Question

The half-life of a radioactive isotope is the amount of time it takes for a quantity of that isotope to decay to one half of its original value. (a) express the half-life of a radioactive isotope in terms of its decay rate. (b) the half-life of radiocarbon or carbon 14 (c-14) is 5230 years. determine its decay rate parameter . (c) carbon dating is a method of determining the age of an object using the properties of radiocarbon. it was pioneered by willard libby and collaborators in 1949 to date archaeological, geological, and other samples. its main idea is that by measuring the amount of radiocarbon still found in the organic mater and comparing it to the amount normally found in living matter, we can approximate the amount of time since death occurred.7 using the decay-rate parameter found in part (b), nd the time since death if 35% of radiocarbon is still in the sample.

Answer 1

Radio active decay reactions follow first order rate kinetics.

a) The half life and decay constant for radio active decay reactions are related by the equation:

$t_{\frac{1}{2}} =$$\frac{ln 2}{k}$

$t_{\frac{1}{2}} = \frac{0.693}{k}$

Where k is the decay constant

b) Finding out the decay constant for the decay of C-14 isotope:

$Decay constant (k) = \frac{0.693}{t_{\frac{1}{2}}}$

$k = \frac{0.693}{5230 years}$

$k = 1.325 * 10^{-4} yr^{-1}$

c) Finding the age of the sample :

35 % of the radiocarbon is present currently.

The first order rate equation is,

$[A] = [A_{0}]e^{-kt}$

$\frac{[A]}{[A_{0}]} = e^{-kt}$

$\frac{35}{100} = e^{-(1.325 *10^{-4})t}$

$ln(0.35) = -(1.325 *10^{-4})(t)$

t = 7923 years

Therefore, age of the sample is 7923 years.