Question 1119970
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*[tex \Large N(t)] is the number of atoms remaining at time *[tex \Large t].  *[tex \Large N_o] is the original number of atoms.  *[tex \Large t_{1/2}] is the half-life.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ N(t)\ =\ N_oe^{\small{\frac{ln(0.5)t}{t_{1/2}}}]


But *[tex \Large \ln(0.5)\ \approx\ 0.693] and the half-life is 5730 years, so


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ N(t)\ =\ N_oe^{\large{\frac{0.693t}{5730}}]


But we are given that 55% of the Carbon-14 remains, so *[tex \Large \frac{N(t)}{N_o}\ =\ 0.55], hence:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ e^{\large{\frac{0.693t}{5730}}}\LARGE\ =\ 0.55]


Solve for *[tex \Large t]
							
								
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
<img src="http://c0rk.blogs.com/gr0undzer0/darwin-fish.jpg">
*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  
								
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