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cutiecomittee123

  • one year ago

That function may be used to model radioactive decay. Q represents the quantity remaining after t years. K is the decay constant for plutonium-240. What is the decay constant if its half life is 6300 years?

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  1. cutiecomittee123
    • one year ago
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    |dw:1433458510363:dw|

  2. whpalmer4
    • one year ago
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    \[Q(t) = Q_0 e^{-kt}\] the half-life is 6300 years, so we know the following: \[Q(0) = Q_0\]\[Q(6300) = \frac{1}2 Q_0\]because\(Q_0\) represents the initial quantity, and after one half-life, there is only half as much left. That means that we can find the value of \(k\) by setting \[Q(6300) = \frac{1}{2} Q_0 = Q_0e^{-k*6300}\]and if we do some cancellation of identical factors from both sides: \[\frac{1}{2}\cancel{Q_0} = \cancel{Q_0}e^{-k*6300}\]\[\frac{1}{2} = e^{-6300k}\]which you should be able to solve for \(k\).

  3. cutiecomittee123
    • one year ago
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    Okay can you show me how to? Im confused

  4. cutiecomittee123
    • one year ago
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    @whpalmer4

  5. whpalmer4
    • one year ago
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    Of course! What happens if we take the logarithm of an exponential?

  6. whpalmer4
    • one year ago
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    \[\log_b b^x = x\]right? and \(\ln\) is just the special case of the logarithm to base \(e\), the base of the natural logarithm, \(e\approx 2.718281828...\)

  7. whpalmer4
    • one year ago
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    That means if we take the natural logarithm of both sides: \[\ln(\frac{1}{2}) = \ln(e^{-6300k})\]\[\ln(\frac{1}{2}) = -6300k\]divide both sides by -6300 and you have your answer.

  8. cutiecomittee123
    • one year ago
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    So when I came up with the answer I went to select it and it said it was actually -ln(0.5)/-6300k

  9. whpalmer4
    • one year ago
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    the decay constant for Pu240 is \[k = -\frac{\ln{(\frac{1}{2}})}{6300} \approx 0.00011\]

  10. whpalmer4
    • one year ago
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    You can check it by plugging it into the equation: \[Q(0) = Q_0*e^{-0.00011(0)} = Q_0*e^0 = Q_0\]\[Q(6300)=Q_0*e^{-0.00011*6300} = Q_0*e^{-0.693} = Q_0*0.5\]so after 6300 years (one half-life), we have half of the initial quantity.

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