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