# How is Alpha Radiation possible?

Alpha radiation would seem to occur when a pair of protons and neutrons are magically plucked from the amorphous (i.e. having no particular structure) nucleus of a heavier atom.

Some of the problems associated with this approach, but which are conveniently ignored, include:

• What happens to the electron pair associated with the removed protons?; Is an anion created?
• how can a positively charged alpha particle manage to pass through heavily populated electron orbitals without disturbing or interacting with them?;
• Why do only four nucleons separate from the nucleus - is this some type of magic number? Why not 1 nucleon (a hydrogen atom or a neutron), 2 nucleons (e.g. deuterium), or 3 (e.g. helium-3) etc. ?

Because of the problems with orbital electrons, it is quite difficult to find a diagram that shows exactly how the alpha radiation process takes place. Most diagrams simply ignore the existence of the orbital electrons; a few show them, but only in an over-simplistic and unconvincing way (see lower figure below).

• The nucleus has structure. Pictures to convey the basic idea are entirely correct. Alpha particles pass through electron clouds all the time in, e.g., Rutherford Backscattering Spectrometry to measure composition profiles. An energetic alpha treats an electron about as well as your car windshield treats a gnat. Apr 8 at 12:47
• Further to @JonCuster’s clarification, you might enjoy this brief overview of the field of nuclear structure.
– rob
Apr 8 at 20:05
• Did you ever think about why radiation is dangerous? The huge energy difference between the nuclear structures (relevant for radiation) and the electron structures (responsible for all chemistry) is essentially exactly that. The two extra electrons that are (eventually, or quite immediately) emitted from the parent atom pale into insignificance compared to the utter mayhem a lone (highly energetic) alpha particle does to the electron structures of everything it comes close to. It effortlessly devastates even the strongest chemical bonds, including DNA (-> cell death, cancer and all that). Apr 11 at 8:18

It's definitely not the case that the electron cloud is undisturbed. The process is quite violent on the scale of the atom undergoing alpha emission.

The electron cloud is excited by the process of the emission of the alpha particle, so the remaining ion isn't in its ground state. Moreover, the remaining nucleus also gets some recoil from the alpha particle, which also adds to disturbance of the electron cloud. The disturbance may in fact lead to some electrons being ejected from the ion.

For example, here is a 1975 paper which reports the "shake-off" of electrons from polonium-210 nuclei during the alpha decay to lead-206. The experiment observed simultaneous alpha particles, emitted electrons, and x-rays from the filling of the "hole" in the lead's electron cloud. Current best results suggest about ten or fifteen electrons are ejected per million polonium alpha decays.

Why it's usually "conveniently ignored" is because chemistry is rarely of interest when discussing nuclear physics.

• +1 Also, the lost electrons generally won't leave with large ionizing energies so they're also not really important when accounting for radiation dose or subsequent cascading reactions - they just drift away, effectively, finding a home somewhere else. They're not like beta radiation, for example, which are electrons (or positrons) with significant kinetic energy.
– J...
Apr 8 at 18:35
• I suspect that this answer (v2) is incorrect. The “sudden approximation” for decays, in which the electron cloud is undisturbed during the nuclear decay, is frequently useful. In polonium-210 (search result), which is famously a nearly pure-alpha emitter, all of the emitted x-rays, Auger electrons, and conversion electrons seem to be associated with the rare alpha decay to an excited nuclear state rather than the primary alpha decay to the ground state. Furthermore,
– rob
Apr 8 at 20:48
• … I’d expect the overlap integral to be largest for alpha particles to scatter from the innermost (K-shell) electrons. For actinide alpha-emitters, this would predict additional alpha energy lines separated by the K-shell electron binding energies of $\sim 0.1\,\rm MeV$, or by the slightly smaller energies of L- or M-shell x-rays. I remember no such features from the limited time I spent studying alpha spectroscopy. The usual lesson is that alpha decays have a monoenergetic two-body final state.
– rob
Apr 8 at 20:49
• @rob but this is counterintuitive: if the potential changes suddenly, the wavefunction of the electron cloud should retain its original form, and it is not an eigenfunction of the new Hamiltonian that lacks two units of attraction in the potential. This is similar to the case of sudden expansion of a quantum well as treated here. What am I missing? Apr 8 at 21:23
• In your linked answer (which is lovely, btw) you change the size of the potential well, so the eigenfunctions are very different. But in a nuclear decay, we change the strength of the attraction without changing its $1/r$ behavior. Consider that sulfide $\rm S^{2-}$, chloride $\rm Cl^-$, argon $\rm Ar$, and the sodium ion $\rm Na^+$ all have “the same” electron configuration $[\mathrm{Ne}] 3s^2 3p^6$. There is a lot of physical chemistry in the scare quotes around “the same.”
– rob
Apr 8 at 22:41

What happens with the electrons is generally ignored because it has no bearing on what happens to the atomic nuclei. By the time any interactions with the electrons might occur, the nucleus has already separated. Two electrons might come along with the alpha particle to form a helium atom, or the end state may be a pair of ions, or some number of electrons may be knocked loose and go elsewhere. Regardless, they have little effect on the alpha particle, and typically the focus of attention and study relating to alpha radiation is on the alpha particle or other atomic nuclei.

As for why it's specifically an alpha particle that separates from the nucleus:

An atomic nucleus is stable because its composition and arrangement form a local minimum energy state. If the nucleus were a macro scale object, classical mechanics would require that in order for it to change, such as by emitting some number of separated nucleons, it must first gain enough energy to climb out of that local minimum and cross a nearby local peak energy state on its way to a new and different local minimum. However, an atomic nucleus is actually small enough that quantum mechanics have strongly relevant effects, and quantum tunneling allows the nucleons to randomly skip past such a peak on occasion.

Even with quantum tunneling, however, energy is still conserved, so the end state has the same overall energy as the initial state. The energy of the remaining large nucleus, plus the energy of the emitted particle(s), thus must be no greater than the initial energy of the pre-emission nucleus.

An alpha particle has an exceptionally low amount of energy for the number of nucleons in it, and emitting one also has minimal effect on the ratio of protons to neutrons in the remaining nucleus. These factors combined make the total energy of an alpha emission end state low enough for it to be reachable from the nucleus's ground state without violating conservation of energy.

Emitting a proton, neutron, deuterium, tritium, or helium-3 nucleus may be theoretically possible, but would require a higher initial energy state. Such high nucleus energy states generally only occur when another particle collides with the nucleus, and electromagnetic forces tend to prevent that in normal Earth environments.

• Just as a question, regarding these other (hypothetical) decay modes: Even if the energy for some less-likely decay were pumped into the nucleus, wouldn't the nucleus also simply emit one or more alphas before that energy could somehow result in one of these other decay products? Apr 8 at 21:45
• @SoronelHaetir, Yes, exactly. There are other modes of cluster decay, including spontaneous fission, yielding products heavier than an $\alpha$, but the $\alpha$-emission is the most energetically preferred. The only exception I'm aware of is the "neutron drip" process in large nuclei with a significant excess of neutrons, when the nucleus sheds an extra neutron. Bound neutrons are slightly repulsive, though not anywhere near the $pp$ Coulomb repulsion, and the strong interaction has a very short range.
– kkm
Apr 8 at 22:10
• “and cross a nearby local peak energy state”—in fact, the decay happens by tunneling, not suddenly gaining enough energy in a fluke to “jump over” the strong interaction barrier. That would have been unrealistically slow/rare.
– kkm
Apr 8 at 22:23
• @SoronelHaetir When you shoot a heavy nucleus with some GeV range particle, it will indeed boil off all kinds of particles. Including, but not limited to, free neutrons, protons and deuterons. This process is called spallation. It's used as a means to produce highly controlled neutron radiation, but also happens due to cosmic rays hitting into oxygen and nitrogen atoms. Apr 9 at 11:23
• Thanks for fixing that. Apr 9 at 17:31

These are exactly the same question I struggled with when I was young, but answers are rather simple.

1. What happens with the two electrons:

Actually, way more than only 2 electrons get involved.

Think about the energy scale of the alpha decay (few MeV) and the energy scale of the binding between the outer electrons and the nucleus (few eV). A single alpha particle can kick millions of electrons out of their home atoms before it slows enough to think about own electrons. This is how alpha particles are detected - by the vast amount of ions they create.

The nucleus that emits the alpha particle gets enough recoil not only to lose a great deal of its own electrons, but to leave a small ionized trace of its own.

The energy scale is linked to the timescale as well. Electrons regroup around the new nuclei (the alpha particle included) much, much later.

1. How the alpha particle manages to escape the atom?

See the energy scale again. If it happens to interact with an electron of the parent atom, the only possible outcome is that the electron is also kicked out, not slowing the alpha particle much.

1. Why only four nucleons?

Well, not always. Different isotopes are known to emit electrons or positrons (see beta decay), neutrons (see the neutron drip line), protons (see the proton drip line), alpha particles (you already know) as well as heavier nuclei all the way to splitting the original nucleus roughly in half (see nuclear fission).

Which one of these mechanisms will happen depends mainly on the total energy (mass) of the decaying nucleus and the possible decay products.

Alpha particle is rather easy to emit because it has rather high binding energy, i.e. it has much lower energy than its constituent nucleons when separated.

Sometimes, more than one decay mode is possible when considering the energies involved. They really do happen (at different rates). In the quantum world, if a process is at all possible, it is obligatory.

With a particle accelerator (and patience, and proper detectors) one can kick whatever particle wants from a nucleus - including, but not limited to, deutrons, He-3 and likes.

1. Unstructured conglomerate of nucleons

Plain wrong approach. Nucleus is pretty much structured - protons and neutrons (separately) occupy orbitals pretty much like the electrons do. They are even called s,p,d,f just like electron orbitals are.

Nucleons are spherical... in a sense. The quarks inside have their own orbitals. Likely s-type because in the nucleons we have at most 2 quarks of the same sort.

• There are some other good answers, but this one is the best. Apr 11 at 14:05
1. The two corresponding electrons are left behind, so the atom changes into an ion after emitting an alpha particle.
2. The strong interaction with its finite range forms a potential well in the Coulomb potential around the core. Alpha radiation can pass through the wall because of quantum tunneling, even if its energy is below that of the wall.
3. An alpha particle has a relativly high binding energy of $$28.511\;\mathrm{MeV}$$.

Alpha particles leave an ion behind. However, it is the nuclear process rather than the ionization one the process that is studied. When an alpha particle passes near an electron orbital, the particle will probably take that electron indeed. That is an important property of alpha particles: they are very electronegative and tend to take any electron near them.

But, why are they Helium nuclei? Why not Oxygen?

In fact, it is because of the stability of the Helium nucleus. If the atom lost a Lithium nuclei this nucleus would get rid of two protons to become more stable, so that you end up with an alpha particle.

• Although appreciated, I have a few problems with it. For instance, Wikipedia (en.wikipedia.org/wiki/Alpha_particle) emphasizes that an alpha is ‘particle identical to a helium-4 nucleus’ and all diagrams I see have no electrons, orbital or otherwise. And, yes, the alpha is a very strong stable particle, but the fact that only helium-4 like alphas are produced suggests that the nucleus may essentially consist of them. With alpha decay being predominant in heavier elements with an excess of neutrons, why not neutron decay? Apr 8 at 10:32
• There is no change in the total number of electrons. As I said they are not considered in the reaction because they are not important. When an alpha particle is emitted, the alpha particle gets a positive charge and the rest of the nucleus has a negative net charge. When I say that the alpha particle takes electrons out of the nucleus I mean of another nucleus when it collides with other objects, not the nucleus where it comes from. Apr 8 at 10:33
• A stable nucleus is more stable than the nucleus divided into alpha particles. It is when there is a disproportion of neutrons and protons that you have that these are emitted. Theyu emit an alpha particle rather than a single neutron because the expulsion of a neutron is energetically less economic as alone neutrons are extremely inestable. Apr 8 at 10:37
• Re, "Aloha particles leave an ion behind." If you are familiar with the most widely known word in the Hawaiian language, then you will understand why your typo makes me smile. Apr 8 at 13:39

Alpha radiation is labeled as "ionizing radiation" for a reason. An alpha particle passing through electrons disturbs the electrons greatly. But the alpha particle has very high mass and kinetic energy compared to the electrons, so the presence of a few electrons doesn't really affect the alpha particle.

So if you are interested in nuclear decay, you can basically ignore the fact that the electrons are even there for all the effect they have on the alpha particle itself. The reverse is not true at all and the electrons are affected dramatically.