Answer :
To determine how long it will take for a quantity of plutonium-240 to decay to 25% of its original amount, we can use the radioactive decay formula:
[tex]\[ Q = Q_0 \cdot e^{-k \cdot t} \][/tex]
where:
- [tex]\( Q \)[/tex] is the remaining quantity.
- [tex]\( Q_0 \)[/tex] is the original quantity.
- [tex]\( k \)[/tex] is the decay constant.
- [tex]\( t \)[/tex] is the time in years.
Given values:
- [tex]\( Q_0 \)[/tex] is the original amount (we can assume this is 1 unit for simplicity).
- [tex]\( Q \)[/tex] is 25% of the original amount, which is [tex]\( 0.25 \cdot Q_0 \)[/tex].
- The decay constant [tex]\( k \)[/tex] is 0.00011.
We need to find the time [tex]\( t \)[/tex] when [tex]\( Q = 0.25 \cdot Q_0 \)[/tex].
Starting with the decay formula, we substitute [tex]\( Q = 0.25 \cdot Q_0 \)[/tex]:
[tex]\[ 0.25 \cdot Q_0 = Q_0 \cdot e^{-0.00011 \cdot t} \][/tex]
Since [tex]\( Q_0 \)[/tex] is not zero, we can divide both sides of the equation by [tex]\( Q_0 \)[/tex]:
[tex]\[ 0.25 = e^{-0.00011 \cdot t} \][/tex]
Next, take the natural logarithm of both sides to solve for [tex]\( t \)[/tex]:
[tex]\[ \ln(0.25) = -0.00011 \cdot t \][/tex]
To isolate [tex]\( t \)[/tex], divide both sides by [tex]\(-0.00011\)[/tex]:
[tex]\[ t = \frac{\ln(0.25)}{-0.00011} \][/tex]
Calculate the natural logarithm of 0.25 and divide by [tex]\(-0.00011\)[/tex]:
[tex]\[ \ln(0.25) \approx -1.3863 \][/tex]
So,
[tex]\[ t = \frac{-1.3863}{-0.00011} \approx 12602.676010180823 \][/tex]
Therefore, it will take approximately 12,602.7 years for a quantity of plutonium-240 to decay to 25% of its original amount.
The correct answer is:
C. 12,600 years
[tex]\[ Q = Q_0 \cdot e^{-k \cdot t} \][/tex]
where:
- [tex]\( Q \)[/tex] is the remaining quantity.
- [tex]\( Q_0 \)[/tex] is the original quantity.
- [tex]\( k \)[/tex] is the decay constant.
- [tex]\( t \)[/tex] is the time in years.
Given values:
- [tex]\( Q_0 \)[/tex] is the original amount (we can assume this is 1 unit for simplicity).
- [tex]\( Q \)[/tex] is 25% of the original amount, which is [tex]\( 0.25 \cdot Q_0 \)[/tex].
- The decay constant [tex]\( k \)[/tex] is 0.00011.
We need to find the time [tex]\( t \)[/tex] when [tex]\( Q = 0.25 \cdot Q_0 \)[/tex].
Starting with the decay formula, we substitute [tex]\( Q = 0.25 \cdot Q_0 \)[/tex]:
[tex]\[ 0.25 \cdot Q_0 = Q_0 \cdot e^{-0.00011 \cdot t} \][/tex]
Since [tex]\( Q_0 \)[/tex] is not zero, we can divide both sides of the equation by [tex]\( Q_0 \)[/tex]:
[tex]\[ 0.25 = e^{-0.00011 \cdot t} \][/tex]
Next, take the natural logarithm of both sides to solve for [tex]\( t \)[/tex]:
[tex]\[ \ln(0.25) = -0.00011 \cdot t \][/tex]
To isolate [tex]\( t \)[/tex], divide both sides by [tex]\(-0.00011\)[/tex]:
[tex]\[ t = \frac{\ln(0.25)}{-0.00011} \][/tex]
Calculate the natural logarithm of 0.25 and divide by [tex]\(-0.00011\)[/tex]:
[tex]\[ \ln(0.25) \approx -1.3863 \][/tex]
So,
[tex]\[ t = \frac{-1.3863}{-0.00011} \approx 12602.676010180823 \][/tex]
Therefore, it will take approximately 12,602.7 years for a quantity of plutonium-240 to decay to 25% of its original amount.
The correct answer is:
C. 12,600 years
