QuestionClick here to watch the video. A scientist determined that the bones from a mastodon had lost $83.8 \%$ of their carbon-14. How old were the bones at the time they were discovered? $\qquad$ $\square$

Studdy Solution

STEP 1

1. The problem involves the decay of carbon-14, a radioactive isotope used in dating.
2. The half-life of carbon-14 is approximately 5730 years.
3. The bones have lost 83.8% of their carbon-14, meaning 16.2% remains.
4. We need to calculate the age of the bones based on the remaining carbon-14.

STEP 2

1. Understand the concept of radioactive decay and half-life.
2. Set up the decay formula.
3. Solve for the time variable to find the age of the bones.

STEP 3

Understand the concept of radioactive decay and half-life.
Radioactive decay follows an exponential decay model, where the amount of a substance decreases over time. The half-life is the time it takes for half of the substance to decay.

STEP 4

Set up the decay formula.
The formula for exponential decay is given by:
$N(t) = N_0 \times \left(\frac{1}{2}\right)^{\frac{t}{T_{1/2}}}$
Where: - $N(t)$ is the remaining quantity of the substance after time $t$ . - $N_0$ is the initial quantity of the substance. - $T_{1/2}$ is the half-life of the substance. - $t$ is the time that has passed.
In this problem, $N(t) = 0.162 \times N_0$ because 16.2% of the carbon-14 remains.

STEP 5

Solve for the time variable to find the age of the bones.
Using the decay formula:
$0.162 \times N_0 = N_0 \times \left(\frac{1}{2}\right)^{\frac{t}{5730}}$
Divide both sides by $N_0$ :
$0.162 = \left(\frac{1}{2}\right)^{\frac{t}{5730}}$
Take the natural logarithm of both sides to solve for $t$ :
$\ln(0.162) = \ln\left(\left(\frac{1}{2}\right)^{\frac{t}{5730}}\right)$
$\ln(0.162) = \frac{t}{5730} \cdot \ln\left(\frac{1}{2}\right)$
Solve for $t$ :
$t = \frac{\ln(0.162)}{\ln\left(\frac{1}{2}\right)} \times 5730$
Calculate $t$ :
$t \approx \frac{-1.817}{-0.693} \times 5730$
$t \approx 2.622 \times 5730$
$t \approx 15022.26$
The age of the bones is approximately:
$\boxed{15022}$ years

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