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a hiker in africa discovers a skull that contains 63% of its original amount of C-14. find the age of the skull to the nearest year. using exponential decay formula???

Guest Jan 8, 2015

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 #3
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According to  a formula that I found online, we have

t = [ ln (N / N0) / (-0.693) ]  * 5730      where t is the age of the fossil, N / N0 is the percent of  C-14 that remains and 5730 is the half-life of C-14....so we have

t = ( ln (.63) / (-0.693)) * 5730 = about 3820 yrs old

 

CPhill  Jan 8, 2015
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 #1
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Pretty simple : $$N_\tau(t)=N_0e^{-\frac{t}{\tau}}$$ where tau is the decaying time. Thus, you know $$N_\tau(t)/N_0 = 63\% \Leftrightarrow ln(0.63) = - \frac{t}{\tau} \Leftrightarrow t = -ln(0.63)\tau$$ the only thing you have to search the internet for is the decay time $$\tau$$

Guest Jan 8, 2015
 #2
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WHAT THE...? PRETTY SIMPLE? i mean yeah... What they said

Guest Jan 8, 2015
 #3
avatar+81077 
+5
Best Answer

According to  a formula that I found online, we have

t = [ ln (N / N0) / (-0.693) ]  * 5730      where t is the age of the fossil, N / N0 is the percent of  C-14 that remains and 5730 is the half-life of C-14....so we have

t = ( ln (.63) / (-0.693)) * 5730 = about 3820 yrs old

 

CPhill  Jan 8, 2015

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