The amount of carbon-14 present obeys an exponential decay law.
A(t) = A(0)e-^kt
Get k from the half-life:
0.5 = 1e-^k(5600)
k = ln(0.5) / (-5600 years) = 1.2378 x 10^-4 years^-1
If only 73% of the carbon-14 remains after time T, then
0.73 = 1e^-kT
T = ln(0.73) / (-1.2378 x 10^-4 years^-1) = 2542.5 years
The tree has been dead for about 2542.5 years.
Following are the calculation when the tree dies:
Given:
Carbon-14 half-life = 5600 years
To find:
The tree dies in years=?
Solution:
Using formula:
[tex]p=p_0 e^{-\lambda\ t}\ \ \ \ \ \ \ \ \ \ \ \ {t_{\frac{1}{2}} = half\ life =5600 \ years}[/tex]
[tex]\therefore \\\\\lambda =\frac{In2}{t_{\frac{1}{2}}}\\\\[/tex]
[tex]=\frac{In2}{5600}\\\\ =0.00012378\\\\[/tex]
[tex]\to p=p_0e^{-0.00012378 \ t}\\\\[/tex]
[tex]=62\% \ of\ P_0 =0.62 p_0\\\\[/tex]
[tex]\to 0.62 p_0 =p_0 e^{-0.00012378 \ t}\\\\\to 0.62 = e^{-0.00012378 \ t}\\\\[/tex]
Using natural log:
[tex]\to {-0.00012378 \ t} = \log 0.62 \\\\\to t = \frac{\log 0.62}{-0.00012378} \ \\\\\to t = 3862.095 \approx 3862 \ years\\\\[/tex]
Therefore, the final answer is "3862 years".
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