According to my textbook:

$\text{decay constant } k = \frac{A}{N}$ ,where $A$ is activity, $N$ is no. of undecayed nuclei

$ = \frac{\text{no. of nuclei decayed per unit time}}{\text{total no. of undecayed nuclei}} $

$ = \text{probability of decay per unit time}$

So, let`s say if a radioactive substance has decay constant $ k = 0.05s^{-1}$

This means $P(\text{a undecayed nuclei to decay in 1s}) = 0.05$

But by using $ N = N_0 e^{-kt}$ (where $N_0$ is original no. of undecayed nuclei, N is no. of undecayed nuclei after time $t$) to find the above probability by putting $t = 1s$:

$ N = N_0 e^{-kt}$

$ \frac{N}{N_0}= e^{-kt}$

$ 1 - \frac{N}{N_0}= 1 - e^{-(0.05)(1)}$

$ \frac{N_0 - N}{N_0} = 1 - e^{-(0.05)(1)}$

$ \frac{\text{no. of decayed nuclei}}{\text{original no. of undecayed nuclei}} = 1 - e^{-(0.05)(1)}$

$ P(\text{a undecayed nuclei to decay in 1s})≈ 0.048770575 ≠ 0.05 ≠ k$

Why does this contradiction happen ?

  • $\begingroup$ Are you familiar with the concept of derivatives? $\endgroup$
    – Photon
    Commented Aug 9, 2021 at 11:38
  • $\begingroup$ In your first calculation, you unwittingly assume that the decays are calculated as if they happened all at once, rather than being distributed in infinitesimal time steps. It's like the difference between simple interest and compound interest. $\endgroup$
    – garyp
    Commented Aug 9, 2021 at 11:39

2 Answers 2


Let $$N(t) \equiv N_0 \exp{(-\kappa (t-t_0))} \tag{1}$$ denote the population of a sample at time $t$, where $\kappa >0$ and $N_0 \equiv N(t_0) >0$. The number of decayed particles at time $t$ is then given by $N_\mathrm{D} (t) \equiv N_0 - N(t).$

Taking the derivative of $N(t)$ shows that

$$ \frac{\mathrm{d}N(t)}{\mathrm{d}t} = -\kappa\, N(t) $$ and thus

$$ \kappa = \frac{\mathrm{d} N_{\mathrm{D}}(t)}{\mathrm{d}t} /N(t) \quad . \tag{2}$$ We can see that $\kappa$ is the decay rate (activity) relative to the existent population $N(t)$. Note that this is basically the definition given in your book. Since by definition $\kappa$ is constant, the right-hand side cannot depend on $t$ and we are free to evaluate it at $t=t_0$.

Then for a very small time interval $\Delta t\equiv t-t_0$ with $ 0 <\Delta t \ll 1$, we can approximate the above relation as follows: $$ \kappa \approx \frac{1}{\Delta t} \frac{\Delta N_\mathrm{D}}{N(t_0)} \overset{(1)}{=} \frac{1}{\Delta t} \frac{\Delta N_\mathrm{D}}{N_0} , \tag{3} $$

where $\Delta N_{\mathrm{D}} \equiv N_{\mathrm{D}}(t_0+\Delta t) - N_{\mathrm{D}}(t_0)$ is the number of particles which decayed in $\Delta t$.

You have used equation $(3)$ in your calculations, which is a suitable approximation only for $t \approx t_0$. If you repeat your calculations with even smaller $\Delta t$, e.g. $\Delta t= 0.001\, \mathrm{s}$, then you will see that the approximation gets better and better. This is due to the fact that in the limit $\Delta t\rightarrow 0$ equation $(3)$ becomes exact.

From another perspective, note that for small $\Delta t$ it holds that $$e^{-\kappa \Delta t } \approx 1- \kappa\, \Delta t + \mathcal O (\Delta t^2) \quad . \tag{4}$$

As an exercise, you can use the above relation in equation $(1)$ (neglecting the quadratic and higher orders) to show that this indeed yields the (approximated) decay constant given in equation $(3)$ and vice versa: That is, if equation $(3)$ is a suitable approximation, then relation $(4)$ is too.


The last formula you have written is precisely the "number of decayed nuclei over the original number of nuclei", and not the "probability of an undecayed nuclei to decay in 1 s". But it would be if in the process of decaying, you added some more nuclei to keep the total number of undecayed nuclei, $N(t)$, equal to the original number of undecayed nuclei, $N_0$.

Think that as more of the nuclei decay, less is the probability of one of them decaying (hence the small difference). In other words, the decay constant is defined as an instantaneous rate of decay, that is, referred to the current number of undecayed nuclei, $N(t)$, which is smaller over time (hence the exponential law), rather than to the initial number $N_0$. Simply put, $N(t)$ is the "initial" number of undecayed nuclei at the time $t$.

As for the contradiction, hoping to have an equality in the last line you have written is, essentially, hoping that $$1 - e^{-kt} = kt, $$ which is asimptotically true for $t \rightarrow 0$, as you can verify writing down the Taylor expansion of the exponential.

Some answers of this question adress the mathematical definition of the exponential decay formula with sufficient clarity.

  • 1
    $\begingroup$ You probably mean $1 - e^{-kt} = kt$ instead of $1 - e^{-kt} = k$. $\endgroup$ Commented Aug 9, 2021 at 12:41
  • $\begingroup$ Yes, thank you very much. $\endgroup$ Commented Aug 9, 2021 at 12:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.