Atomic Physics – Why Atoms Can Only Gain or Lose Electrons, Not Protons

atomic-physicsbinding-energyelectronsnuclear-physicsprotons

I know that an object can become net negative or net positive by losing or gaining electrons, and having more or fewer protons than electrons but why can't protons be transferred too?

Best Answer

The energy required to remove an electron from an atom is called its ionization energy. Typical ionization energies are five or ten electron-volts. A visible-light photon carries an energy somewhere under $\rm3\,eV$ and cannot ionize most free atoms. There is enough ultraviolet light in sunlight that atoms on Earth can be preferentially ionized during the daytime, which drives lots of interesting chemistry. However typical temperatures on Earth $(T=\rm300\,K$, $k_BT = \frac{1}{40}\rm eV)$ are low enough that atoms typically don’t ionize spontaneously. The relative stability of atoms against ionization allows stable molecules to exist.

The energy required to remove a proton from a nucleus is called the proton separation energy. Typical proton separation energies are five or ten million electron-volts. In an environment where proton separation was happening, there would be so much energy kicking around that all of the nuclei would be completely ionized, with no bound electrons at all. If you, a biological person made of molecules like DNA and protein, were to visit such an environment, you wouldn’t be made of molecules any more after your visit, and you would therefore have forgotten your question.

It’s not that protons can’t be transferred. It’s just that if we lived in a place where proton transfer was common, we would have a very different perspective on chemistry.

You may actually be aware of some consequences of one nucleon-transfer reaction. Energetic radiation from outer space can cause spallation when interacting with the Earth, either with the atmosphere or with the heavier nuclei under Earth’s surface. Some of the spallation products are free neutrons, which thermalize and behave like a (very tenuous) component of Earth’s atmosphere. The most common species in the atmosphere is nitrogen-14, which interacts with thermal neutrons by

$$ \rm ^{14}N + n \to p + {}^{14}C $$

The carbon-14 will beta decay back to nitrogen, with a half-life of about 5000 years. So if you find an object made of carbon, you can measure the ratio of carbon-14 versus carbon-12 and learn whether that carbon was recently distilled from the atmosphere. Carbon mostly exits the atmosphere to be concentrated in living plants (and in eaters of recently-living plants), while non-organic carbon does not accumulate new carbon-14: a carbonaceous object with carbon-14 in it was probably alive in the geologically recent past, and figuring out how recent is straightforward.

Empirical Electron Mass

Measuring the mass of an electron historically is a multi-step process. First, the charge is measured with the Millikan oil drop experiment, then the charge-to-mass ratio is measured with a variation of J.J. Thomson's experiment.

Millikan Oil Drop

In 1909, Robert A. Millikan and Harvey Fletcher measured the mass of an electron by suspending charged droplets of oil in an electric field. By suspending the oil droplets such that the electric field cancelled out the gravitational force, the charge of the oil droplet may be determined. Repeat the experiment many times for smaller and smaller oil droplets, and it may be determined that the charges measured are integer multiples of a singular value: the charge of an electron.

$$e = 1.60217662 \times 10^{-19} \, \mathrm{C}$$

J.J. Thomson's Experiments

In 1897 J. J. Thomson proved that cathode rays (a beam of electrons) were composed of negatively charged particles with a massive charge-to-mass ratio (as compared to ionized elements). The experiment began with first determining if cathode rays could be deflected by an electric field. The cathode ray was shot into a vacuumed Crookes tube, within which it'd pass between two plates before impacting an electric screen. When the plates were charged, the beam would deflect and hit the electric screen thereby proving that cathode rays contained a charge.

Later he would perform a similar experiment, but exchange the electric field for a magnetic field. This time though, the magnetic field would induce centripetal acceleration upon the cathode ray and produce circles. By measuring the radius of the circle and the strength of the magnetic field produced, the charge-to-mass ratio ($e/m_e$) of the cathode ray would be obtained.

$$e/m_e = 1.7588196 \times 10^{11} \, \mathrm{C} \cdot \mathrm{kg}^{-1}$$

Multiply this by the elementary charge obtained in the Millikan oil experiment, and account for uncertainty, and the mass of the electrons in the cathode ray is obtained.

$$m_e = \frac{e}{\frac{e}{m_e}} = \frac{1.60217662 \times 10^{-19} \, \mathrm{C}}{1.7588196 \times 10^{11} \, \frac{\mathrm{C}}{\mathrm{kg}}} = 9.10938575 \times 10^{-31} \, \mathrm{kg}$$

Empirical Proton Mass

Ernest Rutherford is credited with the discovery of the proton in 1917 (reported 1919). In that experiment he detected the presence of the hydrogen nucleus in other nuclei. Later he named that hydrogen nucleus the proton, believing it to be the fundamental building block to other elements. Since ionized hydrogen consisted only of a proton, he correctly deduced that protons are fundamental building blocks to the nuclei of elements; however, until the discovery of the neutron, ionized hydrogen and the proton would remain interchangeable. How then, was the proton mass measured? By measuring the mass of ionized hydrogen.

$$m_p = 1.6726219 \times 10^{-27} \mathrm{kg}$$

This is done in one of several ways, only one of which I'll cite here.

J.J. Thomson Variation

Repeat J.J. Thomson's experiment with magnetic deflection; but, swap out the cathode ray for ionized hydrogen. Then you may measure the charge to mass ratio ($e/m$) of the ions. Since the charge of a proton is equivalent to the charge of an electron:

$$m_p = \frac{e}{\frac{e}{m}} = \frac{1.60217662 \times 10^{-19} \, \mathrm{C}}{9.5788332 \times 10^{7} \, \frac{\mathrm{C}}{\mathrm{kg}}} = 1.67262 \times 10^{-27} \, \mathrm{kg}$$

Other variations

Other variations may include the various methods used in nuclear chemistry to measure hydrogen or the nucleus. Since I'm not familiar with these experiments, I'm omitting them.

Empirical Proton to Electron Mass Ratio

So now we've determined: $$m_p = 1.6726219 \times 10^{-27} \, \mathrm{kg}$$ and $$m_e = 9.10938575 \times 10^{-31} \, \mathrm{kg}$$

Using the two values and arithmetic:

$\frac{m_p}{m_e} = \frac{1.6726219 \times 10^{-27} \, \mathrm{kg}}{9.10938575 \times 10^{-31} \, \mathrm{kg}} = 1836$, or $1800$ if you round down.

Theoretical Proton to Electron Mass Ratio

Theoretically, you first need to understand a basic principal of particle physics. Mass and Energy take on very similar meanings in particle physics. In order to simplify calculations and use a common set of units in particle physics variations of $\mathrm{eV}$ are used. Historically this was developed from the usage of particle accelerators in which the energy of a charged particle was $\mathrm{qV}$. For electrons or groups of electrons, $\mathrm{eV}$ was convenient to use. As this extends into particle physics as a field, the convenience remains, because anything develop theoretically needs to produce experimental values. Using variations of $\mathrm{eV}$ thus removes the need for complex conversions. These "fundamental" units, called the planck units, are:

$$\begin{array}{|c|c|c|} \hline \text{Measurement} & \text{Unit} & \text{SI value of unit}\\ \hline \text{Energy} & \mathrm{eV} & 1.602176565(35) \times 10^{−19} \, \mathrm{J}\\ \hline \text{Mass} & \mathrm{eV}/c^2 & 1.782662 \times 10^{−36} \, \mathrm{kg}\\ \hline \text{Momentum} & \mathrm{eV}/c & 5.344286 \times 10^{−28} \, \mathrm{kg \cdot m/s}\\ \hline \text{Temperature} & \mathrm{eV}/k_B & 1.1604505(20) \times 10^4 \, \mathrm{K}\\ \hline \text{Time} & ħ/\mathrm{eV} & 6.582119 \times 10^{−16} \, \mathrm{s}\\ \hline \text{Distance} & ħc/\mathrm{eV} & 1.97327 \times 10^{−7} \, \mathrm{m}\\ \hline \end{array}$$

Now then, what's the rest energies of a proton and electron?

$$\text{electron} = 0.511 \, \frac{\mathrm{MeV}}{c^2}$$

$$\text{proton} = 938.272 \, \frac{\mathrm{MeV}}{c^2}$$

As we did with the experimentally determined masses,

$$\frac{m_p}{m_e} = \frac{938.272 \, \frac{\mathrm{MeV}}{c^2}}{0.511 \, \frac{\mathrm{MeV}}{c^2}} = 1836$$

which matches the previously determined value.

Why?

I'll preface this section by pointing out that "why" is a contentious question to ask in any science without being much more specific. In this case, you may be wondering what causes the proton mass to 1800× larger than the electron. I'll attempt an answer here:

Electrons are elementary particles. They cannot (or at least have never been observed to) break down into "constituent" particles. Protons, on the other hand, are composite particles composed of 2 up quarks, 1 down quark, and virtual gluons. Quarks and gluons in turn are also elementary particles. Here are their respective energies:

$$\text{up quark} = 2.4 \, \frac{\mathrm{MeV}}{c^2}$$

$$\text{down quark} = 4.8 \, \frac{\mathrm{MeV}}{c^2}$$

$$\text{gluon} = 0 \, \frac{\mathrm{MeV}}{c^2}$$

If you feel that something is off, you're correct. If you assume

$$m_p = 2m_{\uparrow q} + m_{\downarrow q}$$

you'll find:

$$m_p = 2m_{\uparrow q} + m_{\downarrow q} = 2 \times 2.4 \, \frac{\mathrm{MeV}}{c^2} + 4.8 \, \frac{\mathrm{MeV}}{c^2} = 9.6 \, \frac{\mathrm{MeV}}{c^2}$$

but

$$9.6 \, \frac{\mathrm{MeV}}{c^2} \ne 938.272 \, \frac{\mathrm{MeV}}{c^2}$$

This begs the question: what happened, why is the proton mass 100 times larger than the mass of its constituent elementary particles? Well, the answer lies in quantum chromodynamics, the 'currently' governing theory of the nuclear force. Specifically, this calculation performed above omitted a very important detail: the gluon particle field surrounding the quark that binds the proton together. If you're familiar with the theory of the atom, a similar analogy may be used here. Like atoms, protons are composite particles. Like atoms, those particles need to be held together by a "force".

For atoms, the Electromagnetic Force binds electrons to the atomic nucleus with photons (who mediate the EM force). For protons, the Strong Nuclear Force binds quarks together with gluons (who in turn mediate the SN force). The difference between the two though, is that photons can exist independently of the electron and nucleus. Thus we can detect it and perform a host of measurements with them. For gluons though, they not only mediate the strong force between quarks, but may also interact with each other via the Strong Nuclear Force. As a result, strong nuclear interactions are much more complex than electromagnetic interactions.

Gluon Color Confinement

This goes further. Gluons carry a property called color. When two quarks share a pair of gluons, the gluon interaction is color constrained. This means that as the quarks are brought apart, the 'color field' between them increases in strength linearly. As a result, they require an ever increasing amount of energy to be pulled apart from each other. Compare this to the EM force. When you try to pull an electron from its atom, it requires enough energy to be plucked from its shell into the vacuum. If you don't, it'll jump up one or more energy levels, then fall back to its original shell and release a photon that carries the difference.

Similarly, if you want to pluck an object from a planet, you need to provide it with enough energy to escape the planet's gravity indefinitely (energy needed to reach escape velocity). Unlike the gravitational force and the electromagnetic force, the force binding gluons to each other grows stronger as they grow apart. As a result, there comes an inevitable point where the it becomes increasingly more energetically favorable for a quark-antiquark pair to be produced than for the gluons to be pulled further. When this occurs, the quark and antiquark bind to the 2 quarks that were being pulled apart, and the gluons that were binding them are now binding the new pair of quarks.

^{This animation is from Wikipedia, courtesy of user Manishearth under the Creative Commons Attribution-Share Alike 3.0 Unported license.}

But wait! Where did those two quarks came from? Recall how pulling the quarks apart requires energy? Well that energy is on the scale of $\mathrm{GeV}$. At these scales, the energy may convert to particles with kinetic energy. In fact in particle accelerators, we typically see jets of color neutral particles (mesons and baryons) clustered together instead of individual quarks. This process is called hadronization but is also referred to as fragmentation or string breaking depending on the context or year. Finally I must point out that this one of the least understood processes in particle physics because we cannot study or observe gluons alone.

Proton Mass

So, now going back to the original question. Earlier we noticed that the empirical proton mass was $938.272 \, \frac{\mathrm{MeV}}{c^2}$; but, theoretically its mass should be $9.6 \, \frac{\mathrm{MeV}}{c^2}$. The $928.672 \, \frac{\mathrm{MeV}}{c^2}$ difference arises from the color constraints that binds the three quarks together. In simpler terms: the nuclear binding energy of the proton.