I understand that some nuclei and their isotopes are not stable and therefore at random intervals bits of the nucleus (i.e. protons & neutrons) break away with differing amounts of energy depending on a few variables. Grossly simplifying things, alpha particles are big and slow, beta are smaller and faster and gamma rays are highly energetic elecromagnetic radiation traveling at the speed of light. The particles seem easy enough to understand on a basic intuitive level, gamma rays less so. While the particles were an "observable" part of the nucleus before they radiated away, the gamma rays were not. However I believe it is correct to say that a by product of the decay is the generation of gamma rays which can also be characterised as photons. X-rays overlap with gamma rays but to be honest I have not spent much time understanding this area.

This is the background that seems logically to lead to my question. These things (alpha, beta, gamma, X-rays) break off from the nucleus. Their energies are such that they are able to knock electrons off other nuclei (hence "ionizing") and - I'm guessing now - there are imbalances in energy levels in the radioactive nuclei which the decay helps resolve.

What's really going on? What is the basic theoretical framework for understanding this phenomenon? And, putting aside the big bang, where does this story start? Is a radioactive nucleus "born" radioactive?

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    $\begingroup$ You ask "What's really going on?" that's not a clear question. We don't know what aspect of radioactivity you are asking about. If you're looking for a full theoretical explanation of the 3 main emissions, their energetic and probabilities. That's too broad. There are people who spend years studying the reasons. $\endgroup$ – Bill N Feb 1 at 21:18
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    $\begingroup$ Lacking knowledge of the specifics I wanted to place the enquiry inside a generalized wrapper to elicit answers about different parts of the processes filling multiple gaps in my knowledge. It's an overused metaphor but the first few strokes on a blank canvas, so-to-speak. However you are absolutely right: simply as a linguistic construction "What's really going on?" is vague and with greater knowledge at my disposal I certainly would have made the question more targetted. $\endgroup$ – Adam Gold Feb 2 at 2:43
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    $\begingroup$ Do you understand how electron orbital transitions work (e.g. how atoms and molecules interact with light)? It's the same with the nucleus, just with nucleon orbitals (which have much higher energy, so the light is usually in the gamma range). And just like you can excite an electron in an atom (e.g. through incident light), you can excite a proton in the nucleus. When the electron/proton drops back to the lowest available orbital, it emits one or more photons. $\endgroup$ – Luaan Feb 3 at 9:01
  • $\begingroup$ I'm about to accept an answer which is a near impossible choice given the quality of responses (and comments)! I would like to thank everyone for their thought and effort and it was a particular bonus to have differents answers speaking to different aspects of the question. I feel genuinely informed by the collective contribution and have a much clearer idea on how to follow up on my interest. Thanks to all. $\endgroup$ – Adam Gold Feb 11 at 3:37

First off, let's stay humble. To begin with, let us remember that no one really knows the answer to your question. That is the nature of the scientific method:

  • We observe macro phenomena.
  • We abstract what we see.
  • We think, "if that is right, then ... is true".
  • We test the extrapolation empirically.
  • We refine our abstraction depending on the result.

We never really know. Revolutions in physics occur when some capable and questioning individual thinks there is a hole in the status quo and proposes a solution. He/she is either proven wrong by experiment or goes on to become a giant, usually post humously.

With that in mind, let us remember that the division of the forces into 'strong', 'weak', and 'electrostatic' is an entirely artificial abstraction that has proven to be useful. It is the current 'standard model' and it is just that, a model. There have been other insights which, as yet, have not revolutionized understanding.

In particular, there is Yukawa's idea of a potential that merges these three forces into a more complex potential. The reasoning is that nature abhors mathematical singularities. Therefore, the electrostatic potential must have a different shape at small radii. Yukawa modelled that idea with a numerator that goes to zero on the inverse radius, but rapidly goes to unity for distances on the order of a nuclear radius. The small radius slope is the 'strong' force, a larger radius transitional slope is the 'weak' force, and the asymptotic slope is the electrostatic inverse square law (the force is the spatial derivative, or slope, of the potential).

With that approach, you get a deep but finite well at the center of a nucleon. With this line of thinking, the prime mover of radioactive decay is entirely electrostatic imbalance. Yukawa's work was, of course, not quite right, but it remains influential. There are many other functions that exhibit the desirable properties, so perhaps we have a case like that of the historical development of the Bose-Einstein statistics underlying black body radiation, where early attempts to reconcile the 'ultraviolet catastrophe' were close, but not quite right.

With regard to the apparent absence of negative charge in the nucleus, it is well to remember that a free neutron will decay into a proton, an electron, and a neutrino wth a half life of about 12 minutes.

This suggests that a neutron can be polarized, since in this line of thinking, it is somehow a combination of an electron and a proton. In fact, this has been tested empirically. I remember attending a presentation of the results of such a test about 15 years ago. At that time, the polarizability of a neutron resisted measurement attempts capable of resolving 10 to the -27 meters, indicating the binding force of that particular combination is very 'strong'. Such a test is 'big' science - it is very expensive.

If you assume these lines of thinking, then in the nucleus the barriers are lower and a electron that unbinds one proton rapidly rebinds another in the stable case. In the unstable case, there are too many neutrons, so there is some non-negligable chance of a beta emission. In the large nucleus case with not enough neutrons there is some non-negligable chance of a whole block breaking off, typically an alpha emission owing to the great stability of that configuration, but in the context of some energetic disturbance, perhaps a fission.

The thinking is that an electron in a neutron can balance more than one nuclear proton owing to the short range shape of the potential.

Finally, radioactive decay is not limited to large nuclei. Get yourself a copy of the isotope tabulation of the periodic chart and spend a few hours (or days or years) examining the relationships between stability, the neutron drip line and the proton drip line. All of that complexity is empirical data and so must be accepted as fact.

Some may criticize the speculative non-standard assumptions underlying this line of thinking. I have the giant shoulders of Yukawa to stand on and help squelch some of that. It is necessary to think beyond the status quo. But more necessarily, one must mathematically quantify the qualitative ideas in a way that predicts known empiricism. That is very hard.

Such boldness will necessarily lead to errors, but it can lead to greatness. Perhaps you are strong enough to answer your own question in your lifetime and become a legend. I know I am not.

What others here have said is a reiteration of the standard model. What they have said is not wrong, but neither does it really address the crux of your question which is rooted in the raw curiosity and wonder surrounding the unknown and unknowable. Kudos for having the courage to wonder!

My favorite quotation of Albert Einstein is:

"It would be sufficient to really understand the electron."

Perhaps the answer to your question is to really understand the neutron, which, in this line of thinking, is the fundamental mechanism governing radioactive decay.

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  • $\begingroup$ Fascinating, thank you and point taken re. the neutron. Also you've highlighted something that has been mentioned in some of the other answers which I really would like to seek clarification about. It appears that the three atomic particles can 'transform' into each other. I don't know if that is true for every combination and it seems not always to be one-to-one as you mention above. How does this happen i.e the 'processes' that initiate and effect the transformation? And also - I'd prefer to refine this question if I knew a good way of doing so - what is happening? $\endgroup$ – Adam Gold Feb 3 at 7:28
  • $\begingroup$ Yes, a neutron can decay into a proton and emit an electron and antineutrino, but that doesn't mean that it's "somehow a combination of an electron and a proton". Protons & neutrons are made of quarks, which bind to each other via gluons. $\endgroup$ – PM 2Ring Feb 7 at 10:07
  • $\begingroup$ @Adam Neutrons & protons can transform into each other, via the weak nuclear force. Such reactions also involve an electron (or positron) and an antineutrino (or neutrino), which ensures that electric charge and lepton number are conserved. $\endgroup$ – PM 2Ring Feb 7 at 10:11

All elements above Iron are created when a star implodes under its own gravity in the form of a supernova. These heavy elements are ejected into space in this process and go on to form parts of planets such as Earth. This includes all their various radioactive isotopes.

From the moment the unstable isotope is created, it wants to radioactively decay so as to attain a more stable configuration. This will continue to happen until the atom reaches a state which has an extremely long half life (such as $\text{Pb}^{82}$). The three most important ways in which an atom can decay are as follows-

1) $\alpha$-Decay: Ejection of Helium$(\text{He}^{2+})$ ion

2) $\beta^-$-Decay: Ejection of electron$(\text{e}^{-}) \text{ and anti-neutrino}(\bar v_e)$

3) $\beta^+$-Decay: Ejection of positron$(\text{e}^{+}) \text{ and neutrino} (v_e)$

These three affect the atomic and mass number in different ways which you can read more about here. All radioactive disintegration follows first order kinetics. The $\gamma$-rays released is just the difference in mass between the two atoms. This happens according to Einstein's famous equation $E=\Delta mc^2$. These are just the basics which should be sufficient for understanding purposes.

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    $\begingroup$ @AdamGold Indeed radio-active decay is confined to a small set of configurations. As described: the main three are: (1) emission of an Alpha particle (Helium nucleus), thus dropping to an isotope two steps down the ladder of atomic numbers, (2) a neutron disappears and a proton appears (plus other emission to conserve charge and other things). This decay mode is available when there is a surplus of neutrons, so a state with less neutrons and more protons is energetically favorable. (3) A proton disappears and a neutron appears. The restrictions are conservation principles. $\endgroup$ – Cleonis Feb 1 at 20:53
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    $\begingroup$ "From the moment the unstable isotope is created, it starts to radioactively decay so as to attain a more stable configuration." Are you talking bout an individual nucleus or a population of nuclei of that nuclide? It's NOT the case for an individual nucleus. Given 2 unstable nuclei, one may transform 1 second after formation and the other may persist for days or years. Halflife values are given for the statistical behavior of a large population. $\endgroup$ – Bill N Feb 1 at 21:15
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    $\begingroup$ I realise that you're trying to keep things simple, but heavy elements aren't just created in supernovae. They may also be created during neutron star mergers, and by the s-process, particular in AGB red giants. $\endgroup$ – PM 2Ring Feb 2 at 3:10
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    $\begingroup$ @Sam There is iron in the Sun (quite a lot more than there is in the Earth), but the Sun didn't produce that iron. The Sun simply isn't hot enough, and never will get hot enough for substantial fusion of heavy elements. And even in stars that are hot enough to fuse silicon and heavier elements, that only happens in the last few days before they go supernova. See en.wikipedia.org/wiki/Silicon-burning_process $\endgroup$ – PM 2Ring Feb 2 at 7:48
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    $\begingroup$ Sam, your first paragraph is not only inaccurate, it's also irrelevant: radioactivity isn't restricted to elements heavier than iron. For example, tritium is a radioactive isotope of hydrogen. Many other errors... What is Pb-72? Lead's atomic number is 82, while the element with 72 protons is hafnium - and both have stable (non-radioactive) isotopes. Radioactivity doesn't "stop" due to a long half-life; U-238's half-life is 4.5 billion years, yet its decay produces 40% of the Earth's internal radioactive heat. $\endgroup$ – Chappo Hasn't Forgotten Monica Feb 6 at 23:23

About the basic theoretical framework:
To an acceptable approximation the total internal energy of the nucleus can be thought of as arising from two forces: the strong nuclear force and the electrostatic force.

The protons in the nucleus all repel each other as they are all positively charged; it's the strong nuclear force that holds the nucleus together. The range of the strong nuclear force is limited. For a large nucleus the strong nuclear force does not extend from one side of the nucleus to another. This is why for nuclei heavier than the nuclei of lead there is no stable isotope.

The wikipedia article about radio-active decay also has the diagram with the table of isotopes

For the lightest elements there is a matching number of neutrons and protons in the nucleus. The higher the atomic number the more neutrons there are in proportion to the protons. The neutrons contribute to the total amount of strong nuclear force holding the nucleus together, but not to electrostatic force, thus supporting stability of the nucleus.

For each atomic number there is an optimum ratio of neutrons to protons. For an isotope with that optimum ratio there is no decay mode available: all states that can be reached through any of the decay modes is a higher energy state than the one that that nucleus is already in.

There is a very interesting video about nuclear decay by the science youtuber Scott Manley. (The target audience is novices, with Scott frequenty giving the "This is of course a gross oversimplification." warning.)

The natural question is: well, if a decay mode is available, why do many isotopes have such a long half life?

Scott Manley gives one example where the only available decay mode requires two beta decays happening simultaneously (or at least within a sufficiently short time window). That makes for a much smaller probability for that decay to occur.

Any radio-active decay has to overcome an energy hump. The usual imagery is one of a marble on a ledge with a rim, next to a long steep hillside. If the marble can make it over the rim it can roll all the way down, but first the marble has to make it over that rim.

For the nucleus of an isotope with a long half life: given enough time at some point in time some part of the nucleus will happen to have the energy to make it over the hump. The probability of making it over the hump is never zero, but the higher the initial hump the smaller the probability.

Acknowledgng the comment by PM 2ring I have scratched out the above two paragraphs. Wrong picture, even when allowing for gross oversimplification.

New attempt: decay mode is not a single process, but a complex process. There is enough intricacy such that depending on specific circumstances the probability of a decay occuring at all can be very low.

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    $\begingroup$ From physics.stackexchange.com/a/488224/123208 "There isn't a potential barrier in beta decay, whether it's beta minus or beta plus decay. In both cases the decay is slow simply because the transition probability is so slow." DMckee goes into more detail: physics.stackexchange.com/a/31517/123208 $\endgroup$ – PM 2Ring Feb 2 at 2:00
  • $\begingroup$ As an FYI, I watched the Scott Manley video. It's about 15 minutes long and for those at my level of knowledge and curiosity, I think it will be very helpful. A good recommendation by @Cleonis. $\endgroup$ – Adam Gold Feb 2 at 2:30

I want to correct your question: you said that $\alpha$ and $\beta$ radiation are "observable" inside the nucleus. If I understood you correctly, you mean that before decaying these particles were a part of the nucleus. For $\alpha$ particles it is correct but not at all for $\beta$:

The nucleus is made out of protons and neutrons, meaning only from quarks (the particles that make them), while a beta particle is an electron or a positron, which isn't a quark (technically, it is a type of lepton). Meaning the beta particle couldn't be observed in the nucleus before the decay because it wasn't there in the first place!

To understand what is going on, you need to understand the nature of $\beta$ decay (which is called "the weak force"). It is quite complex, but as a beginning, you should know about 2 important quarks: the up quark (let's call it 'u') and the down quarks ('d'). The u quark has an electric charge of $+\frac{2}{3} e$ and the d quark has a charge of $-\frac13 e$ (where $-e$ is the charge of an electron, and $e$ is the charge of the proton). A proton is just 2 up quarks and one down quark, so the total charge is $\frac23 e + \frac23 e - \frac13 e = e$, and a neutron is one up quark and 2 down quarks, so the total charge is $\frac23 e -\frac13 e - \frac13 e=0$ (neutral particle).

In beta decay, an up quark becomes a down quark or viceversa (which "converts" a proton: uud , to a neutron: udd , or viceversa). Notice, however, that they have different electric charge. For conservation of charge, what in one moment a down quark "decides" to become an up quark, he will gain an electric charge of $e$ (going from $-\frac13 e $ to $\frac23 e$), meaning something else with a charge of $-e$ needs to be created in order of that to happen: the electron! Meaning there was no electron at all first (the quarks are elementary particles, they don't consist of electrons) and at the moment of decay the beta particle was born. The beta particle couldn't be observed before the decay because it didn't exist yet.

This answer doesn't answer your question, but it is too long for a comment and I think it was worth discussion. I hope it helped a bit to understand more.

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    $\begingroup$ Although this doesn't answer the question, it does add important relevant info, so (hopefully) no one will downvote it. $\endgroup$ – PM 2Ring Feb 2 at 4:00
  • $\begingroup$ I completely agree, it is a missing piece that, as the OP, I would not have inferred. $\endgroup$ – Adam Gold Feb 9 at 5:52

Your question deals with two related concepts: radioactive decay and ionizing radiation. Any charged particle with mass can ionize atoms if it has enough kinetic energy, so alpha and beta radiation is ionizing. Gamma radiation is ionizing too, because it consists of photons with significantly more energy than the practical thresholds generally set at between 10 and 100 eV. So, radioactive decay produces ionizing radiation.

However, radioactive decay also results in other effects such as neutron emission and electron capture. Your question seems to be focused on why radioactive decay happens, not its consequences for other matter nearby, eg. whether or not it results in "ionizing" radiation.

As noted in the other answers, radioactive decay is the spontaneous collapse of a nuclear structure which is metastable for complicated quantum mechanical reasons which are answered directly in this question. A nuclear physics textbook would probably be a better resource than Stack Exchange for understanding this topic.

To answer the last part of your question, a radioisotope is "born radioactive," because it forms part of a decay chain (or was created in some other nuclear reaction), and will itself "die" from radioactive decay at some random point in the future.

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  • $\begingroup$ Indeed, that's correct to spilt out the concepts and it's a constructive clarification. Without wanting to be wise after the fact, the question was somewhat intended to blend the concepts together, if nothing else because this would be the place for someone to break them down and add overall clarity to my enquiry. Your response has done that succinctly and provided some pointers to follow up on the different 'chunks' of the question. Thank you. $\endgroup$ – Adam Gold Feb 2 at 2:26

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