Exponential Decay: need help pease

[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: 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.

 

[soroban]

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: .\(\displaystyle A\:=\e^{rt}\) . 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, \(\displaystyle A\) is half of the initial amount, \(\displaystyle P.\)
. . \(\displaystyle \text{That is, when }t = 140,\;A = \tfrac{1}{2}P\)

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

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

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

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

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

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

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


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

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

. . . . . . . . . \(\displaystyle =\;15e^{-9.297093077}\) .**

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

. . . . . . . . . \(\displaystyle =\; 11.14495717\)

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


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

**

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

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

 

[flajoton]

Thank you very much. This was a ton of help. I totally understand it now.

 

[galactus]

If I may give my two cents.

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

But, we are told that T=140.

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

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

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