# Can lightly-ionized atoms be accelerated to relativistic speeds with current technology?

I know there're lots of various particle accelerators, which can accelerate particles to TeV's of energy, but it seems they all work on electrons or nuclei or other elementary of tightly-bound particles.

Is it possible, using current technology, to accelerate lightly-ionized atoms, like e.g. $\text{Ca}^+$ to speeds of the order of $\frac34c$? If yes, then is it done somewhere? If no, what are limiting factors?

Apart from just possibility to accelerate, would such ions be left intact after acceleration? Or would the accelerating field strip off or considerably excite some of the remaining electrons?

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Why is it of interest whether they're lightly ionized? –  Ben Crowell Aug 19 '14 at 4:00
@BenCrowell because then they'll not lose that much in their size compared to highly ionized state. It'd be a step towards doing experiments with more macroscopic objects than just nuclei or electrons — in the context of special relativity. –  Ruslan Aug 19 '14 at 5:14

In relativistic heavy-ion collisions, you're usually interested in reactions with much higher energy than the kilo-eV scale of a few dozen bound electrons. A typical RHI collision will have something like 10,000 particles in the final state, so a few dozen bound electrons in the initial state won't make much difference either.

What does make a difference is the charge-to-mass ratio. A massive, charged particle in an electric field will have an acceleration $a$ that obeys $$F = ma = qE,\quad\text{or}\quad a = E\frac qm.$$ By "stripping" all the electrons from a nucleus you get a much larger $q/m$, and so you get more energy per ion for the same electric field $E$. That's what the high-energy, quark-gluon plasma folks are after.

However, there are other applications where you might want a singly-ionized nucleus — for instance, to use for atomic or molecular spectroscopy, perhaps loaded into a Penning trap. In that case, between your ion source and your accelerator you have to have some magnetic steering of the beam, which separates the beam by species with different $q/m$. If you went to a nuclear structure laboratory and asked for a beam of singly-ionized calcium at $\frac34c$, they would probably say "calcium-40? -44? -48? or one of the rare or unstable calciums?"

Searching the web for "ion beam spectroscopy" finds

• an old review article which says that "fast ion beams" typically have energies of 10–105 eV.

• at the recently-closed Holifield rare-isotope accelerator an ion spectroscopy facility "… with negatively- as well as positively-charged ions … of exotic nuclei like, e.g., 79Cu and 85Ga … [at] energies of about 200 keV." This facility was probably for nuclear spectroscopy, rather than electronic transitions, but the inner-shell electrons in heavy atoms are fast and can have a big effect on nuclear properties.

• a recent conference proceeding describes spectroscopy of a $\lambda \approx 600\,\mathrm{nm}$ transition in Ar+ ions in a 20 keV beam.

The argon-ion beam should appeal to your interest in special relativity. The ion beam wasn't dramatically relativistic, having $\gamma-1 = \frac{\text{kinetic energy}}{mc^2} \approx \frac12\times10^{-6}$. But the transition frequency was measured with a precision ${\Delta\nu}/{\nu} \approx 6\times10^{-10}$. A change in beam energy of 1 eV, about one part in 104, changed the frequency of the resonance by nearly fifty times the experimental uncertainty. This is the power of precision experiments, that there are complementary ways to access the stranger corners of the world: by going there directly, where the strange effects are obvious, or by looking very carefully where the strange effects are subtle. The argon-ion paper has some very interesting references.

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I was hoping someone would address this low ionization part of the question. I wish I knew more about physics at those energy scales, but there is never enough time for everything. –  dmckee Aug 18 '14 at 23:21
@rob Apart from just possibility to accelerate, would such ions be left intact after acceleration? Or would the accelerating field strip off or considerably excite some of the remaining electrons? –  Ruslan Aug 19 '14 at 8:53
@Ruslan Looks like I was editing when you commented. Let me know if you still have questions. –  rob Aug 19 '14 at 10:03

Not only can it be done, it is being done at several facilities. See e.g. http://alicematters.web.cern.ch/?q=lhc-heavy-ion-program-begins. LHC trumps RHIC's energy by orders of magnitude. It can accelerate heavy ions to 2.76 TeV per nucleon pair, compared with RHIC’s 200 GeV. Since a nucleon has a rest mass of approx. 1GeV, these nuclei are highly relativistic.

Having said that, they are not accelerating Ca, but e.g. lead and gold, i.e. much heavier nuclei.

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the ions in these accelerators are fully ionized, not lightly (for lead, this would be Pb 82+), the force exerted on the fully ionized ions is much larger than on the singly ionized while the mass (to be accelerated) of both is more or less the same. –  Andre Holzner Aug 18 '14 at 20:30
You are right, that's why heavy ion colliders specify energy per nucleon pair, rather than particle energy. To get the total energy in the heavy ion, you have to multiply the 2.76TeV per nucleon pair with the number of nucleons. Using a collider like the LHC with heavy ions, rather than protons packs quite a punch... in this case, almost two orders of magnitude more, to be exact! I find that very cool... almost the same machine, but a totally different physics regime! –  CuriousOne Aug 18 '14 at 20:33
They have also collided lighter nuclei, search for RHIC beam energy scan program –  Hydro Guy Aug 19 '14 at 3:39
@user23873: Thanks for the suggestion. I will take a look! –  CuriousOne Aug 20 '14 at 3:23

The (average) mass of the Calcium ion $\text{Ca}^+$ is around $m_{ca} = 40 amu = 40 \times 1.661 \times 10^{−24} g = 66.44 \times 10^{−24} g$.

That means that to accelerate (from rest) an ion of Calcium to a speed of $\frac{3}{4}c$ the energy needed would be:

$$\Delta E_{ca} = (\frac{1}{\sqrt{1-(\frac{3}{4})^2}}-1)m_{ca}c^2 = 0.512m_{ca}c^2$$

which is around (taking $c=299792458 m/s$):

$$\Delta E_{ca} = 3.057 \mu Joules$$

Now, assuming that the CERN LHC accelerator is the current state-of-the-art in particle accelerators/colliders. The LHC can provide energies in the range of 0.64 microjoules to 92.0 µJ (per particle) depending on type of particles.

As one can see from the previous crude approximations, the acceleration (even to half the speed of light) of (a sufficient amount of particles of) ionised Calcium is out of the ability of current state-of-the-art technology.

Per an error in getting the correct calcium mass and the context of @ACuriusOne's answer. The acceleration of calcium ions is within the range of the LHC collider.

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And why the downvote exactly? –  Nikos M. Aug 18 '14 at 19:38
Because your calculation is wrong. Please fix it. You can look at the state of the art at the CERN website: 2.76TeV per nucleon pair, which is over a thousands times more kinetic energy than rest mass energy of the nucleons. You have simply inserted the wrong mass for Ca. It's not 40gram(/mol) but 40 atomic units, which is roughly the mass of a proton, each. –  CuriousOne Aug 18 '14 at 19:42
@CuriousOne, yes i was looking for the mass of calcium in grams, in this case you are right. i was still looking for the calcium mass to re-validate my answer when the downvote appeared. –  Nikos M. Aug 18 '14 at 19:46
If you correct your answer, I can up-vote it, if you like. –  CuriousOne Aug 18 '14 at 19:47
@CuriousOne, its ok, i can be wrong, but i am still looking for the calcium mass. However reading also the references in your answer, if this is so, since gold and lead are heavier elements than calcium, it is wrong whether i find the correct mass or not. But i want to think a little upon it. –  Nikos M. Aug 18 '14 at 19:50