To determine how long it will take for 27 grams of Plutonium-240 to decay to 9 grams, we can use the exponential decay formula [tex]\( Q(t) = Q_0 e^{-kt} \)[/tex], where:
- [tex]\( Q_0 \)[/tex] is the initial amount of Plutonium-240, which is 27 grams.
- [tex]\( Q(t) \)[/tex] is the amount remaining after time [tex]\( t \)[/tex], which is 9 grams.
- [tex]\( k \)[/tex] is the decay constant, given as 0.00011.
We need to solve for [tex]\( t \)[/tex] in the equation. Starting with [tex]\( Q(t) = Q_0 e^{-kt} \)[/tex], we can substitute the known values:
[tex]\[ 9 = 27 e^{-0.00011 t} \][/tex]
To isolate the exponent, we first divide both sides by 27:
[tex]\[ \frac{9}{27} = e^{-0.00011 t} \][/tex]
[tex]\[ \frac{1}{3} = e^{-0.00011 t} \][/tex]
Next, take the natural logarithm of both sides to get rid of the exponential:
[tex]\[ \ln\left(\frac{1}{3}\right) = \ln\left(e^{-0.00011 t}\right) \][/tex]
Using the property of logarithms that [tex]\(\ln(e^x) = x\)[/tex], we get:
[tex]\[ \ln\left(\frac{1}{3}\right) = -0.00011 t \][/tex]
Now solve for [tex]\( t \)[/tex] by dividing both sides by -0.00011:
[tex]\[ t = \frac{\ln\left(\frac{1}{3}\right)}{-0.00011} \][/tex]
Calculate [tex]\(\ln\left(\frac{1}{3}\right)\)[/tex]:
[tex]\[ \ln\left(\frac{1}{3}\right) = \ln(1) - \ln(3) = 0 - \ln(3) = -\ln(3) \][/tex]
Since [tex]\(\ln(3) \approx 1.0986\)[/tex]:
[tex]\[ \ln\left(\frac{1}{3}\right) = -1.0986 \][/tex]
Now we can substitute this back into the equation for [tex]\( t \)[/tex]:
[tex]\[ t = \frac{-1.0986}{-0.00011} = \frac{1.0986}{0.00011} \approx 9987.3844 \][/tex]
Thus, the time [tex]\( t \)[/tex] is approximately 9987.3844 years. Rounding this to the nearest 10 years, we get:
[tex]\[ t \approx 9990 \text{ years} \][/tex]
Therefore, the amount of time it takes for 27 grams of Plutonium-240 to decay to 9 grams, to the nearest 10 years, is:
A. 9,990 years.
