(a) Since we're dealing with half-life, the initial equation is
f(t) = C(1/2)^(t/5730)
Now, 1/2 = e^(ln 1/2) = e^(-ln2)
So,
f(t) = C*(e^-ln2)^(t/5730) = C*e^(-ln2/5730 t) = Ce^(-0.000121t)
(b) 1/5 = e^(-0.000121t)
ln(1/5) = -0.000121t
Now finish it off
May someone help me with my HW problem (b)?
Radiocarbon dating can be used to estimate the age of material, such as
animal bones or plant remains, that comes from a formerly living organism.
When the organism dies, the amount of carbon-14 in its remains decays
exponentially. Suppose that f(t) is the amount of carbon-14 in an organism
at time t (in years), where t = 0 corresponds to the time of the death of the
organism. Assume that the amount of carbon-14 is modelled by the
exponential decay function
f(t) = Ce^kt (t >= 0);
where C is the initial amount of carbon-14 and k is a constant.
The half-life (or halving period) of a radioactive substance like carbon-14 is
the time that it takes for the amount of the substance to decrease to half of
its original level.
(a) Given that the half-life of carbon-14 is 5730 years, show that the value
of the constant k is -0.000121, correct to three significant figures.
(b) Suppose that some ancient animal bones have been found where the
amount of carbon-14 is known to have decreased to 1/5 of the level that
was present immediately after the death of the animal. How much time
has elapsed since the death of this animal? Give your answer to the
nearest whole number of years.
2 answers
Thanks for your help
1 answer