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November 17th, 2017, 04:49 PM   #1
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How do I find the half-life

In 2 years, 20% of a radioactive element decays. Find its half-life rounded to 2 decimal places.

I think k=LN(0.20)/2, but I'm unable to figure out how to solve for the half-life.
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November 17th, 2017, 04:57 PM   #2
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$0.8 = \left(\dfrac{1}{2}\right)^{2/h}$

solve for the half-life, $h$
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November 17th, 2017, 05:42 PM   #3
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I prefer to write any problem involving "half life" as powers of 1/2. Since skeeter beat me to that, here is how to do it in a more "formulaic" method.

Any such problem (constant rate of decay) can be written in the form $S(t)= Ce^{kt}$.
When $t = 0$, $S(0) = C$, the initial amount.
When $t = 2$, $S(2) = Ce^{2k}$ and we are told that is $C - 0.2C = 0.8C$, so $e^{2k}= 0.8$.
Hence $2k= \ln(0.8)$ or $k= \frac{\ln(0.8)}{2}$.
Hence the formula is $\displaystyle S(t)= Ce^{\frac{\ln(0.8)t}{2}}= C\left(e^{\ln(0.8)}\right)^{t/2}= C(0.8^{t/2})$.
"Half life" is the value of $t$ that makes that $C/2$.
We need to solve $C(0.8^{t/2})= C/2$ or $0.8^{t/2}= \frac{1}{2}$.
That is the same as skeeter's solution.

Last edited by skipjack; November 17th, 2017 at 07:09 PM.
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