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I was wondering if it is possible to move the atom nucleus and leave behind the electrons? I can imagine that the electrons will follow the nucleus. But what if the speed of the nucleus is almost the same as the speed of the electrons or faster. where will the electrons go?

If it is not possible, do we have a theory I can read to explain what could happen?


(Edit: as of the comments, "core" actually refers to "nucleus" -changed)

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Yes, I have a theoretical article about this: arxiv.org/abs/0806.2635 . If you transfer sufficiently large momentum to the nucleus, the atom gets excited including a probability to get ionized. This is described with the second atomic form-factor $f_{nn'}(q)$. –  Vladimir Kalitvianski Oct 6 '12 at 10:22
    
The word "core" usually mean nucleus plus inner shell electrons, and these are hard to move together, independent of the outer shell electrons. This is a phase space difficulty, and it is central to understanding phenomena in crystals--- certain things you can dream up just never happen, like knocking out an atom completely valence shell ionized (a "core", in standard terminology). You are using core to mean nucleus, and this is nonstandard. –  Ron Maimon Oct 7 '12 at 16:09
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3 Answers

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Not only is it possible to remove the nucleus from an atom, but the RHIC does it every day!

The RHIC collides heavy nuclei like gold to measure the properties of nuclear matter at high densities. Gold atoms have their electrons stripped off in the Tandem van de Graaff accelerator. The atoms are subjected to such strong electric fields that the positive nuclei and negative electrons are pulled apart.

Response to comment:

See http://isnap.nd.edu/html/research_FN.html for a few more details on how the atoms can have their electrons stripped off (this is a different accelerator from the one at the RHIC). You start with singly ionised atoms. These are easily made e.g. by shining ultraviolet light on the atoms. The singly ionised atoms are accelerated to a high speed than crashed into a very thin carbon sheet. The heavy nuclei plough straight through while the electrons are scattered, and the nuclei are then accelerated away with a second electric field.

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What is the theory behind that assuming I want to learn given that I only know the Physics we studied @college. –  Mahmoud Fayez Oct 6 '12 at 8:14
    
Hi Mahmoud, I've edited my answer to address your comment. I hope this helps. I'm afraid I don't know any more details as this isn't my area. –  John Rennie Oct 6 '12 at 8:23
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To access the nucleus separately from the atom, you need to affect it by a "high-energy probe", one that has a good enough spatial resolution. For example, you may hit the nucleus with another nucleus or X-rays etc. When you do so, it's like hitting it with a small bullet, and if you hit the nucleus with a small bullet, the electrons will continue in their motion almost undisturbed. However, because the nucleus will be kicked away rather quickly, the electrons find out they are no longer parts of a bound state, the atom. So the atom will be ionized: the electrons will be "liberated". It's almost an inevitable consequence of the high-energy manipulation because the energies needed to manipulate with nuclei are multiples of 1 MeV or so, about 1 million times greater than atomic bound state energies that are comparable to 1 eV.

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I am not a physicist but this just came to my mind so I thought to ask. maybe we can use that to generate free electrons and electricity, can we? –  Mahmoud Fayez Oct 6 '12 at 7:58
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It's tempting to think that electricity = electrons, but this isn't true. Electricity is the motion of electrons. There are lots of nearly free electrons in your average piece of metal, so getting free electrons isn't the problem. It's making them move that's the problem. Generating free electrons by stripping them off atoms wouldn't be a useful way to generate electricity. –  John Rennie Oct 6 '12 at 8:12
    
thanks this is really helping me understand more about electricity. –  Mahmoud Fayez Oct 6 '12 at 8:50
    
Let me say John's comment in different words. "Electricity" is an ambiguous word because it may mean the "electric phenomena" in general, "electric charge", or "electric energy". It's the latter that you pay to your power utility for, it's the latter why we need power plants. Energy may be converted from one form to another but the total energy is always conserved: you can't "produce energy out of nothing". And if you strip electrons off atoms, you ionize them, and you actually have to spend energy to do so – you gain a negative amount. You need to spend it to "dig the electrons from holes". –  Luboš Motl Oct 6 '12 at 8:57
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There is a nucleus and there is a core--- they mean two different things. You can knock a nucleus out of an atom by smashing it with a heavy particle, or with a fast neutron, or if it undergoes fission, or in many other ways. This is adressed in the other answers.

But the core of an atom is the nucleus plus all the electrons in inner shells. If you were to remove the core of the atom, you would leave behind only the valence electrons, It might seem that it is possible to knock the core out of an atom, because the inner electrons are more tightly bound than the outer electrons by a factor of order unity.

But there is no real mechanism for doing so suddenly without disturbing the atom. The reason is that a fast particle can knock out the nucleus, but it is very unlikely to knock out the remaining core electrons so that they move along with the nucleus--- the phase space is very small.

This has a practical effect--- it means that the energy required to knock a nucleus out of a heavy-atom crystal by collision is much larger than the energy required to completely ionize away the outermost electrons of the atom in vacuum. This means that K-shell vacancies in a heavy metal can only knock out H-isotopes from their lattice positions, where there is no core, not heavy atoms, because to knock out the heavy atom nucleus requires making two K-shell (innermost S-shell) vacancies, and this is at least twice the energy of the K-shell excitation. So although a vacant K-shell has enough energy to knock out the core of the heavy metal atom, the coherent process which does this is unavailable since it requires a conspiracy which simultaneously knocks out the nucleus and the two K-shell electrons at the same time into nearly the same k, and this is a small corner of the outgoing phase space.

This is not really said in the literature, but it is certainly understood in atomic physics. The knock-out interactions are taken particle by particle, and do not involve the atomic cores.

In your question you use "core" to mean "nucleus", so the other answers are more appropriate, but I think it is interesting to adress the question as stated.

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