Math  /  Calculus

Question11. A cloth fragment is found in an ancient tomb. It contains 70%70 \% of the carbon-14 (half-life is 5700 years) that it is assumed to have had initially. How old is the cloth? Round your answer to the nearest integer.

Studdy Solution

STEP 1

1. The initial amount of carbon-14 in the cloth is considered to be 100%100\%.
2. The remaining carbon-14 in the cloth is 70%70\%.
3. The half-life of carbon-14 is 57005700 years.
4. The decay of carbon-14 follows an exponential decay model.

STEP 2

1. Recall the formula for exponential decay.
2. Set up the equation using the given information.
3. Solve for the age of the cloth.
4. Round the answer to the nearest integer.

STEP 3

Recall the formula for exponential decay:
N(t)=N0ekt N(t) = N_0 \cdot e^{-kt}
where: - N(t) N(t) is the remaining quantity of carbon-14 at time t t , - N0 N_0 is the initial quantity of carbon-14, - k k is the decay constant, - t t is the time elapsed.

STEP 4

Set up the equation using the given information. We know that N(t)=0.7N0 N(t) = 0.7N_0 and the half-life t1/2=5700 t_{1/2} = 5700 years. The decay constant k k is given by:
k=ln(2)t1/2 k = \frac{\ln(2)}{t_{1/2}}
Substitute k k into the decay formula:
0.7N0=N0eln(2)5700t 0.7N_0 = N_0 \cdot e^{-\frac{\ln(2)}{5700} \cdot t}

STEP 5

Solve for t t . First, divide both sides by N0 N_0 :
0.7=eln(2)5700t 0.7 = e^{-\frac{\ln(2)}{5700} \cdot t}
Take the natural logarithm of both sides:
ln(0.7)=ln(2)5700t \ln(0.7) = -\frac{\ln(2)}{5700} \cdot t
Solve for t t :
t=ln(0.7)5700ln(2) t = -\frac{\ln(0.7) \cdot 5700}{\ln(2)}
Calculate t t :
t0.356757000.6933011 t \approx -\frac{-0.3567 \cdot 5700}{0.693} \approx 3011

STEP 6

Round the answer to the nearest integer:
t3011 t \approx 3011
The age of the cloth is approximately:
3011 years \boxed{3011} \text{ years}

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