I've spent over an hour researching Carbon-14 decay for a Calculus problem, but I have one main problem when solving them: how do you solve for the k value (decay constant)?
Here is the problem
The method of carbon dating makes use of the fact that all living
organisms contain two isotopes of carbon, carbon-12, denoted 12C (a
stable isotope), and carbon-14, denoted 14C (a radioactive isotope).
The ratio of the amount of 14C to the amount of 12C is essentially
constant (approximately 1/10,000). When an organism dies, the amount
of 12C present remains unchanged, but the 14C decays at a rate
proportional to the amount present with a half-life of approximately
5700 years. This change in the amount of 14C relative to the amount of
12C makes it possible to estimate the time at which the organism
lived.A fossil found in an archaeological dig was found to contain 20% of the original amount
of 14C. What is the approximate age of the fossil?
So, I'm not completely lost. I'm aware that the equation I need is:
$$\frac{[\ln\frac{N}{No}]}{k} * t_{1/2}$$
And I find many websites that insert -.693 for k when referencing Carbon-14 problems, but I have no idea why they use that value. I assume that the "approximately 1/10,000" part of the problem is significant, but I don't understand why.
Can someone please help me with understanding how to calculate this k value that some places have as -.693 and some sites have as .0001..., both referencing Carbon-14 problems?
Thanks!
Best Answer
I do not get the $-0.693$ value, but perhaps my answer will help anyway.
If we assume Carbon-14 decays continuously, then $$ C(t) = C_0e^{-kt}, $$ where $C_0$ is the initial size of the sample. We don't know this value, but we don't need it. Since it takes 5,700 years for a sample to decay to half its size, we know $$ \frac{1}{2} C_0 = C_0e^{-5700k}, $$ which means $$ \frac{1}{2} = e^{-5700k}, $$ so the value of $C_0$ is irrelevant.
Now, take the logarithm of both sides to get $$ -0.693 = -5700k, $$ from which we can derive $$ k \approx 1.22 \cdot 10^{-4}. $$