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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 plutonium240. What is the decay constant if its half life is 6300 years?
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 plutonium240. What is the decay constant if its half life is 6300 years?

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cutiecomittee123
 one year ago
Best ResponseYou've already chosen the best response.1dw:1433458510363:dw

whpalmer4
 one year ago
Best ResponseYou've already chosen the best response.0\[Q(t) = Q_0 e^{kt}\] the halflife 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 halflife, 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 \(k\).

cutiecomittee123
 one year ago
Best ResponseYou've already chosen the best response.1Okay can you show me how to? Im confused

whpalmer4
 one year ago
Best ResponseYou've already chosen the best response.0Of course! What happens if we take the logarithm of an exponential?

whpalmer4
 one year ago
Best ResponseYou've already chosen the best response.0\[\log_b b^x = x\]right? and \(\ln\) is just the special case of the logarithm to base \(e\), the base of the natural logarithm, \(e\approx 2.718281828...\)

whpalmer4
 one year ago
Best ResponseYou've already chosen the best response.0That 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.

cutiecomittee123
 one year ago
Best ResponseYou've already chosen the best response.1So when I came up with the answer I went to select it and it said it was actually ln(0.5)/6300k

whpalmer4
 one year ago
Best ResponseYou've already chosen the best response.0the decay constant for Pu240 is \[k = \frac{\ln{(\frac{1}{2}})}{6300} \approx 0.00011\]

whpalmer4
 one year ago
Best ResponseYou've already chosen the best response.0You 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 halflife), we have half of the initial quantity.
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