How many milligrams of 14C should be in the bone fragment 10,000 years later? Round to the nearest tenth.

1. Break Down the Problem

To determine how much of the original $^{14}C$ remains in a bone fragment after 10,000 years, we need to use the concept of radioactive decay. The half-life of $^{14}C$ is approximately 5,730 years.

2. Relevant Concepts

The formula for radioactive decay is given by:

$N(t) = N_0 \left( \frac{1}{2} \right)^{\frac{t}{t_{1/2}}}$

Where:

• $N(t)$ = remaining quantity of the radioactive substance after time $t$
• $N_0$ = initial quantity of the substance
• $t$ = time elapsed (in years)
• $t_{1/2}$ = half-life of the substance

3. Analysis and Detail

We are looking for the amount of $^{14}C$ remaining after $t = 10,000$ years.

1. Given $t_{1/2} = 5730$ years.
2. Calculate how many half-lives have passed in 10,000 years:

$\text{Number of half-lives} = \frac{10,000}{5730} \approx 1.747$

3. Substitute $t$ and $t_{1/2}$ into the decay formula. We won't use a specific initial value yet, but let's assume $N_0 = 1 \, \text{mg}$ for simplicity.

$N(10,000) = 1 \left( \frac{1}{2} \right)^{1.747}$

4. Calculate $\left( \frac{1}{2} \right)^{1.747}$:

$\left( \frac{1}{2} \right)^{1.747} \approx 0.297$

Thus, the remaining quantity after 10,000 years is:

$N(10,000) = 1 \times 0.297 \approx 0.297 \, \text{mg}$

4. Verify and Summarize

Since we assumed an initial amount of 1 mg for simplicity, if the actual initial amount was different, the remaining amount can be scaled proportionally. However, the decay factor given a half-life of $5730$ years remains constant.

Final Answer

The bone fragment will contain approximately 0.3 mg of $^{14}C$ after 10,000 years.