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Imagine if you were to heat liquid mercury until it reaches a plasma state. Would it be possible to completely separate the free electrons (ions) from the positively charged nuclei (cations)? Next, how would a pure ion plasma react to the presence of

(a) an electrical field and

(b) a magnetic field?

Also, how would the pure cation plasma react to the presence of

(c) an electrical field and

(d) a magnetic field?

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    $\begingroup$ You could certainly ionized Hg or any other element if given enough energy and you could even start to separate the electrons and ions. The problem is that there would come a point where the applied field no longer outweighed the ambipolar field created by the separated charges. Assume we somehow managed to isolate just ions, the problem then would be one of containment and preventing recombination, neither of which are trivial. $\endgroup$
    – honeste_vivere
    Commented May 22, 2018 at 14:27

2 Answers 2

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As was already said by @honestre_vivere in the comment, it is definitely possible to fully ionize your ions, you "just" need to apply enough energy. In fusion plasmas, for example, even high-Z elements are often fully ionized (and can lead to significant losses via bremsstrahlung-radiation, and this is why high-Z impurities are to be avoided in a fusion plasma).

The second part of your question deals with non-neutral plasmas. They do exist and are studied in the lab. Compared to neutral plasmas ("normal" plasmas) they are actually very easy to confine with static electric (and magnetic) fields. Penning traps provide an efficient and easy way to confine them.

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A 100% degree of ionization is not possible as the ions and electrons will be recombining again. So whatever amount of energy you are putting into the gas, there will always be some neutral and only partially ionized atoms present.

You can calculate the degree of ionization in a state of ionization/recombination equilibrium from the equation $$(1)\:\:n_p^2 \alpha = \sigma F n_n$$

where $n_p$ is the plasma density, $\alpha$ the recombination coefficient, $\sigma$ the ionization cross section, F the ionizing flux (e.g. radiation) and $n_n$ the density of neutrals (or only partially ionized) atoms. Now if we define $$N =n_p+n_n$$ we have $$n_p =N q$$ $$n_n = N (1-q)$$ where q is the degree of ionization. Inserting this into (1) gives a quadratic equation for q which has the solution

$$(2)\:\:q = \frac{\sigma F}{2 \alpha N} ( \sqrt{ 1+\frac{4 \alpha N}{\sigma F}} -1)$$

The figure below plots the resulting degree of ionization as a function of the ionizing flux $F$ (assuming all other constants =1).

dgree of ionization

As is obvious, the degree of ionization can never be 100% but only approaches this value asymptotically if $F$ is increased further. The asymptotic behaviour is obtained by expanding the square root in (2) into a Taylor series up to second order, which gives $$q\approx 1-\frac{\alpha N}{\sigma F}$$

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