# Exponential Decay Problem

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!

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 $$k \approx 1.22 \cdot 10^{-4}.$$
• Austin, thanks for the answer. Plugging that k value into my equation yields an incredibly large age (-9.17 * 10^7). Am I missing something here? Should I be using the first equation in your answer? Dec 3, 2012 at 2:38
• please check your Math. It seems you calculated the natural log of .5 to be -.301, which is incorrect. It seems to be -.693 Dec 3, 2012 at 2:47