Question

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?

Answer 1

`Q(t) = Q_0e^(-(ln(2))/5t)`

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:

`(dQ)/(dt) = - kQ`

This is a nice, easy and seperable diff eq:

`int (dQ)/(Q) = -k int dt`

`ln(Q) = - kt + C`

`Q(0) = Q_0`

`ln(Q_0) = C`

`implies ln(Q) = ln(Q_0) - kt`

`ln(Q/Q_0) = -kt`

Raise each side to exponentials:

`(Q)/(Q_0) = e^(-kt)`

`Q(t) = Q_0e^(-kt)`

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

Let half life be denoted by `tau`.

`Q(tau) = Q_0/2 = Q_0e^(-ktau)`

`therefore 1/2 = e^(-ktau)`

Take natural logs of both sides:

`ln(1/2) = -ktau`

`k = - (ln(1/2))/tau`

For tidiness, rewrite `ln(1/2) = -ln(2)`

`therefore k = ln(2)/tau`

`k = ln(2)/(5)yr^(-1)`

`therefore Q(t) = Q_0e^(-(ln(2))/5t)`