QuestionMA 107 Test 3 (Livengood) Problem 2 Carbon-14 is a radioactive isotype of carbon. Assuming that modern-day carbon and ancient carbon contain the same amount of carbon-14, we can measure the amount of carbon-14 present in ancient artifacts to estimate how old they are.
An ancient painting is found in a cave. One of the pigments in the painting is charcoal, which contains carbon-14. Therefore, we can measure the carbon-14 present in the painting and compare it to the carbon-14 present in the living tree that the wood for the charcoal came from.
The charcoal in this painting contains $20 \%$ of the carbon-14 that modern-day charcoal contains. The half-life of carbon-14 is about 5730 years. Using the continuous exponential decay formula $A(t)=A_{0} e^{k t}$ , determine how old the charcoal used to make the painting is.

Studdy Solution

STEP 1

1. The continuous exponential decay formula is given by $A(t) = A_{0} e^{kt}$ .
2. The half-life of carbon-14 is 5730 years.
3. The charcoal in the painting contains 20% of the carbon-14 that modern-day charcoal contains.
4. We need to determine the age of the charcoal, i.e., $t$ .

STEP 2

1. Determine the decay constant $k$ using the half-life information.
2. Set up the exponential decay equation using the given percentage of carbon-14.
3. Solve for $t$ , the age of the charcoal.

STEP 3

Determine the decay constant $k$ .
The half-life formula is given by:
$A(t) = \frac{1}{2} A_{0}$
Using the exponential decay formula:
$\frac{1}{2} A_{0} = A_{0} e^{k \cdot 5730}$
Divide both sides by $A_{0}$ :
$\frac{1}{2} = e^{k \cdot 5730}$
Take the natural logarithm of both sides:
$\ln\left(\frac{1}{2}\right) = k \cdot 5730$
Solve for $k$ :
$k = \frac{\ln\left(\frac{1}{2}\right)}{5730}$

STEP 4

Set up the exponential decay equation using the given percentage of carbon-14.
Given that the charcoal contains 20% of the original carbon-14, we have:
$A(t) = 0.2 A_{0}$
Substitute into the decay formula:
$0.2 A_{0} = A_{0} e^{kt}$
Divide both sides by $A_{0}$ :
$0.2 = e^{kt}$

STEP 5

Solve for $t$ .
Take the natural logarithm of both sides:
$\ln(0.2) = kt$
Substitute $k$ from Step 1:
$\ln(0.2) = \left(\frac{\ln\left(\frac{1}{2}\right)}{5730}\right) t$
Solve for $t$ :
$t = \frac{\ln(0.2)}{\frac{\ln\left(\frac{1}{2}\right)}{5730}}$
Calculate $t$ :
$t \approx \frac{-1.6094}{-0.000121}$
$t \approx 13,573 \text{ years}$
The age of the charcoal is approximately:
$\boxed{13,573} \text{ years}$

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