So today my Math teacher and I got into a discussion about Carbon Dating and exponential decay. The book gave one equation and I (thinking back to chemistry) got another. They both had the same answer (which was great, my teacher was proud) but we could not figure out why. Here it is:(adsbygoogle = window.adsbygoogle || []).push({});

4. [Suppose that drilling into what was once a lake bottom produces a piece of wood which, according to its mass, would have contained 5 nanograms of carbon-14 when the wood was alive. Use the fact that this radioactive carbon decays continuously at a rate of about 1.2% per century to analyze the sample.]

a. How much of the carbon would be expected to remain 2 centuries later?

Now, I took this and got the following equation: 5 * 10^-9 / 1.012^2

My logic was that 5*10^-9 was the 5 nanograms, and 1.012^2 was the two centuries with the exponential decay. The answer that I got was 4.88 nanograms, which was the same as the book. However, the books equation was:

5(.988)^2 Which is the equation for exponential decay that we learned ( a(b)^2 )

Now for the part we argued a bit on. Why would dividing work in the equation that I got, when you multiply in the book equation? I tried to reason through it and only came up with this:

Because I used 1.012 instead of 0.012, you would have to divide because 1 is the 5 nanograms and 0.012 is the decay. So drawn out I got:

5*10^-9 / (5*10^-9 + 0.012%)^2

Sorry if this is confusing, it is my first shot at trying to work my way through a problem that even my teacher has no answer for. Thanks for any help though!!

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Carbon Dating and a problem that came up in Math class - NOT HOMEWORK HELP

**Physics Forums | Science Articles, Homework Help, Discussion**