# The half-life of cobalt 60 is 5 years. How do you obtain an exponential decay model for cobalt 60 in the form Q(t) = Q0e^−kt?

Aug 6, 2016

$Q \left(t\right) = {Q}_{0} {e}^{- \frac{\ln \left(2\right)}{5} t}$

#### Explanation:

We set up a differential equation. We know that the rate of change of the cobalt is proportional to the amount of cobalt present. We also know that it is a decay model, so there will be a negative sign:

$\frac{\mathrm{dQ}}{\mathrm{dt}} = - k Q$

This is a nice, easy and seperable diff eq:

$\int \frac{\mathrm{dQ}}{Q} = - k \int \mathrm{dt}$

$\ln \left(Q\right) = - k t + C$

$Q \left(0\right) = {Q}_{0}$

$\ln \left({Q}_{0}\right) = C$

$\implies \ln \left(Q\right) = \ln \left({Q}_{0}\right) - k t$

$\ln \left(\frac{Q}{Q} _ 0\right) = - k t$

Raise each side to exponentials:

$\frac{Q}{{Q}_{0}} = {e}^{- k t}$

$Q \left(t\right) = {Q}_{0} {e}^{- k t}$

Now that we know the general form, we need to work out what $k$ is.

Let half life be denoted by $\tau$.

$Q \left(\tau\right) = {Q}_{0} / 2 = {Q}_{0} {e}^{- k \tau}$

$\therefore \frac{1}{2} = {e}^{- k \tau}$

Take natural logs of both sides:

$\ln \left(\frac{1}{2}\right) = - k \tau$

$k = - \frac{\ln \left(\frac{1}{2}\right)}{\tau}$

For tidiness, rewrite $\ln \left(\frac{1}{2}\right) = - \ln \left(2\right)$

$\therefore k = \ln \frac{2}{\tau}$

$k = \ln \frac{2}{5} y {r}^{- 1}$

$\therefore Q \left(t\right) = {Q}_{0} {e}^{- \frac{\ln \left(2\right)}{5} t}$