Does the process of decaying in radioactive elements occur every second? The equation consists of $\rm e$, so it must mean that decay due to radiation must occur every second, right?

  • $\begingroup$ Certainly many isotopes have half-lives much longer than a second, so it is not clear what you mean. Please clarify. $\endgroup$
    – Jon Custer
    Commented Jul 17, 2020 at 15:28
  • $\begingroup$ @JonCuster I am not talking about half lives ,if we take a radioactive element say strontium 90 ,it takes about 28 years to lose half of its quantity, that's ok but how does it do so,after 28.79 yrs approx does the half of it disappear into thin air(or) does it lose it's quantity every instant and after 28.79 years it happens to have half of its initial mass? $\endgroup$
    – user794763
    Commented Jul 17, 2020 at 15:33
  • $\begingroup$ Decay is an ongoing statistical process. $\endgroup$
    – Jon Custer
    Commented Jul 17, 2020 at 15:34
  • $\begingroup$ @JonCuster pls elaborate $\endgroup$
    – user794763
    Commented Jul 17, 2020 at 15:36
  • 2
    $\begingroup$ Re, "every second." Nature doesn't know anything about seconds. Nor does it know anything about years or months or any other time scale we might choose to talk about radioactive half lives. Whatever is happening, it happens continuously. $\endgroup$ Commented Jul 17, 2020 at 16:38

7 Answers 7


The process of radioactive decay occurs at every instant in which th substance exists. The rate at which the decay happens is proportional to the number of active particles at the given instant. It isn't stroboscopic. But if you solve the differential equation, you will get a fair approximation of how many radioactive particles remain after some amount of time.


does the half of it disappear into thin air...?

Note! It doesn't disappear. It becomes something else. You gave, as an example, 90Sr. When at atom of 90Sr decays, it emits a beta particle, and it becomes an atom of 90Y (Yttrium).

Beta emission happens in a nucleus that has too many neutrons to be stable. One of the neutrons spontaneously becomes; a proton, an energetic electron (a.k.a., "beta particle"), and an antineutrino. Because the nucleus now has one more proton than previously, the atomic number goes up by one.

90Y has a half-life of only a few days, before it decays (again by beta emission) and it becomes a stable 90Zr (Zirconium) atom.



No it doesn't. It decays at a rate depending on its half-life. As @Jon Custer said its a statistical process and the nuclei will randomly decay in some period. The decay rate, decay per unit time is given by A=λN where A is decay rate, λ = decay constant and N is number of Nuclei in sample.


When we talk about half-time we deal with a collection of many atoms (macroscopic quantity, i.e. of the order of Avogadro number). Every atom has a probability $p=\frac{1}{\tau}$ to decay per unit time, whereas the average number of atoms that have not decayed is given by $N_0e^{-t/\tau}$, where $N_0$ is the initial quantity of isotopes.


What really happens

It really depends. If you have a moderate value of decay rate ($\sim 10^{10} \:\rm dps$), then the decay of multiple particles every second is inevitable. However, if the value of decay rate is small enough ($\sim 2 \: \rm dps$), then it's quite possible that you might find a particular second where none of the particles decayed. In the cases where the decays rate $≤1\:\rm dps$, it's highly likely to find such time intervals.

Why does it happen?

The reason why this is the way it is, is because of the quantum mechanical nature of nuclear decay. You can't really predict whether a certain particle will have definitely decayed by a certain time $t$, because of the quantum mechanical nature of the process of nuclear decay. However, the probability of particle being decayed by a time $t$ increases as the time $t$ increases, so eventually it will be highly probable that the particle would decay. However since we are dealing with probabilities, we can never be sure that a particle decayed.

If you're into quantum mechanics, you might also know that the state of a nucleus is a superposition of decayed and undecayed. And you can't know whether it decayed until you collapse its wavefunction by making a measurement, and checking if it decayed. (see the last section for the continuation of this line of thought)

How come we write a continuous fuction describing decay?

The function that we write can be regarded as "empirical", or more appropriately, an approximation of the reality rather than the absolute truth. Thus if you were to perform a nuclear decay experiment, it's highly likely that you would find measurements which are quite close (though not exact) to the mathematical equation of nuclear decay:

$$N(t)=N_0\mathrm e^{-\lambda t}\tag{1}$$

Now, no matter how many times you repeat the experiment, or how precise you make your experimental procedure, you'd always get results which are closer to equation $(1)$, but you'd never be able to do the experiment and get the exactly same results as predicted by the equation. And no, it's not because of the experimental errors that might have crept in, it's because of the uncertain and probabilistic nature of nuclear decay.

Many-Worlds interpretation

This is a bit of an off-shoot section, so it isn't strictly related to the question, but it's worth reading :-)

Now if you believe in the many-worlds interpretation of quantum mechanics, then you'd be fascinated to hear what it predicts during nuclear decay. It predicts that our universe branches multiple (gazillion times) during the process of nuclear decay, with every universe being different. In other words, whenever there are two possibilities for a nucleus, decay or not decay, our universe branches into two other universes, where in one of the universes, the nucleus has decayed while in the other one, it hasn't. And this is true for all the nuclei.

This line of logic implies an extremely mind-boggling result. It implies that there might be a universe where none of the particles have decayed in the first second of nuclear decay (despite the nuclear decay having a rate of $10^{10}\:\rm dps$). At first, this sounds to be unreal, thus thinking us into doubting the many-worlds interpretation.

However, if we think further, then we can notice that the probability of us being in such a universe is extremely low (I mean extremely extremely extremely low). Why? Because everytime a universe branches into two new universes, we have almost equal probability of being in one or the other. But to reach to a universe where no particles have decayed in the first second, you'd alwyas have to end up in the "undecayed" version of the universe, every single time, for $10^{10}$ branchings. And this is extremely unlikely, thus we never find ourselves conducting such an experiment. But do note that if we were to perform our experiment $2^{10^{10}}$ times, we might have a chance to witness that "special" decay where nothing decays in the first second.


The poster may be asking if radioactive decay is a continuous process or a discrete process.

It is a continuous process. For any given increment in time, there is a probability that a certain amount of material will decay. This probability is called the "decay constant" and usually denoted with the greek symbol "$\lambda$".

In the form of a differential equation, the "rate of change" of the nuclide concentration is proportional to the decay constant times the concentration itself. $$ \frac{dN}{dt}=-\lambda \, N $$ The solution of the differential equation makes it clear that it is a continuous process $$ N(t)=N(0) \, \exp(-\lambda t) $$

The decay constant isn't a very intuitive variable since the units are "per time". Therefore, the decay constant usually converted to a "half-life", which is more intuitive to understand because it has units of time. You can use the previous equations to show that the half-life is related to the decay constant by the expression $$ t_{h} = \frac{\ln(2)}{\lambda} $$

The confusion may be that the half-life has units of time (e.g. seconds), but it is still refers to a continuous process.


In order to understand radioactive decay, we have to see what is in the nucleus of an atom and how its parts interact.

As you may know, all nuclei, no matter what kind of atom (element), consist of protons and neutrons (nucleons). A proton is a positively charged heavy body, and a neutron is just slightly heavier than a proton and has no charge. The neutron can be thought of as a proton with an electron (or, more accurately, beta particle) bound to it by the weak nuclear force.

The protons of a nucleus repel one another very strongly through the electric force of repulsion between like-charged bodies. However, the attractive strong nuclear force is much stronger than the electric force at such small distances, and so the strong nuclear force overcomes the electrostatic (Coulomb) repulsion and holds the protons and neutrons together.

This union results in a ball of protons and neutrons (to a certain approximation) shaking back and forth violently within the nucleus, but held together with the strong nuclear force. Sometimes, the nucleus' (or, isotope's) configuration (shape) is always energetically "stable," and will never break apart no matter how much time passes - like an intact water balloon (so long as something strong enough doesn't come along and break it). Other nuclei (of a different isotope), every so often in this violent vibration, assume a shape that cannot be held by the volumetric and surface tension. These are called "unstable" nuclei, or "radioactive" nuclei. When this happens, a piece of the nucleus, a particle, breaks off. This is called "decay," or "radioactive decay." When the nucleus decays, it does not disappear. It simply breaks into multiple pieces.

We cannot say, for any given "unstable" nucleus, exactly when it will assume a configuration leading to decay. But, if we have a large number of nuclei (say, $N = 10^{23}$; around the number of atoms when amassed you can see with your naked eye) we can say approximately that the number of decaying nuclei at time t, dN(t), must be proportional to the number of nuclei present at time t, N(t). Additionally, we can say that dN(t) should be proportional to the time that passes over a small enough duration, dt. Note that in these proportionalities N is considered continuous rather than discrete. This is an approximation - or, really, an error - since we cannot really have a fraction of a radioactive particle, by definition.

Continuing, nonetheless, we have said $dN \propto N dt$.

Rearranging, and introducing a proportionality constant, $\lambda$ (the "decay constant"), we see

$\frac{dN(t)}{dt} = - \lambda N(t)$

The constant $\lambda$ is considered positive by definition, so the negative is introduced to capture that the change in the number of nuclei is negative over time period dt.

Rearranging the equation gives

$\frac{dN(t)}{N(t)} = -\lambda dt$

The solution to this equation through basic calculus is

$N(t) = N(t=0) e^{-\lambda t}$

This is where the "e" comes from.

Now, your question is "does decay happen every second?" The problem is, the question assumes there is a yes or no answer. Also, it is useful to explain the half-life.

The half-life, by definition, is the time during which half of the nuclei in the sample should have decayed (become something else... not disappeared).

We can calculate this time by using the equation above as

$\frac{N(t_{1/2})}{N(t=0)} \equiv 1/2 = e^{-\lambda t_{1/2}}$

where $t_{1/2} = \text{half-life}$. Solving the right side, we get

$-ln(2) = -\lambda t_{1/2}$


$t_{1/2} = ln(2)/\lambda$

This shows the relationship between the half-life and the decay constant. Over the time $t_{1/2}$, half of the original nuclei will have decayed, leaving N(t=0)/2 nuclei in their original state. After another half-life, half of the remaining undecayed nuclei will have decayed, with N(t=0)/4 remaining originals. Generally, after H half-lives, $N(t=0)/2^{H}$ nuclei will remain undecayed.

The problem with the above derivation, as mentioned, is that it calculates an average behavior of a large number of nuclei (or, more accurately, a proportion of total nuclei) that, as far as the equation is concerned, is continuous. This is called the "classical" approach.

In order to derive the actual decay of a number of nuclei N, we should start with the statistical representation, which correctly treats the number of nuclei as discrete, but - since we only know the probability that any given nucleus will decay in a certain duration - gives a probability distribution of final outcomes instead of a deterministic result. Therefore, the answer to your question is: "In each second, there is a probability P that decay will happen, and there is a probability (1-P) that decay will not happen." Of course, once all nuclei have decayed, the probability P is zero.

It is possible, albeit exceedingly improbable for large numbers of nuclei, that in a sample of radioactive nuclei, all of them will decay in the same short time dt. We can call this outcome #1. There is only one way that this outcome can be achieved. If the probability that a nucleus decays in time dt is p, then the probability it does not decay in the same time is (1 - p). The probability of outcome #1 is $p^{N}$.

There are N ways in which outcome #2 occurs, where all but one nuclei decay in time dt. This means the probability of outcome #2 is $(\frac{N!}{(N-1)!})(1-p)p^{N-1} = N(1-p)p^{N-1}$.

Outcome #3 is that all but 2 nuclei decay in time dt. The probability of outcome #3 is $\frac{N!}{(N-2)!2!}(1-p)^{2}p^{N-2}$.

In general, Outcome #k, that all but k-1 nuclei decay in time dt has probability $\frac{N!}{(N-(k-1))!(k-1)!}(1-p)^{k-1}p^{N-(k-1)}$.

One of all of these N+1 outcomes must be fulfilled at the end of time dt, so the summation of all these probabilities is one.

It is useful for the purpose of answering your question to mention that Outcome #N+1 is that no decay occurs over the time period under consideration, and the probability of this event is $(1-p)^{N}$.

The irony here is that I've slipped a "classical" model of the nucleus itself into the discussion - we could call it the "liquid drop" model - when, in fact, the nucleus itself is just a superposition of a great number of possible configurations or accessible states based on its internal energy, and we can, in principle, count these number of possible states to arrive at the statistical, and correct, model of the nucleus as well.

We can also relate the two approaches. If we set dt = 1, then in the first equation this corresponds to t=1 (at this point, it doesn't matter what we choose for units, but we do know whatever units we choose are consistent with $\lambda$), leading to equation

$\frac{N(1)}{N(0)} = e^{-\lambda} = \frac{\text{Number of nuclei not decayed at time 1}}{\text{Number of nuclei at time 0}} = 1 - p$

where, admittedly there was a bit of a sleight of hand since p was for a single atom while N corresponds to a population of atoms; but, since p is derived from a population of atoms, it's okay.


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