The activity of a radioactive source is the rate at which hat source decays and it is defined using:

$A = \lambda N$         (1)

where
$A$  is the activity
$\lambda$ is the decay constant
$N$ is the number of undecayed nuclei

Since the activity is the rate at which undecayed nuclei decay, we can also write;

$A = \frac{dN}{dt}$         (2)

where
$dN$ is the change in the number of undecayed nuclei
$dN$ is the change in time (in which $dN$ decay over)

Equating equations (1) and (2) gives us;

$\frac{dN}{dt} = - \lambda N$

(the negative sign arises because the number if undecayed nuclei is decreasing over time.)

Rearranging this so as to integrate;

$dN = -\lambda N dt$

$\frac{dN}{N} = -\lambda dt$

This becomes:

$\int \frac{dN}{N} = \int -\lambda dt$

Since we are integrating, limits are required;

• We are integrating from an initial time $t = 0 s$ to some point in time $t s$ later.
• The number of undecided nuclei will change from an initial amount of $N = N_{0}$ to a new amount some time ($t$) later, $N(t)$.

$\int_{N_{0}}^{N(t)} \frac{dN}{N} = \int_{0}^{t} -\lambda dt$

This becomes;

$\ln N(t) - \ln N_{0} = -\lambda t - - \lambda 0$

$\ln N(t) - \ln N_{0} = -\lambda t$

Using log rules whereby  $\ln X - \ln Y = \ln \frac{X}{Y}$ , this becomes;

$\frac{\ln N(t)}{\ln N_{0}} = -\lambda t$

Taking the exponential of both sides gives;

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

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

Rearranging for $N(t)$ gives:

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

where
$N(t)$ is the amount of undecayed nuclei remaining after time $t s$
$N_{0}$ is the initial amount of undecayed nuclei at $t = 0 s$
$\lambda$ is the decay constant (which is dependent on the type of nuclei material)
$t$ is the chosen time frame in which the nuclei are decaying over

Since $A = \lambda N$

Rearranging for $N$ gives;

$N = \frac{A}{\lambda}$

The maximum activity will occur when there is the maximum number of undecided nuclei (because there are more unstable nuclei and therefore a possibility for more to decay per unit time), so we can write;

$N_{0} = \frac{A_{0}}{\lambda}$

Substituting this in to  $N(t) = N_{0} e^{(-\lambda t)}$ gives;

$\frac{A(t)}{\lambda} = \frac{A_{0}}{\lambda} e^{(-\lambda t)}$

Multiplying both sides by $\lambda$ gives;

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