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2022-01-29

How do you Find exponential decay rate?

Rylee Marshall

Explanation:
Usually, exponential decays begin with a differential equation of the following form:

In other words, the rate of population decay is directly inversely proportional to the current population at a given time. $t$. So we can introduce a proportionality constant:

We will now solve the equation to find a function of $N\left(t\right)$
$\to \frac{dN}{N}=-\alpha d\left(t\right)$

$\to N\left(t\right)=A{e}^{\alpha t}$ 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 $\left\{{e}^{-x}\left[-1.465,3.9,-0.902,1.782\right]\right\}$
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 $2.865×{10}^{11}$ atoms left. Work out, $A$ and $\alpha$ . When will the plutonium have only $5×{10}^{8}$ 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: ${10}^{6}years$
$N\left({10}^{6}\right)={10}^{24}{e}^{-\alpha \left({10}^{6}\right)}=2.865×{10}^{11}$
Rearrange to get:
$\alpha =-\frac{1}{{10}^{6}}\mathrm{ln}\left(\frac{2.865×{10}^{11}}{{10}^{24}}\right)\approx 2.888×{10}^{-5}yr\left\{-1\right\}$
So $N\left(t\right)={10}^{24}{e}^{-2.888×{10}^{-5}t}$
For the next part:
$N\left(t\right)=5×{10}^{8}={10}^{24}{e}^{-2.888×{10}^{-5}t}$
Rearrange to get t:
$t=-\frac{1}{2.888×{10}^{-5}}\mathrm{ln}\left(\frac{5×{10}^{8}}{{10}^{24}}\right)\approx 1.22×{10}^{6}yr$
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:

= -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.

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