If a bone is found to have 20% of its normal amount of carbon 14, how old is the bone?
To determine the age of the bone based on the amount of carbon-14 present, you can use the concept of radioactive decay. Carbon-14 is a radioactive isotope that decays over time, and its half-life is approximately 5730 years. This means that after 5730 years, half of the carbon-14 originally present in a sample will have decayed.
In this case, if the bone is found to contain 20% of its normal amount of carbon-14, we need to calculate how many half-lives have occurred to reach that amount.
First, let's assume the "normal" amount of carbon-14 in the bone is 100. If it now contains 20% of that amount, the current level of carbon-14 is 20.
Since each half-life reduces the amount by half, we can set up the equation:
100 * (1/2) ^ n = 20
Where 'n' is the number of half-lives that have occurred.
To solve for 'n', we can take the logarithm of both sides of the equation:
log(100 * (1/2) ^ n) = log(20)
Simplifying further using logarithm properties:
log(100) + log((1/2) ^ n) = log(20)
2 + n * log(1/2) = log(20)
Next, we'll isolate 'n':
n * log(1/2) = log(20) - 2
Now, we can calculate 'n' by dividing both sides of the equation:
n = (log(20) - 2) / log(1/2)
Evaluating this equation provides us with a value of approximately 2.73.
Since each half-life is 5730 years, multiplying 'n' by the half-life will give us the age of the bone:
Age = n * half-life
Age = 2.73 * 5730
Age ≈ 15,633 years
Therefore, the bone is estimated to be around 15,633 years old based on the 20% remaining carbon-14.