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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)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?

-Should I be doing $0.20=0.5^{t/5700}$ to get a final result of $13235$ years or using the equation $t=[\ln0.20/(-.693)]\cdot5700$?

b) The Dead Sea Scrolls are approximately 2000 years old. What percent of the original 14C remains in them?

-I'm very unsure of which parts of the given question to use for this part & what equation I should use for the calculation in general. Maybe A=Ce^kt? With k=.0001 amd t=2000?

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1 Answer 1

Natural logarithms need to be used in this sort of problem ONLY when you're dealing with instantaneous rates of change. You've got $$ \left(\frac 12\right)^{t/5700} = 0.2 = \frac15. $$ So $$ 2^{t/5700} = 5. $$ Hence $$ \frac{t}{5700} = \log_2 5, $$ and the bottom line is $$ t = 5700\log_2 5. $$

This is the same thing as $$ e^{(-\ln2)t/5700} = \frac15, $$ etc. So you've got $e^{-kt}$, where $k=-(\ln2)/5700$. You need that form if you want to talk about how fast it's changing at a particular instant in time. Here's a commonplace clumsy mistake: Put in some rounded value of $(\ln2)/5700$ and go from there, and end up with an approximate answer, where use of $1/2$ instead of $e^{-\ln2}$ would yield an exact answer. So bringing in $e$ and natural logarithms adds complications. It should be used when needed, but it's not needed to answer questions like the one posed here.

For part (b), you've got $\left(\frac12\right)^{2000/5700}$.

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Thanks! In your opinion would the way I did part A be wrong? I got a very large # ~14,000. Also, for part b that's all I'd have to do? With that calculation I get .784107 remains, that just seems too easy. –  Anna Apr 24 '13 at 16:14
    
What you did is correct if you have $\ln0.5$ where you've got the rounded value $-0.693$. It is usually best not to round more than absolutely necessary until the bottom line. That means find $(\ln0.2)/\ln0.5$ first and then round, rather than rounding first and then dividing. Also, it wouldn't hurt to be aware that $(\ln0.2)/\ln0.5)$ $=(\ln(1/5))/\ln(1/2)$ $=(-\ln 5)/(-\ln2)$ $=(\ln5)/\ln2$ $=\log_2 5$. Using logarithms to bases other than $2$ in this problem is an extra complication. You need them when you think about instantaneous rates of change, but you don't need them here. –  Michael Hardy Apr 24 '13 at 19:41

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