A sample of charcoal is found to have only 1/8th of Carbon-14. Half-life of Carbon-14 is 5730yrs, what is the age of the charcoal?
k = 0.693/t1/2
Substitute and solve for k.
ln(No/N) = kt
Pick any convenient number for No, such as 100.
Then N = 1/8 that or 12.5
k = from above.
Solve for t in years if half life is in years.
716
To find the age of the charcoal, we can use the concept of exponential decay and the half-life of Carbon-14.
The half-life of Carbon-14 is given as 5730 years, which means that after each 5730-year period, the amount of Carbon-14 in a sample will be reduced by half.
In this case, we are told that the sample of charcoal has only 1/8th of the original amount of Carbon-14. Let's assume the original amount is 'A' (which is the total amount of Carbon-14 in the sample when it was alive). Therefore, the current amount of Carbon-14 in the sample is 1/8 * A.
Now, let's start by finding out how many half-lives it took for the Carbon-14 to reduce from the original amount 'A' to 1/8 * A.
Each half-life is 5730 years, so let 'n' be the number of half-lives:
(A / 2^n) = (1/8) * A
Now, let's solve for 'n':
2^n = 8
Take the logarithm (base 2) of both sides to get the value of 'n':
n * log2(2) = log2(8)
Since log2(2) = 1, we have:
n = log2(8)
Using a calculator, we find that log2(8) ≈ 3, as 2^3 = 8.
This means that it took 3 half-lives for the Carbon-14 amount to reduce from 'A' to 1/8 * A.
Since each half-life is 5730 years, the age of the charcoal can be calculated by:
Age = 3 * 5730 years
Therefore, the age of the charcoal is approximately 17190 years.