How do you Find exponential decay rate?

framtalshg
2022-01-29
Answered

How do you Find exponential decay rate?

You can still ask an expert for help

Rylee Marshall

Answered 2022-01-30
Author has **6** answers

Explanation:

Exponential decays typically start with a differential equation of the form:

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

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:

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

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.

Exponential decays typically start with a differential equation of the form:

That is, the rate at which a population of something decays is directly proportional to the negative of the current population at time

We will now solve the equation to find a function of

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

graph

Perhaps an example might help?

Consider a lump of plutonium 239 which initially has

We are told the lump has

Now at 1 million years:

Rearrange to get:

So

For the next part:

Rearrange to get t:

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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what I have done so far is I put the nubmer and x,y and z in matrix form:

$\left[\begin{array}{ccc}2& -3& 4\\ 6& 4& -2\\ 1& 5& 4\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}-19\\ 8\\ 23\end{array}\right]$

what I have done so far is I put the nubmer and x,y and z in matrix form:

$\left[\begin{array}{ccc}2& -3& 4\\ 6& 4& -2\\ 1& 5& 4\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}-19\\ 8\\ 23\end{array}\right]$

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$\begin{array}{|cc|}\hline t& N=N(t)\\ 10& 16.9\\ 20& 22.5\\ 30& 49.5\\ 40& 52.6\\ 50& 55.2\\ 60& 55.6\\ \hline\end{array}$

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