22. A normal adult jawbone contains 200 mg of Carbon-14 in a living person. If scientists found a jawbone that only had 50mg of Carbon-14, how old is the bone? (The half-life of C-14 is 5730 years).
To find the age of the bone, we can use the formula for exponential decay:
Amount remaining = Initial amount * (1/2)^(time elapsed / half-life)
In this case, the initial amount is 200 mg, the amount remaining is 50 mg, and the half-life is 5730 years. We can plug these values into the formula and solve for time elapsed:
50 = 200 * (1/2)^(time elapsed / 5730)
(1/4) = (1/2)^(time elapsed / 5730)
Taking the natural logarithm of both sides to solve for time elapsed:
ln(1/4) = (time elapsed / 5730) * ln(1/2)
-1.3863 = -0.6931 * (time elapsed / 5730)
2 = time elapsed / 5730
time elapsed = 2 * 5730 = 11460 years
Therefore, the bone is approximately 11460 years old.