Exponential Decay: need help pease

flajoton

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The radioactive element polonium has a half life of 140 days. If 15 mg. of this substance decays exponentially, how much of the substance would remain after 60 days?
I am not sure on how to do this problem. i think i have to use the formula: A=Pe^rt but I am not sure as to where these values go. The answer to the problem is already given to me. It is 11.15 mg. I just cannot figure it out. Thank you for your help. :D
 
Hello, flajoton!

The radioactive element polonium has a half life of 140 days.
If 15 mg. of this substance decays exponentially,
. . how much of the substance would remain after 60 days?

I am not sure on how to do this problem.
i think i have to use the formula: .\(\displaystyle A\:=\:pe^{rt}\) . Yes!
. . but I am not sure as to where these values go.
The answer: 11.15 mg. . . um... not quite

"Half-life is 140 days."
This means: after 140 days, A\displaystyle A is half of the initial amount, P.\displaystyle P.
. . That is, when t=140,  A=12P\displaystyle \text{That is, when }t = 140,\;A = \tfrac{1}{2}P

Substitute into the formula:   12P  =  Pe140r\displaystyle \text{Substitute into the formula: }\;\tfrac{1}{2}P \;=\;Pe^{140r}

\(\displaystyle \text{Switch sides: }\;\;Pe^{140r{ \;=\;\tfrac{1}{2}P\)

Divide by P:    e140r  =  12\displaystyle \text{Divide by }P:\;\;e^{140r} \;=\;\tfrac{1}{2}

Tale logs:     ln(e140r)  =  ln(12)  =  ln(21)  =  ln(2)\displaystyle \text{Tale logs: }\;\;\ln\left(e^{140r}\right) \;=\;\ln\left(\tfrac{1}{2}\right) \;=\;\ln\left(2^{-1}\right) \;=\;-\ln(2)

. . Then:     140rln(e)This is 1  =  ln(2)\displaystyle \text{Then: }\;\;140r\underbrace{\ln(e)}_{\text{This is 1}} \;=\;-\ln(2)
We have:   140r  =  ln(2)r  =  ln(2)140\displaystyle \text{We have: }\;140r \;=\;-\ln(2) \quad\Rightarrow\quad r \;=\;-\frac{\ln(2)}{140}

We are told that:   P=15\displaystyle \text{We are told that: }\;P = 15

Substitute into the formula:     A  =  15eln(2)140t\displaystyle \text{Substitute into the formula: }\;\;\boxed{A \;=\;15e^{-\frac{\ln(2)}{140}t}}


The question asks: "What is A when t=60?\displaystyle \text{The question asks: "What is }A\text{ when }t = 60\:?

We have:   A  =  15eln(2)140(60)\displaystyle \text{We have: }\;A \;=\;15e^{-\frac{\ln(2)}{140}(60)}

. . . . . . . . . =  15e9.297093077\displaystyle =\;15e^{-9.297093077} .**

. . . . . . . . . =  15(0.742997145)\displaystyle = \; 15(0.742997145)

. . . . . . . . . =  11.14495717\displaystyle =\; 11.14495717

. . . . . . . . .   11.14 mg\displaystyle \approx\;11.14\text{ mg}


~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

**

If they rounded this to 0.297\displaystyle -0.297 (a bad idea),

. . they would get:   A=15e(0.297)(60)  =  11.14566019    11.15\displaystyle \text{they would get: }\;A \:=\:15e^{-(0.297)(60)} \;=\;11.14566019 \;\approx\;11.15

 
If I may give my two cents. :D

Half-life is given by T=ln(2)k\displaystyle T=\frac{-ln(2)}{k}, where k is the constsnt of proportionately in the formula A=A0ekt\displaystyle A=A_{0}e^{kt}

But, we are told that T=140.

140=ln(2)kk=ln(2)140.00495105129\displaystyle 140=\frac{-ln(2)}{k}\Rightarrow k=\frac{-ln(2)}{140}\approx -.00495105129

So, since there is 15 mg initially, we have A=15eln(2)140(60)=1523711.144957\displaystyle A=15e^{\frac{-ln(2)}{140}(60)}=15\cdot 2^{\frac{-3}{7}}\approx 11.144957

I agree with Soroban. They are off by a small amount. But when dealing with radioactive material, this may be significant :wink: .
 
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