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???
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 $$\tau$$
+128474 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
