I feel like the answer to this question is so simple it must be staring me in the face, but I just cannot find it. If the protrons in the nucleus of an atom are positively charged, what keeps the negatively charged electrons in their orbit and stops the atom, and all matter, from collapsing?
I’m sure someone with more knowledge will come along and make my explaination look like a first grader’s science project (think vinegar-and-baking-soda volcano) compared to a particle accelerator, but here it is anyway:
Electrons, like photons, are both particles and waves. The different orbitals that electrons can take up correspond to the positions where the electrons can form a standing wave. Electorns can’t orbit any closer than the inner s orbital because their wavelength is too long.
Or something like that.
<head in sand>They’re in orbit!</head in sand>
Yeah, what Joe said, iirc.
Though can’t neutrons in a nucleus emit electrons? And hence presumably protons absorb them? When does an electron have to ‘orbit’, and when can it just go straight in? Never mind.
Quantum physics explains why this doesn’t happen. Energy levels of electrons in atoms can only take specific values. The Pauli exclusion principle states that no two electrons can be in the same quantum state at the same time. Together, those principles account for the complicated electron configurations of atoms, and also explain why electrons don’t spiral into the nucleus.
Hmmsies, this is probably wrong, since I dont really have to recall things like this for A-Level physics, but enjoy reading up on them anyway.
According to Richard Feynmann’s quantum theory, a particle travels through space from A to B by every possible path. There are a huge amount of possible electron orbits, but not all of these are seen around the nucleus. To explain this, the orbitting electrons are modelled as waves, with a wavelength corresponding to it’s momentum by the de Broglie equation; [lambda] = h/mv[sup]2[/sup]. Some electron orbits have a circumferunce equal to n[lambda], where n is a whole number. In these, on each orbit of the nucleus, the electron wave would meet other waves and be in phase with them. This leads to constructive interference, and these orbits are orbits where electrons are found; Bohr’s allowed orbits.
The remaining electrons are in orbits which do not have a circumference equal to n[lambda]. Each time they go around the nucleus, they meet other electrons out of phase, and eventually, they are all cancelled out by destructive interference. Electrons are not found in these orbits.
As you can see, electrons cannot really exist in orbits other than Bohr’s. Therefore, if for some reason one is deflected toward the nucleus, it will be cancelled out. This is why electrons do not crash into the nucleus.
Cheers, Harry
wanders off to find a source of information on quantum physics, hoping to prove myself correct
Yes, the Pauli exclusion principle is exactly what’s in operation and can be simply stated for this purpose as " different particles cannot occupy the same space".
You can force an electron to strike a proton, but at the energy levels required, the proton doesn’t behave as a simple proton, but rather as a trio of quarks.
The Pauli eclusion principle can be overcome, for example in Neutron stars where protons and electrons are forced together to create neutrons.
Or a dense cloud of virtual quark-antiquark pairs, gluons, whatever.
And on preview, MC Master of Cermonies provides another good example of how it can happen.
Some clarifications/corrections/etc.:
-
Electrons can coexist with the protons and neutrons in the nucleus. The simplest case is hydrogen, with its one proton and one electron. The electron’s position can only be discussed probabilistically (until it interacts), and the probability of it being smack on top of the proton is non-zero. This is true for any electron that is in a state with zero angular momentum – the so-called S states. All atoms have some electrons in S states (in particular, the two lowest-energy electrons.)
-
Taking the example of hydrogen, the reaction you are asking for is:
[sup]1[/sup]H = e[sup]-[/sup] + p --> [symbol]n[/symbol][sub]e[/sub] + n
where [symbol]n[/symbol][sub]e[/sub] is an electron-flavor neutrino. The system to the right of the arrow has more energy than the system to the left (recall that the neutron weighs more than the proton by about 1 MeV/c[sup]2[/sup]), so the reaction does not spontaneously occur. -
For heavier elements, we have to consider the energy levels of the nucleus itself. We can use carbon-12 as an example. It has 6 neutrons and 6 protons. The Pauli exclusion principle applies to the protons and the neutron separately, so the energy levels look something like:
^
| --empty-- --empty--
| --empty-- --empty--
| --empty-- --empty--
| ----n---- ----p----
energy ----n---- ----p----
| ----n---- ----p----
| ----n---- ----p----
| ----n---- ----p----
| ----n---- ----p----
| (neutrons) (protons)
After the reaction, one of the protons becomes a neutron, but the Pauli exclusion principle forces that new neuton to sit above the other six. The resulting isotope has energy levels:
^
| --empty-- --empty--
| --empty-- --empty--
| ----n---- --empty-- (E7)
| ----n---- --empty-- (E6)
energy ----n---- ----p----
| ----n---- ----p----
| ----n---- ----p----
| ----n---- ----p----
| ----n---- ----p----
| (neutrons) (protons)
Given that neutrons and protons have similar energy levels, this system has (E7-E6) more energy – about 13 MeV. Thus, the reaction will not proceed. (And this is why boron-12 is very unstable, with a half-life of 20 milliseconds.)
Note that the Pauli exclusion principle applies not to the electrons in the atom but to the nucleons in the nucleus. Also, I have ignored spin in the above discussion, but it would only serve to complicate the picture; the results are the same.
- There are nuclear configurations where it would be energetically favorable to change a proton into a neutron. For example, you might expect the nitrogen-12 nucleus (with 7 protons and 5 neutrons) to be higher in energy than a carbon-12 nucleus. It is, in fact, and nitrogen-12 decays via electron capture to carbon-12 with a half-life of 11 milliseconds.
In summary, it usually isn’t energetically favorable to change a proton into a neutron because the Pauli exclusion principle would force the new neutron into a high (nuclear) energy level. When it is favorable, the isotope can decay via electron capture. (Electron capture is a.k.a. K-capture, as the lowest electron level is sometimes called the K level.)
Actually, let me add something… Note that the mere fact that the electron is attracted to the proton is not enough to cause problems with orbit decay. The earth, after all, is attracted to the sun, and we don’t see it spiralling in to a cloud of flaming doom or anything.
Roughly speaking, the reason that the earth isn’t spiralling into the sun can be thought of as the effects of centrifigual force. That’s of course hand-wavy, since centrifugal force doesn’t exist, but it captures the flavor of the explanation simply enough that I think it’s worth putting it in this way.
Now, by the same logic, you could have an electron orbiting a nucleus without worrying about quantum mechanics. The proton would be pulling the electron towards it, but centrifugal force would be keeping this inwards pull from actually doing anything interesting. There’s only one problem with this, which is that when electrons orbit something, classically speaking they have to emit light, and in doing so, they lose energy, and the orbit decays. All that QM has to do is explain why some orbits are stable, and that is where Bohr’s work comes in.
Note also that it’s misleading to say that the Pauli principle is what keeps the orbits stable. The Pauli principle is responsible for the periodic table in that it tells us that we can’t just put all the electrons in the 1s orbit, but it can’t tell us that the 1s orbit is stable. For that, we need energy quantization. Not that anyone actually said otherwise, but in case someone was confused about legion’s point, which was correct as it stands.