# How do you Find exponential decay rate?

Jan 12, 2018

See below.

#### Explanation:

$\frac{\mathrm{dN}}{\mathrm{dt}} \propto - N \left(t\right)$

That is, the rate at which a population of something decays is directly proportional to the negative of the current population at time $t$. So we can introduce a proportionality constant:

$\frac{\mathrm{dN}}{\mathrm{dt}} = - \alpha N \left(t\right)$

We will now solve the equation to find a function of $N \left(t\right)$:

$\to \frac{\mathrm{dN}}{N} = - \alpha \mathrm{dt}$

$\to \int \frac{\mathrm{dN}}{N} = \int - \alpha \mathrm{dt} \to \ln \left(N\right) = - \alpha t + C$

$\to N \left(t\right) = A {e}^{- \alpha t}$ where $A$ is a constant.

This is the general form of the exponential decay formula and will typically have graphs that look like this:

graph{e^-x [-1.465, 3.9, -0.902, 1.782]}

Perhaps an example might help?

Consider a lump of plutonium 239 which initially has ${10}^{24}$ atoms. After one million years have elapsed years the plutonium now has $2.865 \times {10}^{11}$ atoms left. Work out, $A$ and $\alpha .$ When will the plutonium have only $5 \times {10}^{8}$ atoms left and what is the decay rate here?

We are told the lump has ${10}^{24}$ atoms at $t = 0$ so:

$N \left(0\right) = A {e}^{0} = {10}^{24} \to A = {10}^{24}$

Now at 1 million years: ${10}^{6}$ years:

$N \left({10}^{6}\right) = {10}^{24} {e}^{- \alpha \left({10}^{6}\right)} = 2.865 \times {10}^{11}$

Rearrange to get:

$\alpha = - \frac{1}{{10}^{6}} \ln \left(\frac{2.865 \times {10}^{11}}{10} ^ 24\right) \approx 2.888 \times {10}^{- 5} y {r}^{-} 1$

So $N \left(t\right) = {10}^{24} {e}^{- 2.888 \times {10}^{- 5} t}$

For the next part:

$N \left(t\right) = 5 \times {10}^{8} = {10}^{24} {e}^{- 2.888 \times {10}^{- 5} t}$

Rearrange to get $t$:

$t = - \frac{1}{2.888 \times {10}^{- 5}} \ln \left(\frac{5 \times {10}^{8}}{{10}^{24}}\right) \approx 1.22 \times {10}^{6} y r$

Now for the last part, the decay rate is already defined a way back at the very start, simply evaluate it at the given time:

$\frac{\mathrm{dN}}{\mathrm{dt}} = - \alpha t = - 2.888 \times {10}^{- 5} \left(1.22 \times {10}^{6}\right)$

$= - 35.23$ atoms per year.

The idea is to start with differential equation above, which gives the decay rate, and solve it to get the population at any given time.