(Radiocarbon dating) Carbon-14 is an isotope of carbon. Living things contain a

Question

(Radiocarbon dating) Carbon-14 is an isotope of carbon. Living things contain a fixed percentage of C-14 because they are in equilibrium with the atmosphere; but once they die, they stop taking in new C-14 and the amount present begins to decay. The rate of decay of C-14 is proportional to the quantity of C-14; that is: C 0 = kC where C is the amount of C-14 present. (a) The half-life of a substance is the amount of time it takes for half of the original quantity to decay. The half-life of C-14 is 5600 years. Use this fact to determine the value of the rate constant k. (b) Suppose the equilibrium level of C-14 is C0. Solve the IVP C 0 = kC, C(0) = C0 (c) Suppose that corn kernels found in an archaeological site contain 20% of the equilibrium level of C-14. Determine how many years ago the corn was harvested

Explanation / Answer

Let y(t) be the mass (in grams) of 14C present at time t. Then, we have

y'(t) = ky(t)

and as we have already seen, y(t) = Ae^kt .

The initial condition is y(0) = 50, so that
y(0) = Ae0 = A
and y(t) = Ae^kt .

To find the decay constant k, we use the half-life:

0.50 = y(5600) = e^5600k .

apply ln both sides

ln0.50=5600k

-0.6931472=5600k

k=-0.0001238

so if it will be 0.20

we will have 0.20 = 1e^(-0.0001238)*t

we will apply ln both sides

ln(0.2)=-0.0001238*t

t=13,000 years

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