the proton. Of course, then the question is why does it ? This brings us to nuclear physics and the quark model. The neutron composition is $uud$ and the proton has $udd$. Quark rest energies do not explain this. $d$ is actually $2.5~MeV/c^2$ heavier than $u$. Most of the nucleon rest energy comes from the gluons and so does the n-p difference. For some nuclei the energy actually goes down upon transforming a p into a n so that electron capture can occur.

Then why, in hydrogen, does the electron not go closer to the nucleus than 1 angstrom? The reason is that in quantum mechanics the kinetic energy is proportional to the squared gradient of the orbital and increases faster than its potential energy decreases when it is shrunk.

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    $\begingroup$ Thanks for your reply. In @jacob1729 's answer, in my understanding it was established that the electron actually does enter for extremely short periods of time. That contradicts your statement that it doesn't go closer than 1 angstrom. So simply put, what is going on here? Does it get in there? $\endgroup$ Jan 16, 2019 at 15:38
  • $\begingroup$ @DuckHunterZx I meant that the expectation value of the distance to the proton is not smaller than half an angstrom. $\endgroup$
    – my2cts
    Jan 16, 2019 at 16:38
  • $\begingroup$ @my2acts Thanks for your clarification. So this doesn't contradict the other answer, right? $\endgroup$ Jan 16, 2019 at 16:41

All the references you refer to use a sort of semi-classical semi-quantum language, and you yourself admit you are confused about quantum mechanics. So it might be helpful to have a 'pure' quantum answer (although it might not directly address your question):

Quantum particles do not possess a property called 'position'.

This sounds strange, since you can measure where the particle is (eg by constructing a box that tells you if there is a particle inside of it, then waiting until it beeps to signify the particle is in the box). Measurement is a complicated (and I think poorly understood?) process where an object so complicated that you need to resort to a classical system (the measuring device) interacts with a system you are trying to treat quantum mechanically. The result is tainted by the classical measurement device and you end up with classical notions like position in your output.

So the simple answer to the question is that no, an electron cannot be in the nucleus, because it cannot be anywhere. It does have a finite chance to be found in the nucleus if you were to measure it (again, imagine putting a box around the nucleus that tells you if there is an electron there, there is a finite chance it says yes). The fact that electron capture occurs means that the full quantum solution means there is a chance that the atom will later be found in a state which describes one less proton and electron and one more neutron. The semi-classical interpretation is that the electron spends a small fraction of its time in the nucleus (corresponding to whatever the probability of your magic measuring box saying it is there) and so there is a chance to "tunnel" in to the nucleus, fuse with an attractive proton and create a neutron. These semi-classical pictures can give accurate answers but they are not the full quantum picture.

Edit 1: Uncertainty Principle The argument with the uncertainty principle is I believe being misapplied. Leaving aside any issues of interpretation, the uncertainty principle argument runs that if you locate the particle to be in the nucleus, then you must only know its momentum to some uncertainty:

$$\Delta x \Delta p \sim h$$

Where $h$ is Planck's constant and $\Delta x$ is the uncertainty in our position measurement. If we are claiming to have measured the electron to be "in the nucleus" then we must have $\Delta x$ roughly the radius of the nucleus. (It's been pointed out in the comments that the notion of radius is not commensurate with positions not being quantum mechanical. That's okay, we're treating the nucleus classically here anyway.) This forces $\Delta p \sim \frac{h}{\Delta x}$ and so the uncertainty in the particle's momentum becomes infinitely large as we try to confine the particle to a given location.

There are two things to comment on here: first there is the rather boring statement that the uncertainty in the momentum is not actually the momentum itself. Clearly if $\Delta p$ is very large we expect $p$ to have a probability distribution peaked around some large value that is at least $\Delta p$ but it doesn't preclude $p=0$ directly. Secondly (and more importantly), the mechanism for stopping electrons "crashing into the nucleus" is that as their momentum increases due to the uncertainty principle, they spend less time in the nucleus and the amount of time they do spend is not enough to fuse with the nucleus (i.e. they go through it too fast to fuse usually). It is NOT that the electron starts moving faster than the speed of light $c$ because the usual formula $p=mv$ stops working at relativistic speeds in such a way that the speed remains bounded below $c$ even as $p$ becomes arbitrarily large.

  • $\begingroup$ So could we more accurately describe the question by saying both the nucleus and electron are two "boxes" and then saying that whenever they intersect or the electron box is in the nucleus box we get a beep? Or we don't even get to describe an electron as that box? Also wouldn't the uncertainty principle prohibit it from entering the box in the first place? Thank you for your reply, and your read is accurate, i am still coming to terms with these quantum concepts (all i've done so far is classical mechanics). $\endgroup$ Jan 15, 2019 at 22:12
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    $\begingroup$ How do you reconcile your statement that quantum particles do not have positions with e.g. the value of the proton radius of less than a femtometer? $\endgroup$
    – my2cts
    Jan 15, 2019 at 22:15
  • $\begingroup$ @DuckHunterZx the boxes symbolize measurements. When you're not actively measuring, the nucleus and electron simply don't have positions. The situation is like this: in classical physics you can describe a particle via its position and momentum $(\vec{q},\vec{p})$ and that gives the state of the particle. The quantum state does not have a way of getting a position out (it is sufficient to work out the probability the electron being in the box though). $\endgroup$
    – jacob1729
    Jan 15, 2019 at 22:16
  • $\begingroup$ @my2cts I don't know anything about nuclear physics, is your statement qualitatively the same as the statement that a hydrogen atom has a radius less than an angstrom? If so, I feel it doesn't invalidate my point and that this could be phrased in terms of a measurement process (that outputs a sharply peaked value for radius at somewhere less than a femtometre). $\endgroup$
    – jacob1729
    Jan 15, 2019 at 22:19
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    $\begingroup$ @DuckHunterZx I don't follow the faster than light argument. Is the reasoning "if it is confined to the size of the nucleus then its momentum has an uncertainty corresponding to more than light speed, and so this is impossible"? If so it's bogus. $\endgroup$
    – jacob1729
    Jan 15, 2019 at 22:35

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