The half-life of carbon 14 is 5,730 years. Approximately, how old is a bone that has 70% of its original carbon 14?
Question
The half-life of carbon 14 is 5,730 years. Approximately, how old is a bone that has 70% of its original carbon 14?
Solution
To solve this problem, we need to use the formula for exponential decay which is N = N0 * (1/2)^(t/h), where:
- N is the final amount of the substance
- N0 is the initial amount of the substance
- t is the time that has passed
- h is the half-life of the substance
In this case, we know that N/N0 = 70/100 = 0.7, and h = 5730 years. We want to find t.
So, we have 0.7 = (1/2)^(t/5730).
To solve for t, we'll need to use logarithms. We can take the natural log (ln) of both sides:
ln(0.7) = (t/5730) * ln(1/2).
Now, solve for t:
t = 5730 * ln(0.7) / ln(1/2).
Using a calculator, we find that t ≈ 2556 years.
So, the bone is approximately 2556 years old.
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