An archaeologist finds the Carbon-14 in a sample of 3.10 g of material to be decaying at 107 counts per second. A modem 1.00 g sample of the same...

Friday, May 15, 2009

An archaeologist finds the Carbon-14 in a sample of 3.10 g of material to be decaying at 107 counts per second. A modem 1.00 g sample of the same...

Hello!

Denote the mass of the older sample as $\displaystyle{m}_{{1}}$ and of the modern as $\displaystyle{m}_{{2}}$ . The older sample contains $\displaystyle{x}_{{1}}$ grams of Carbon-14 and the modern $\displaystyle{x}_{{2}}$ . Also denote the half-life as T and the age of the older sample as E years. And denote the number of counts per second as $\displaystyle{n}_{{1}}$ and $\displaystyle{n}_{{2}}.$

After the time T half of Carbon-14 atoms remains, after 2T 1/4 remains and after time t $\displaystyle\frac{{1}}{{{2}}^{{\frac{{t}}{{T}}}}}$ remains (this is true for t less than T also).

Therefore E years ago there was $\displaystyle{x}_{{1}}\cdot{{2}}^{{\frac{{E}}{{T}}}}$ atoms in older sample, and because materials are the same,

$\displaystyle\frac{{{x}_{{1}}\cdot{{2}}^{{\frac{{E}}{{T}}}}}}{{x}_{{2}}}=\frac{{m}_{{1}}}{{m}_{{2}}}.$

Now use the number of counts per second. Denote one second in years as s. Then

$\displaystyle{x}_{{1}}{\left({1}-{{2}}^{{-\frac{{s}}{{T}}}}\right)}={n}_{{1}},{x}_{{2}}{\left({1}-{{2}}^{{-\frac{{s}}{{T}}}}\right)}={n}_{{2}}.$

Thus $\displaystyle\frac{{x}_{{1}}}{{x}_{{2}}}=\frac{{n}_{{1}}}{{n}_{{2}}}$ and from the first equation

$\displaystyle{{2}}^{{\frac{{E}}{{T}}}}=\frac{{m}_{{1}}}{{m}_{{2}}}\cdot\frac{{n}_{{2}}}{{n}_{{1}}}$ , or

$\displaystyle{E}={T}\cdot{\log}_{{2}}{\left(\frac{{m}_{{1}}}{{m}_{{2}}}\cdot\frac{{n}_{{2}}}{{n}_{{1}}}\right)}.$

In numbers it is approximately 12200 years.

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Friday, May 15, 2009