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2022-01-29

How do you Find exponential decay rate?

Rylee Marshall

Beginner2022-01-30Added 6 answers

Explanation:

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

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

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:

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

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

$\to \frac{dN}{N}=-\alpha d\left(t\right)$

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

$\to N\left(t\right)=A{e}^{\alpha t}$ 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 $\left\{{e}^{-x}[-1.465,3.9,-0.902,1.782]\right\}$

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 $2.865\times {10}^{11}$ 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:

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

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:

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

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:

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

= -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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