anonymous
  • anonymous
The half-life of a radioactive substance is the time it takes for half of the material to decay. Phosphorus-32 is used to study a plant's use of fertilizer. It has a half-life of 14.3 days. Write the exponential decay function for a 50-mg sample. Find the amount of phosphorus-32 remaining after 84 days.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
The half-life formula is: |dw:1389736938636:dw|
anonymous
  • anonymous
A is the final amount , Ao is the initial amount, h is the half-life of the substance, and t is the time
anonymous
  • anonymous
just plug the variables into the formula to solve for A.

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anonymous
  • anonymous
how do I do the rest?
anonymous
  • anonymous
Ao is 50 mg, h is 14.3, t is 84 plug it into the formula and solve for A
anonymous
  • anonymous
wouldn't the answer be 146.85?
anonymous
  • anonymous
@coolsday
anonymous
  • anonymous
by plugging those in?
dape
  • dape
So the half-life formula can also be written as \(2^{-\lambda t}\), where \(\lambda\) is the 'decay constant', or just \(1/h\), where h is half life. So the exponent is \(84/14.3\approx5.874\), which says that the sample will have time to halve in size about 5.9 times in the 84 days (this is an easy way to remember the formula). Putting this in, we have that \(2^{-84/14.3}\approx1.7\%\). So about 1.7% of the sample will remain. Starting with 50 mg this means that about \(50\times1.7\%=0.85\) mg of the sample will remain.
dape
  • dape
Oh, and the 'exponential decay function' we just get by putting all this together, so \[P(t)=50\times2^{-t/14.3}\] Where t is time in days.
anonymous
  • anonymous
So that's the answer?
anonymous
  • anonymous
@dape
anonymous
  • anonymous
hellloooo?
anonymous
  • anonymous
0.85 mg will remain after 84 days if you use the formula.

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