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

  1. Mar 31, 2010 #1
    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:

    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!!
  2. jcsd
  3. Apr 1, 2010 #2
    I'm not sure how much math you have, but here's the explanation.

    The general equation for exponential decay is N = N0 * exp(-k*t), where N is the amount remaining, N0 is the original amount, exp(-k) is shorthand for e-k, where e is the Euler constant, k is the decay rate constant (units of 1/time), and t is the elapsed time. If you apply this formula,

    N = 5 * exp(-0.012*2) = 4.881429 ng, which is the answer you got.

    This formula assumes that the decay occurs continuously during the elapsed time. However, the formula in your book assumes that the decay occurs once per century. The two formulas actually give different answers, if you go to the third decimal place:

    N = 5 * (1-0.012)2 = 4.88072 ng

    Now, how come both formulas give you similar answers? The reason is that the formula in your book is a simplification. It stems from the fact that exp(-k) can be expressed as a Taylor series as:

    exp(-k) = 1 - k/1! + k2/2! - k3/3! + ...

    If you drop the higher order terms (exponents larger than 1), exp(-x) approximates 1 - k/1! = 1 - k.

    Plugging this back into the original equation, you get

    N = N0 * exp(-kt) = N0 * (1 - k) * t

    which is the formula in the book.

    This problem is similar to the compound interest problems, but in the opposite direction. Instead of decay, in compound interest you see growth. Compound interest can be computed in 2 different ways: if the interest is compounded continuously, the equation is

    P = C * exp(rt)

    where P is the future value, C is the initial deposit, r is the interest rate, and t is the time.

    However, if the interest is compounded once per year, the equation simplifies to

    P = C * (1 + r)t

    I hope this helps.
  4. Apr 1, 2010 #3


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    I was just going to say that....:redface:
  5. Apr 1, 2010 #4
    Thank you! I didn't understand all of it, but it's starting to make sense. I'll show this to my teacher and see what she can make of it. Thanks again!
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