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Scientists found a fossilized bone from an organism at an excavation site in North Dakota. When they took the bone back to the lab, they realized that the bone had only 12.5% of the total carbon-14 left. Based on the amount of carbon-14 left in the bone how old is the bone if the half-life of carbon-14 is 5730 years?

Respuesta :

desirayjt
t = 3⋅5,730 = 17,190 years

Answer: 17328.7 years

Explanation:

Half-life of sample of carbon-14 = 5730 years

[tex]\lambda =\frac{0.693}{t_{\frac{1}{2}}}=\frac{0.693}{5730}=1.2\times 10^{-4} years^{-1}[/tex]

[tex]N=N_o\times e^{-\lambda t}[/tex]

N = amount left = 12.5 g

[tex]N_0[/tex]= initial amount = 100 g

[tex]\lambda[/tex]= disintegration constant= [tex]1.2\times 10^{-4} years^{-1}[/tex]

t= ?

[tex]12.5=100\times e^{-1.2\times 10^{-4}\times t}[/tex]

[tex]t=17328.7years[/tex]

Thus the bone is 17328.7 years old.

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