# Question #2720a

Mar 1, 2017

Decay constant: $\lambda \approx 2.3 \times {10}^{- 4}$

#### Explanation:

If the half-life is 3000 years, we can say that the amount of activity follows this rule:

$A \left(t\right) = {A}_{o} {\left(\frac{1}{2}\right)}^{\frac{t}{3000}}$

So for $t = 3000$ we have $A \left(3000\right) = {A}_{o} {\left(\frac{1}{2}\right)}^{1} = {A}_{o} / 2$

This is the most literal formulation of the idea of a half-life .

The decay constant is slightly more abstract, it requires us to move to using natural logs:

$A \left(t\right) = {A}_{o} {\left(\frac{1}{2}\right)}^{\frac{t}{3000}} = {A}_{o} {\left(2\right)}^{\textcolor{red}{-} \frac{t}{3000}} = \textcolor{b l u e}{{A}_{o} {e}^{- \lambda t}}$, where $\lambda$ is the decay constant.

So we have now calculus-friendly exponential decay.

We can then say that:

${2}^{- \frac{t}{3000}} = {e}^{- \lambda t}$

Or:

$\ln {2}^{- \frac{t}{3000}} = \ln {e}^{- \lambda t}$

$\implies - \frac{t}{3000} \ln \left(2\right) = - \lambda t$

$\implies \lambda = \frac{\ln \left(2\right)}{3000} \approx 2.3 \times {10}^{- 4}$

So we test this again using our new definition:

$A \left(3000\right) = {A}_{o} {e}^{- 2.3 \times {10}^{- 4} \cdot 3000} = 0.501 \approx \frac{1}{2}$