## 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?

1. cutiecomittee123

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2. whpalmer4

$Q(t) = Q_0 e^{-kt}$ 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

Okay can you show me how to? Im confused

4. cutiecomittee123

@whpalmer4

5. whpalmer4

Of course! What happens if we take the logarithm of an exponential?

6. whpalmer4

$\log_b b^x = x$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

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

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

the decay constant for Pu240 is $k = -\frac{\ln{(\frac{1}{2}})}{6300} \approx 0.00011$

10. whpalmer4

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.