Describing the statistical nature of the decay process, which is correct theoretically, does not actually explain what goes on realistically on the level of a radioactive nuclei. When a nuclei becomes excited it can reach a more stable state by emitting particles. However, the nuclei is subject to the interplay of a number of forces - strong and weak nuclear force and electromagnetic. This interplay is not evenly distributed to all parts of a nuclei and cannot be calculated for individual nucleons, hence the statistical description which fits very nicely. This actually means not all probabilities are the same but we don't know which ones so we it is described by the statistical process of all the nuclei as a whole which equates to all nuclei having the same chance of decaying after one half life. Some theoreticians do, however, discard this physical interpretation as being of little practical use. It is worth noting that theoretical explanations can be very good at explaining a process and may give great predictions but what actually happens may be quite different. For example the wave equation is fantastic as a theoretical tool to describe the evolution of a particles at some time, t, but problems arise when one thinks the wave equation is actually real (refer to Baysian probability).

Radioactive decays are completely random. Its impossible to forecast the moment of decay for an individual atom. The highest probability is immediatly after t=0, but it can also need infinite time. The decay law holds only for big assembles and tells us, that a big assemble of nuclei decays to 50% within the next halflife and induring the next halflife 50% of these 50% and so on. The reasons for different halflifes of nuclei are described by nuclear physics and QM.

The instability of a nucleaus corresponds to a probability that this nucleus decays in a certain time. This probablity can be described by quantum processes. Thus, the decay of a single atom can not be predicted, only if there is a bigger number of nuclei, it is possible to perform statistics.

Imagine this: a region in space where a large number of bombs have been set up to explode at time T by an equal number of military engineers. Now imagine that each bomb timer has been set up by a different man, according to his own watch and also that not all timers function precisely the same way. Some of the watches of the engineers setting them up are precise, some are slow, some are even slower (for the sake of the argument we will consider the fastest watch as the precise one). Also, the bomb timers have the same problems: some precise, some slow. As a result, although the time has been set up at T, some will blow up at T, as expected, some at T+dt and dt has a lot of values. If you study how all the watches and timers are showing the time in respect to a given reference (say the fastest watch), than you can predict how many bombs will blow up each second and how many will remain after a certain time, but because you have no idea which timer was set by who, you cannot predict at which time a certain bomb will blow. That is a very rough description of what goes on in the radioactive material. Of course, the true quantum mechanism behind that is quite different, but at least this gives you an idea of why not all nuclei will disintegrate at the same time.

An alternative illustration.

Imagine a lottery with a 1:14M chance of winning the jackpot, and 14M people playing. On average you would expect a jackpot every draw, sometimes you may get no jackpots, other times 2, 3 or more ... but, on average one jackpot each draw. You can't predict in advance who will win the jackpot each time, but you know the average number per draw.

Now, imagine that there are no new players in the lottery, and every time someone wins the jackpot they stop playing. Over time, the number of players decreases. The odds on any particular combination of numbers coming out of the machine stays the same, and so the average number of jackpots per draw would reduce over time. In fact, the number of jackpots would follow an exponential decay, with a half-life exactly as we see in radioactive decay. If we have very frequent draws in our lottery then the number of jackpots (on average) will decrease more rapidly with time, less frequent draws and the half life will be longer.

Another way to think of it is to imagine a room with 100 people standing each holding a coin. Every 10 seconds they toss the coin. If it lands on heads they sit down (a decay). If it lands on tails they keep standing up (no decay). They process continues. The half life is 10 seconds and the number of people standing at time t decreases exponentially.

Radioactivity is a statistical process. Each atom has a probability of decaying which is the decay constant of the material. The process is governed by the fact that the rate of decaying atoms is proportional to the number of atoms present. The equation in which a function is equal to its derivative (except of some multiplication constant) is always solved by an exponential, which is the only function that equals its derivative. The constant mentioned above is the decay constant of the material. And there is no way to tell which of the atoms that will be those that decay in the next time period.

Describing the statistical nature of the decay process, which is correct theoretically, does not actually explain what goes on realistically on the level of a radioactive nuclei. When a nuclei becomes excited it can reach a more stable state by emitting particles. However, the nuclei is subject to the interplay of a number of forces - strong and weak nuclear force and electromagnetic. This interplay is not evenly distributed to all parts of a nuclei and cannot be calculated for individual nucleons, hence the statistical description which fits very nicely. This actually means not all probabilities are the same but we don't know which ones so we it is described by the statistical process of all the nuclei as a whole which equates to all nuclei having the same chance of decaying after one half life. Some theoreticians do, however, discard this physical interpretation as being of little practical use. It is worth noting that theoretical explanations can be very good at explaining a process and may give great predictions but what actually happens may be quite different. For example the wave equation is fantastic as a theoretical tool to describe the evolution of a particles at some time, t, but problems arise when one thinks the wave equation is actually real (refer to Baysian probability).

Disintegration of atoms of a radio active material is completely probabilistic in nature,no one knows which at atom at what time it is going to be disintegrate,Thats why we define half life,and disintegration is directly proportional to number of atoms present in that substance.As per the Quantum mechanics there is a infinite probability of the particles to escape from the nuclear barrier.So the disintegration is totally Quantum phenomenon.

Very interesting and great ideas, explanations and imaginations were written here. However our friend who wrote the question, started from the idea which contain its self- describtion (the half life t1/2) which tell us that half of the numbers of nuclei of this radioactive isotope which exist in our sample will decay by the end of this time . It does not tell us that the age of these nuclei is equal to t1/2. At this point and before we go far in complicating any idea we must put it in its correct environment , of course we in nuclear physics always have to be carefull in mixing classical images and the modern concepts of Q.M views. However at the end the decay in nuclei is statistical phenomena and depends on the probability which has no meaning for individual or single nucleus. Again and if we recall that nuclear decay does not depend on surrounding conditions (temp. pressure ,....etc), that will lead us to assume that the decay of any nucleus will be the out come of the internal forces and their interactions.

Spontaneous emission of a photon from an atom can be considered as stimulated emission of a photon from an atom by a photon from the vacuum field. This makes the emission of light less messy but just moves the probability source to the fluctuation in the vacuum field.

I have always assumed that the same is true of radioactive decay.

Putting all the probabilistic bits into one basket tidies up the physics.

@Anthony Dymoke-Bradshaw:

That is a really interesting view.

However, an explanation why different nuclei decay with different probalities can be given on the base of the specific nuclear properties. So the reason for different decay times and decay modes lies (deep) within the nuclear structure; there is no simple single parameter which could be related to the interaction with the vacuum field. So I guess you won't get everything in one basket with this approach.

I agree completely with E. Strubs arguements.

Some very bright physicists have found the probalistic nature of quantum processes unsettling and sought for a "hidden variable" which removes the randomness. Fluctuations in the vacuum field would, to me, seem to fall into the category of "hidden variable" explanations for quantum processes. As noted, it doesn't even remove the random element from the description, just moves it to another place.

As intelligent people struggle with understanding quantum systems, and (for examplw) how they relate to General Relativity they may come up with a better description. But for now we're left with the roll of the dice.

Be nice to have all the dice rolling in one place though.

Been following the Q&A here, though I am not an expert in the field. Very interesting, but what I wonder is; If a radioactive isotope is decaying in your body, when is it the most dangerous?

@Cara,the most dangereous decays are alpha decays, because alpha particle release their complete kinetic energy in a typical distance of about 10-30 mikrom. That´s the diameter of a single cell. If the cell doesn´t perfect an apoptosis you can probably cause cancer.

It's an interesting question, what is more dangerous.

As Hanno noted alpha particles release all their energy into one cell, or maybe two. In the vast majority of cases this results in cell death. A dead cell is not a problem, unless of course the source is so active that very large numbers of cells in a small location are killed. It is the very small number of cells that are damaged but not killed that could potentially cause cancers.

Beta particles have a much greater penetration depth, and so deposit small amounts of energy into hundreds or thousands of cells. In most cases, this energy is insufficient to cause irreperable damage (cells have quite good repair mechanisms) but in a very small number of cells the damage may not be correctly repaired and this could potentially cause cancers.

What process is more important? The small number of cells where alphas don't deposit enough energy to kill the cell, or the small number of cells where the beta deposits enough energy to cause irreperable damage?

I think the question can be regarded from two points of view:

1. the LET: in this case clearly the alpha particles have the highest LET and will lead to to higher effective doses, taking into account the same activity and the same organ - the nuclides with the higfhest radiotoxicity are alpha emitters.

2. the response of the organism: whilst the above remains, we have to take into account also the adaptive response of the organism. This is not taken into account in the LNT assumption, but it has a role to play in how the organism reacts to a dose. If we take out of the question the double strand breakes in the DNA, when it is more likely that the cell will die at mitosys, the alphas will produce deviant cells in a more concentrated area than the betas, so I assume this will increase the risk of cancer - which again leads to the conclusion that alpha emitters are more radiotoxic than betas.

But, besides that, there are also other things to be taken into account for the actual committed dose: solubility of the radionuclide, biological half time, chemistry, etc.

...but this turns into a totally different discussion...

Interesting analysis by Alan! Indeed there is no problem if some body is irradiated by alphas percutaneously because alphas are completely stopped in the first cells of the scallus skin. If alphas are incorporated by breathing they distroy epithelian cells of the lungs and are more dangereous.

The penetration depth of betas is greater but the ionisation density is smaller than that from alphas and is uniformely distributed within the cell plasma and not in the DNA. To estimate the stochastic risks (cancer, heriditary defects) the energy dose is therefore multiplied with weighting factors wR. The weighting factor of alphas is 20, the factor for betas 1.

My resume is: alphas are more dangereous than betas for the same energy dose.

@ Radu, I just read your answers and I agree with your arguements. Your are right, thats a completely different discussion!

Sorry to divert the question/ answer line. Thanks for your interesting responses.

Don´t mind Cara, was a pleasure. Greetings Hanno

Cara, I believe your question can be answered with an analogy. In your country, all humans living at present will die at some point in the future. When, we don't know for sure. Some will die later than others (every day some will die). But at the end, all will be death.

Or, another example. you buy a dozen oranges to eat next week. Every day you eat one or two of them, but the oranges don't know when they will be eaten. However, at the end of the week, none will be left. Ernie.

That's an interesting question, but a difficult one to answer. but think of this,... may be this will give you an answer.... many researchers have seen the radioactive decay law but only you got this question your mind...Why .

There is a perplexing question when all the unstable atoms have decayed and only one atom remains. Will it decay or not?

It is a rather hipothetical question, since one cannot tell when only pne remained. But the last nucleus will sometime decay, since is still non stable, and has a probability to decay of Lambda (ln(2) devided by the half life), which is a finite value. One cannot tell when will that happen, but it will happen.

The remaining single atom has the highest probability to decay immediately, that´s the rule for all instable atomic nuclei.  But high probability does not mean, the atom obeys. The reason is the stochastic character of radioactivity. And for 1 single atom, take a lot of time, could be this atoms survives your life span. ;-)

Disintegration of a nucleus is considered to have a quantum indeterminism nature. Since its birth, the life time of an unstable nucleus is calculated as a random number taking values in an exponential law.

There is a particle model which considers each (massive or massless) 'elementary particle' as a 'bound state' of 2x3=6 so-called 'hypotrons'. The 'up-quark' and the 'down-quark' are considered as 'bound states' of two hypotrons. Thereby (and finally), the neutron can be considered as a 'bound state' of (proton)+(electron)+(anti-neutrino), and the anti-neutron can be considered as a 'bound state' of (anti-proton)+(positron)+(neutrino). By means of a minimum principle concerning a (non-additive) quantity, which I called 'supercharge' and which is related to the charge of the hypotrons in a non-linear manner, it is possible to reproduce the 'valley of stable nuclei' provided that it is assumed that anti-neutrons are contained in nuclei instead of neutrons. For a preliminary research note see http://kreuzer-dsr.de/kdsr/bulletin/KDSR_HypotronTheory_Flyer.pdf and http://kreuzer-dsr.de/kdsr/test/KDSR-PRS-02/KIPS/START-AND-RUN.htm

The fact that a decaying nucleus can eject a neutron can be considered as a result of a REACTION (anti-neutron)+(neutrino)-->(neutron)+(anti-neutrino) INSIDE THE NUCLEUS which occurs 'randomly' and which can be considered TO CAUSE THE DECAY. The 'randomness' can be considered as a consequence of (up to now) unknown (highly chaotical or quasi-stochastic) internal dynamical properties of the neutrino-'core', considered as a bound system of 6 hypotrons, and the anti-neutron-'core', considered as a bound system of 3x6=18 hypotrons.

Note that whether or not a particle and anti-particle annihilate under all (spatial) circumstances should be considered as an OPEN question. What can be observed in collision experiments (proton)->++