In my curriculum, the decay constant is "the probability of decay per unit time"

To me, this seems non-sensical, as the decay constant can be greater than one, which would imply that a particle has a probability of decaying in a time span that is greater than 1.

Can someone explain this?


3 Answers 3


You're missing two things. First, that the decay constant is the probability of decay per unit time. That part is important. The actual decay probability over a short time period is equal to the probability per unit time, multiplied by the time period:

$$P = \lambda\Delta t$$

$\lambda$ can be as large as you like, but for a small enough interval $\Delta t$, you'll still have $P < 1$. So there's no contradiction there.

The other thing you're missing is that $\lambda$ is only the probability per unit time given that the nucleus has not already decayed. That's also important. You have to start with an undecayed nucleus.

So let's say you have an undecayed nucleus at $t = 0$.

\begin{align} P_0(\text{decayed}) &= 0 & P_0(\text{undecayed}) &= 1 \end{align}

After some short time $\Delta t$, the probability that it will have decayed is $\lambda\Delta t$, as above.

\begin{align} P_1(\text{decayed}) &= \lambda\Delta t & P_1(\text{undecayed}) &= 1 - \lambda\Delta t \end{align}

Now consider the next time interval, from $t = \Delta t$ to $t = 2\Delta t$. If the nucleus didn't decay in the first time interval, it has a probability $\lambda\Delta t$ of decaying in this second interval. But if the nucleus did decay in the first time interval, the probability that it will have decayed by the end of the second time interval is 1. So overall, the probability that it has decayed by $t = 2\Delta t$ is

\begin{align} P_2(\text{decayed}) &= P_1(\text{undecayed})\lambda\Delta t + P_1(\text{decayed})(1) \\ &= (1 - \lambda\Delta t)\lambda\Delta t + \lambda\Delta t \\ &= (2 - \lambda\Delta t)\lambda\Delta t \\ P_2(\text{undecayed}) &= P_1(\text{undecayed})(1 - \lambda\Delta t) \\ &= (1 - \lambda\Delta t)^2 \end{align}

You can probably see the pattern from here:

\begin{align} P_3(\text{decayed}) &= P_2(\text{undecayed})\lambda\Delta t + P_2(\text{decayed})(1) \\ &= (1 - \lambda\Delta t)^2\lambda\Delta t + (2 - \lambda\Delta t)\lambda\Delta t \\ &= \bigl(3 - 3\lambda\Delta t + (\lambda\Delta t)^2\bigr)\lambda\Delta t \\ P_3(\text{undecayed}) &= P_2(\text{undecayed})(1 - \lambda\Delta t) \\ &= (1 - \lambda\Delta t)^3 \end{align}

In particular, at $t = n\Delta t$,

$$P_n(\text{undecayed}) = (1 - \lambda\Delta t)^n$$

Now, in the limit where $\Delta t$ is short, and $n$ is large, as it must be if $T = n\Delta t$ is going to be a normal-scale time interval, you may recognize this as an exponential:

$$\lim_{n\to\infty}P_n(\text{undecayed}) = \lim_{n\to\infty}(1 - \lambda\Delta t)^n = \lim_{n\to\infty}\biggl(1 - \frac{\lambda T}{n}\biggr)^n = e^{-\lambda T}$$

So the equation for exponential decay emerges naturally from the fact that the decay constant is the decay probability per unit time for an undecayed nucleus. (Or of course the same argument applies to any other system that undergoes exponential decay, not just nuclei.)

  • $\begingroup$ Why is it that $lim_{n \to \infty} \left( 1+\frac{x}{n}\right)^n=e^x$ still holds even when $x$ is a function of $n$? ($T=T(n)$ in our case). $\endgroup$ Aug 12, 2020 at 16:06
  • 2
    $\begingroup$ @floccinaucinihilipilificator It doesn't. But in this case $T$ is not the thing you should think of as a function of $n$. It's $\Delta t$ which is a function of $n$, specifically $\Delta t(n) = T/n$, where $T$ is a constant. $\endgroup$
    – David Z
    Aug 12, 2020 at 17:07

Decay constant isn't a probability.

As written here, it's proportionality constant between size of population of radioactive atoms and the rate at which the population decreases due to decay: $$-\frac{dN}{dt}=\lambda N,$$ where $\lambda$ is decay constant and $N$ is population.

Looking at the solution of the equation above we can say: $N=N_0 \exp(-\lambda t)$, so here we can see that $\lambda$ is inverse of time needed for population to decrease by a factor of $e$.

If you read the article your question links to after editing by @Qmechanic, you'll see another examples of exponential decay, not only for radioactive decay.


The probability that a single nucleus decays in a short period of time is approximately given by the decay constant multiplied by the length of the time period. This is only a first order approximation of the exponential decay. If the decay constant is larger than one, the time unit it is given in is obviously too large for the first order approximation too be valid at that time scale.

  • 1
    $\begingroup$ What does it mean for a dimensional constant to be larger than one (which is dimensionless)? $\endgroup$
    – Ruslan
    Jan 3, 2014 at 20:49

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