Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?


share|cite|improve this question
up vote 0 down vote accepted

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 $$ k \approx 1.22 \cdot 10^{-4}. $$

share|cite|improve this answer
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? – David Dec 3 '12 at 2:38
So either the answer is that ridiculously big number (9.17e7) or 30,476 years, being calculated with the equation I provided and the first equation in your answer, respectively. I'm leaning toward the latter. – David Dec 3 '12 at 2:41
Like you said, the equation I'm using was for a different type of problem. Thanks for your help! You made these kinds of problems incredibly simple! – David Dec 3 '12 at 2:43
please check your Math. It seems you calculated the natural log of .5 to be -.301, which is incorrect. It seems to be -.693 – David Dec 3 '12 at 2:47
@David You are correct. It appears I mistakenly hit the "log base 10" button on my calculator. Now the mystery is solved! – Austin Mohr Dec 3 '12 at 2:58

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.