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Emily778

  • one year ago

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.

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  1. coolsday
    • one year ago
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    The half-life formula is: |dw:1389736938636:dw|

  2. coolsday
    • one year ago
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    A is the final amount , Ao is the initial amount, h is the half-life of the substance, and t is the time

  3. coolsday
    • one year ago
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    just plug the variables into the formula to solve for A.

  4. Emily778
    • one year ago
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    how do I do the rest?

  5. coolsday
    • one year ago
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    Ao is 50 mg, h is 14.3, t is 84 plug it into the formula and solve for A

  6. Emily778
    • one year ago
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    wouldn't the answer be 146.85?

  7. Emily778
    • one year ago
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    @coolsday

  8. Emily778
    • one year ago
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    by plugging those in?

  9. dape
    • one year ago
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    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.

  10. dape
    • one year ago
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    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.

  11. Emily778
    • one year ago
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    So that's the answer?

  12. Emily778
    • one year ago
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    @dape

  13. Emily778
    • one year ago
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    hellloooo?

  14. coolsday
    • one year ago
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    0.85 mg will remain after 84 days if you use the formula.

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