Question #2033

The [sup]14[/sup]C/[sup]12[/sup]C ratio in fossil is only 1/16th part that in a living plant. How old is the fossil? (t[sub]1/2[/sub] of [sup]14[/sup]C = 5720 years)

Expert's answer

The law of half-life decay is

<img src="/cgi-bin/mimetex.cgi?N%20=%20N_0%20%5Ccdot%202%5E%7B-%5Cfrac%7Bt%7D%7B%5Ctau%7D%7D" title="N = N_0 \cdot 2^{-\frac{t}{\tau}}">

As N/N0 = 1/16,

<img src="http://latex.codecogs.com/gif.latex?2%5E%7B-%5Cfrac%7Bt%7D%7B%5Ctau%7D%7D%20=%202%5E%7B-4%7D%5C%5C%20t%20=%204%20%5Ctau%20=%204%5Ccdot%205720%20=%2022,880%20%5C%20years" title="2^{-\frac{t}{\tau}} = 2^{-4}\\ t = 4 \tau = 4\cdot 5720 = 22,880 \ years">

<img src="/cgi-bin/mimetex.cgi?N%20=%20N_0%20%5Ccdot%202%5E%7B-%5Cfrac%7Bt%7D%7B%5Ctau%7D%7D" title="N = N_0 \cdot 2^{-\frac{t}{\tau}}">

As N/N0 = 1/16,

<img src="http://latex.codecogs.com/gif.latex?2%5E%7B-%5Cfrac%7Bt%7D%7B%5Ctau%7D%7D%20=%202%5E%7B-4%7D%5C%5C%20t%20=%204%20%5Ctau%20=%204%5Ccdot%205720%20=%2022,880%20%5C%20years" title="2^{-\frac{t}{\tau}} = 2^{-4}\\ t = 4 \tau = 4\cdot 5720 = 22,880 \ years">

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