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Emily778

  • 11 months 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
    • 11 months ago
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    The half-life formula is: |dw:1389736938636:dw|

  2. coolsday
    • 11 months 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
    • 11 months ago
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    just plug the variables into the formula to solve for A.

  4. Emily778
    • 11 months ago
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    how do I do the rest?

  5. coolsday
    • 11 months 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
    • 11 months ago
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    wouldn't the answer be 146.85?

  7. Emily778
    • 11 months ago
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    @coolsday

  8. Emily778
    • 11 months ago
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    by plugging those in?

  9. dape
    • 11 months 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
    • 11 months 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
    • 11 months ago
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    So that's the answer?

  12. Emily778
    • 11 months ago
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    @dape

  13. Emily778
    • 11 months ago
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    hellloooo?

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

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