# B Split an atom into each and every nucleon

1. Dec 16, 2016

### Albertgauss

Has anyone ever tried to split an atom up into each of its individual nucleon? If not, what's the closest anyone has come to? For example, if Uranium has 238 nucleons, has anyone ever tried to split it up into each and every last proton and neutron? Can anyone even do it for a light element like Carbon or Oxygen? I'm not sure how easy or hard what I'm asking is in this question.

2. Dec 16, 2016

### Staff: Mentor

You would have to do it in multiple steps, catching the fission products each time: if you try to do it in a single collision, it has so much energy that you will destroy many of the protons and neutrons and create new particles - the LHC does this with lead collisions (to study the new particles produced). Completely impractical, and there is no point in it.

3. Dec 16, 2016

### stoomart

My understanding is the lighter the element, the harder it is to separate the nucleons due to their binding energy, and is "completely impractical" as mfb said.

EDIT: I assume this happens to all atoms during spaghettification when captured by a black hole, before their subatomic particles are destroyed by the singularity.

4. Dec 16, 2016

Staff Emeritus
What would be the point? What would you learn from such an enterprise?

5. Dec 17, 2016

### ohwilleke

I get it.

Clearly a lot of the readers in this forum don't understand why anybody would climb a mountain or swim across the English channel either. Not all things you can do in science are utilitarian.

6. Dec 17, 2016

### snorkack

Baryon numbers are conserved and all baryons save nucleons have half-lives under a nanosecond.
So, when you collide two lead 208 nuclei with each other, what are you going to have a nanosecond after? 416 lone nucleons, plus some nucleon/antinucleon pairs? Or will some nucleons remain stuck in small and tightly bound fragment nuclei, like a few alpha particles, even while most nucleons acquire kinetic energy far in excess of their binding energy?

7. Dec 17, 2016

### vanhees71

Well in heavy-ion collisions a nanosecond is like eternity. The usual scale in this physics are femto-meters (Fermi) $1 \; \mathrm{fm}=10^{-15} \; \mathrm{m}$. In ultrarelativistic heavy-ion collisions such as performed at the CERN SPS, LHC and BNL RHIC you collide two heavy nuclei like Pb or Au. Usually they collide with some finite impact parameter, i.e., you have a non-central collision. You can think of this in terms of the Glauber model, i.e., just by geometry, a part of the nucleons really collide (or at this energies the corresponding partons within these nucleons). They are called participants, and the rest are the spectators.

In this collision some very hot and dense lump of matter is created, and that's the purpose to investigate these collisions. The reason is that there is overwhelming evidence that this matter is consisting of partonic degrees of freedom (at least in the sense of quasi particles). We know from lattice-QCD calculations at finite temperature and vanishing baryo-chemical potential (i.e., a system with as many bayrons as anti-baryons), which is the situation at very high collision energies reached at RHIC and LHC, that at temperatures of about $T=150-160 \; \text{MeV}$ there is a cross-over transition from a hot and dense hadron gas to such a matter consisting of partons, which is called the Quark Gluon Plasma (QGP).

The difficulty, however, is that we cannot directly observe this (QGP) or even have some box with hot and dense QCD matter at a given temperature and baryo-chemical potential at hand, but the just produced QGP rapidliy expands and cools down, and all we can measure are hadrons (and also leptons and photons) with our detectors, and this gives a snapshot at two stages of this rapid fireball evolution: When the fireball cools down and becomes dilute enough at a certain point the inelastic interactions cease, and the "chemistry" of the fireball is fixed (modulo decays of resonances, which survive longer). Although this is a highly dynamical medium one can describe this chemistry with an astonishing accuracy with a thermal model, assuming that directly at chemical freeze-out the medium is described by an equilibrated medium consisting of the known hadrons, i.e., a hadron-resonance-gas model, and this works over a wide range of collision energies (even down to the low energies at GSI in Darmstadt). At the high energies the chemical freeze-out temperatures are close to the pseudocritical temperature of the cross-over transition between a QGP and a hot and dense hadron-resonance gas. Then the now hadronized fireball cools down and dilutes out further, but elastic collisions are still going on until also these collisions become inefficient (i.e., the mean free path of the hadrons becomes larger than the extension of the fireball), and thus the particles move more or less freely to the detector. Through measuring the energy and (transverse-)momentum spectra of these hadrons one can infer on the state at this thermal freezeout, described by a temperature, various chemical potentials, and a flow field of the medium. At RHIC and LHC you get typical thermal-freezeout temperatures of around 100 MeV.

We infer all this from the fact that the evolution of the bulk of the fireball can be pretty well described by relativistic hydrodynamics, assuming an astonishing short formation time of the plasma (i.e., when the matter reaches local equilibrium and hydrodynamics becomes applicable) of $\lesssim 1 \; \text{fm}/c$, and the shear-visocisity-over-entropy-density ratio is close to the bound $\eta/s \geq 1/(4 \pi)$. In fact the NY times ones titled "The most sloshy liquid" about the findings on the QGP at RHIC.

8. Dec 17, 2016

### Staff: Mentor

More from the experimental side: The most central collisions typically produce something like 10000 hadrons. There are 416 more baryons than antibaryons, but that is a small asymmetry. As an example, here is a prediction for the antiproton to proton ratio (solid red line), for high energies it is extremely close to 1. And even the remaining slight asymmetry is not from surviving nucleons: all those nucleons are newly produced when the fireball cools down.

In less central collisions, some parts of the nuclei don't collide directly. They fly away, disassembling quickly, but those fragments are too close to the beam direction to be measured in the detectors.
There is a difference between "swimming across the English channel" and "swimming across the English channel, but only in 100-meter steps that are either precisely South or precisely East, while carrying a rubber duck".

9. Dec 17, 2016

### snorkack

The threshold for nucleon-antinucleon pair creation from proton-proton collision is 6 rest energies of proton (5,7 GeV).
The total binding energy of lead nuclei is just over 1,6 GeV.
If you have 2 lead nuclei colliding with total energy of 6 GeV, that should be enough energy to produce 3 nucleon-antinucleon pairs.
It also is just 15 MeV per nucleon.
So, what´s the energy where a colliding pair of lead 208 nuclei produces, not an average of 5200 nucleons and 4800 antinucleons, but an average of 417 nucleons and 1 antinucleon?

10. Dec 17, 2016

### Staff: Mentor

The first number is for a fixed-target experiment, the second number applies to the center of mass system, mixing two different frames is misleading.

Pion production just needs 150 MeV, excited states of (edit) nucleons start at ~300 MeV.

If you collide the nuclei with just 10-50 MeV per nucleon pair to avoid pion production, you'll probably get several fragments with much more than 1 nucleon each, flying away at high speed. You will certainly not get 416 separate nucleons.

Last edited: Dec 18, 2016
11. Dec 18, 2016

### snorkack

In the centre of mass of an at all massive nucleus, the energy of an incoming proton is close to its energy in the stationary frame.
So, what´s the threshold energy of antiproton production for a proton colliding with a nucleus? If the nucleus could take the recoil, it would be just 2 GeV. But the nucleus is not bound as strongly as that...
Um? They start at 7 eV!
Above pions, you have muon pairs (about 220 MeV), delta resonances (from 300 MeV) and hyperons (starting from about 700 MeV - about 180 MeV difference between proton and lambda mass and about 490 MeV for the kaon to carry away the weirdness).
Do you also get a lot of hypernuclei?

12. Dec 18, 2016

### Staff: Mentor

p->stationary heavy ion could produce an antiproton/proton pair with just ~2 GeV proton energy, but that is an extremely unlikely process.
Sorry, meant the individual nucleons, not nuclei.
Probably not at that energy, but certainly at higher energies.

13. Dec 19, 2016

### Albertgauss

Obviously, as can be seen from the other replies, there would be much difficulties in getting such a scenario to work if there is even the technology to accomplish such.

One basic reasoning is to try to understand how the energy is distributed throughout the atomic nucleus. I thought it might be useful to figure out how the energy adapts to each nucleon as it comes into the atom. If you think of the atomic nucleus as one giant stable quark gluon plasma, it might be a way to study how the quarks respond to small increments of quarks coming in at a time (obviously in the form of one nucleon at a time) or leaving at a time.

Also, I think it’s known that most light elements are formed from stars but the heavier atoms must come from something like a supernova. But I think this is something vague in that, you know what pressures and temperatures are needed to make heavy nuclei, but it is not known the exact steps such processes. For example, we know that fusion is simply adding to protons together to make a helium and this is the basis of all energy emanating from the sun. But it seems to be not that simple as fusion is still very hard to make work even after 60 years in the labs that can attain millions of degrees. Also it seems weird to me that when heavy elements radioactively decay to get to a more stable nucleus, they only do so along very strict rules. You can eject a helium atom; you can have a beta decay, but why are not other nuclei possible? Why can’t you eject a carbon nucleus? Why not eject a bismuth nucleus? I feel like there is much going on inside the nucleus of an atom we do not understand yet.

Maybe the other experiment would be more useful, at least is finding out how heavy elements are indeed made. Maybe the other experiment would be to add one nucleon at a time to get up to some heavy nuclei. That might lend more inference with the details of the processes of how heavy elements are actually made. Based on what we know about the atom, it might not be as simple as adding one nucleon at a time. It may be like the reverse of radioactive decay, where to build up a nucleus you can only do it with certain nuclei.

I admit this is mostly all imagination right now, but it’s fun to think about such things.

14. Dec 19, 2016

Staff Emeritus
Energy levels are already known, nucleus by nucleus. Building them up one nucleon at a time would add no new information.

It's not. We only discuss conventional science here.

They are possible. Not just possible, but observed. There is neutron emission, proton emission, double proton emission, cluster decays and mid-weight nuclear emission, usually called spontaneous fission.

Your argument that because you don't understand something that nobody understands it is fallacious.

15. Dec 19, 2016

### Staff: Mentor

Do not mistake "You don't know it" for "no one knows". All the processes you mention are well understood. A project to split a specific uranium atom into 238 individual nucleons would not improve our knowledge there in any way. There are tons of experiments studying a huge range of nuclei. You don't have to take a single nucleus and split it all the way into individual nucleons, you can simply study every isotope you can produce (~3000 so far, only 10% of them occur in nature).
It is not a quark-gluon plasma.
The fusion part has been figured out decades ago, we know precisely how much fusion will happen under which conditions. The remaining issues are not about fusion, they are about creating a plasma large, hot, dense and long-living enough to make it interesting.
Carbon is possible but rare (cluster decay), bismuth is so heavy that it is the main decay product, not the thing that gets emitted, but of course you have decays producing a bismuth nucleus.

16. Dec 19, 2016

### Albertgauss

All right, I think I'm good here. I did not realize decay by the other nuclei was already established. I'm satisfied with the responses to this post.

Sorry about that "I don't know" --> "No one knows". I'll be more careful about not putting these kinds of comments next time.

17. Dec 19, 2016

### snorkack

... only the long-lived states.
If we take a lead or bismuth or uranium nucleus, with merely 1600 or 1800 MeV binding energy in total, and hit it with a 2000 MeV particle, how will that energy get distributed around the nucleus?
I suspect it might get distributed quite unevenly. Like, the 2000 MeV particle knocks out 2 nucleons that get 1000 MeV between them (so escape with a 500 MeV), then of the remaining energy 500 MeV is given to 4 nucleons (escaping with a 125 MeV each) then from the rest another 250 MeV goes to 8 nucleons (escaping with 31 MeV each)... and at that point, the next 125 MeV to 16 nucleons is not enough to meet their several binding energy. 14 nucleons have carried out 1750 MeV and the remaining 250 MeV distributed across the remaining 194 nucleons means any single nucleon has just 1,3 MeV free energy, not enough to escape.
If a heavy nucleus is excited with 10 or 20 MeV energy, which should be enough to unbind one nucleon, but which is spread evenly among 200+ nucleons, with just 50 or 100 keV per nucleon, how does the energy get back together to unbind one specific nucleon? How long does this process take?

18. Dec 20, 2016

### Staff: Mentor

A few nucleons will get ejected at high energy. The remaining nucleus can get into an excited state, emitting protons, photons or (rarely) neutrons after a while, possibly undergo beta decay later, and become a stable or at least long-living nucleus again.