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

A = \lambda N         (1)

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)

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)}

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)}