Radioactivity is the spontaneous disintegration of atoms of certain elements into atoms of other elements with the emission of smaller particles.
Elements, such as uranium, which spontaneously emit energy without the absorption of energy are said to be naturally radioactive.
Imagine having a very small sample of 238U, say 1 mg. This small mass contains many trillions of atoms. Now consider just one of these atoms. When will this particular atom undergo radioactive decay? The answer is that it is impossible to predict when, if ever, an atom will decay. Fortunately, the chemist is rarely concerned with the behaviour of a single particle. 1 mg of 238U contains 2.5 x 1018 atoms. With such large numbers of atoms it is possible to make precise predictions about rates of decay.
Experiments show that radioactive decay rates are directly proportional to the number of atoms present:
rate = k x N
where N = the number of atoms.
Different radioisotopes decay at different rates, so how can we compare their decay rates?
|90Sr undergoes beta decay and has a half-life of 28 years. Given 8 g of this isotope, 4 g would remain after 28 years, 2 g would remain after 56 years, and 1 g after 84 years.|
The table below gives the half-lives of some radioisotopes, all of which undergo beta decay:
|14C||5.73 x 103 years|
It is the time required for one-half of a given quantity of a reactant to react. Half-life is commonly used as a measure of the rate of first-order reactions. It indicates the kinetic stability of a reactant: the longer the half-life, the greater the stability.
The decomposition of dinitrogen pentoxide in tetrachloromethane is a typical first-order reaction:
N2O5(sol) ® 2NO2(sol) + ½O2(g)
Its rate equation is:
rate = k[N2O5(sol)]
where k is a constant called the rate constant.
It can be deduced that:
k = 0.693/t½
The half-life may be evaluated from the rate constant or vice versa. The shorter the half-life, the faster the reaction.
|The half-life for the decomposition of N2O5 in tetrachloromethane at 30 ºC is 2.4 hours. If we start with 10.0 g of N2O5 at t = 0 (q0), then, after a period of 2.4 hours, 5.0 g will remain. After a second period of 2.4 hours (4.8 hours in total), 2.5 g remain. After a total of 7.2 hours, 1.25 g remain, and so on.|
For 3 half-lives, the amount of N2O5 remaining at time t (qt) is:
qt = 10 g x (0.5)3 = 1.25 g
Generalising, we have
qt = q0(0.5)t/t½
in which t/t½ gives the number of half-lives.