Answer :
Sure, let's determine the age of the fossilized leaf given that it contains 40% of its normal amount of carbon-14, and knowing that the half-life of carbon-14 is 5600 years. We'll use the exponential decay formula:
[tex]\[ A = A_0 e^{kt} \][/tex]
where:
- [tex]\( A \)[/tex] is the remaining amount of carbon-14.
- [tex]\( A_0 \)[/tex] is the initial amount of carbon-14.
- [tex]\( k \)[/tex] is the decay constant.
- [tex]\( t \)[/tex] is the time in years.
We are given:
- [tex]\( A = 0.40 A_0 \)[/tex] (since it contains 40% of the normal amount).
- [tex]\( t \)[/tex] is what we need to find.
- The half-life of carbon-14 is 5600 years.
First, we need to determine the decay constant [tex]\( k \)[/tex]. The decay constant can be found using the half-life formula:
[tex]\[ k = \frac{\ln(0.5)}{t_{1/2}} \][/tex]
where [tex]\( t_{1/2} \)[/tex] is the half-life of carbon-14. Plugging in the given half-life:
[tex]\[ k = \frac{\ln(0.5)}{5600} \][/tex]
Next, we use the exponential decay formula to solve for [tex]\( t \)[/tex]:
[tex]\[ 0.40 A_0 = A_0 e^{kt} \][/tex]
Dividing both sides by [tex]\( A_0 \)[/tex] and substituting the value of [tex]\( k \)[/tex]:
[tex]\[ 0.40 = e^{kt} \][/tex]
[tex]\[ \ln(0.40) = kt \][/tex]
[tex]\[ t = \frac{\ln(0.40)}{k} \][/tex]
Substitute [tex]\( k \)[/tex] from the first step:
[tex]\[ t = \frac{\ln(0.40)}{\frac{\ln(0.5)}{5600}} \][/tex]
Evaluating this yields:
[tex]\[ t \approx 7402.797331369229 \][/tex]
Rounding this to the nearest year, we find:
[tex]\[ t \approx 7403 \][/tex]
Therefore, the fossil is approximately 7403 years old.
[tex]\[ A = A_0 e^{kt} \][/tex]
where:
- [tex]\( A \)[/tex] is the remaining amount of carbon-14.
- [tex]\( A_0 \)[/tex] is the initial amount of carbon-14.
- [tex]\( k \)[/tex] is the decay constant.
- [tex]\( t \)[/tex] is the time in years.
We are given:
- [tex]\( A = 0.40 A_0 \)[/tex] (since it contains 40% of the normal amount).
- [tex]\( t \)[/tex] is what we need to find.
- The half-life of carbon-14 is 5600 years.
First, we need to determine the decay constant [tex]\( k \)[/tex]. The decay constant can be found using the half-life formula:
[tex]\[ k = \frac{\ln(0.5)}{t_{1/2}} \][/tex]
where [tex]\( t_{1/2} \)[/tex] is the half-life of carbon-14. Plugging in the given half-life:
[tex]\[ k = \frac{\ln(0.5)}{5600} \][/tex]
Next, we use the exponential decay formula to solve for [tex]\( t \)[/tex]:
[tex]\[ 0.40 A_0 = A_0 e^{kt} \][/tex]
Dividing both sides by [tex]\( A_0 \)[/tex] and substituting the value of [tex]\( k \)[/tex]:
[tex]\[ 0.40 = e^{kt} \][/tex]
[tex]\[ \ln(0.40) = kt \][/tex]
[tex]\[ t = \frac{\ln(0.40)}{k} \][/tex]
Substitute [tex]\( k \)[/tex] from the first step:
[tex]\[ t = \frac{\ln(0.40)}{\frac{\ln(0.5)}{5600}} \][/tex]
Evaluating this yields:
[tex]\[ t \approx 7402.797331369229 \][/tex]
Rounding this to the nearest year, we find:
[tex]\[ t \approx 7403 \][/tex]
Therefore, the fossil is approximately 7403 years old.
