Question
The amount of radiocarbon present after t years is given by y=ae^-.0001216t .
find the half live of 14
find the half live of 14
Answers
Steve
ln(1/2) = -0.69314718
So, -.0001216 = 0.00017543ln(1/2) = ln2/5700
e^(-.0001216t)
= e^(ln(1/2)*t/5700)
= (1/2)^(t/5700)
So, the half-life is 5700 years
So, -.0001216 = 0.00017543ln(1/2) = ln2/5700
e^(-.0001216t)
= e^(ln(1/2)*t/5700)
= (1/2)^(t/5700)
So, the half-life is 5700 years