The Rate of Radioactive Decay

Submitted by ChemPRIME Staff on Thu, 12/16/2010 - 16:00



We have labeled all isotopes which exhibit radioactivity as unstable, but radioactive isotopes vary considerably in their degree of instability. Some decay so quickly that it is difficult to detect that they are there at all before they have changed into something else. Others have hardly decayed at all since the earth was formed.

The process of radioactive decay is governed by the uncertainty principle, so that we can never say exactly when a particular nucleusThe collection of protons and neutrons at the center of an atom that contains nearly all of the atoms's mass. is going to disintegrate and emit a particle. We can, however, give the probability that a nucleus will disintegrate in a given time interval. For a large number of nuclei we can predict what fraction will disintegrate during that interval. This fraction will be independent of the amount of isotope but will vary from isotope to isotope depending on its stability. We can also look at the matterAnything that occupies space and has mass; contrasted with energy. from the opposite point of view, i.e., in terms of how long it will take a given fraction of isotope to dissociate. Conventionally the tendency for the nuclei of an isotope to decay is measured by its half-lifeIn chemical kinetics, the time it takes for one half of the limiting reactant to be consumed. In nuclear chemistry, the time for half of a sample to undergo radioactive decay., symbol t1/2.

This is the time required for exactly half the nuclei to disintegrate. This quantity, too, varies from isotope to isotope but is independent of the amount of isotope.

Figure 1 shows how a 1-amol (attomole) sample of 12853I, which has a half-life of 25.0 min, decays with time. In the first 25 min, half the nuclei disintegrate, leaving behind 0.5 amol. In the second 25 min, the remainder is reduced by one-half again, i.e., to 0.25 amol. After a third 25-min periodThose elements from a single row of the periodic table., the remainder is (½)3 amol, after a fourth it is (½)4 amol, and so on. If x intervals of 25.0 min are allowed to pass, the remaining amount of 12853I will be (½)x amol.

This example enables us to see what will happen in the general case. Suppose the initial amount of an isotope of half-life t1/2 is n0 and the isotope decays to an amount n in time t, we can measure the time in terms of the number of t1/2 intervals which have elapsed by defining a variable x such that


x=\frac{t}{t_{\text{1/2}}}      (1)


Figure 1 Radioactive decay of  12853I. In the course of each 25-min period, the amount of the isotope decreases by one-half.


Thus after time t our sample will have been reduced to a fraction (½)x of the original amount. In other words

\frac{n}{n_{\text{0}}}=\left( \frac{\text{1}}{\text{2}} \right)^{x}      (2)


Taking natural logarithm of each side, we then have


ln\frac{n}{n_{\text{0}}} = ln \left( \frac{\text{1}}{\text{2}} \right)^{x} =  x ln \frac{\text{1}}{\text{2}} = {-{0.693}x}\,


Substituting from Eq. (1) we thus obtain


ln \frac{n}{n_{\text{0}}} = \frac{-\text{0.693}t}{t_{\text{1/2}}}      (3)



EXAMPLE 1 What amount of 12853I will be left when 3.65 amol of this isotope is allowed to decay for 15.0 min. The half-life of 12853I is 25.0 min.


SolutionA mixture of one or more substances dissolved in a solvent to give a homogeneous mixture. Substituting in Eq. (3) we have


ln \frac{n}{n_{\text{0}}} = \frac{-\text{0}\text{.693}}{t_{\text{1/2}}}t = \frac{-\text{0}\text{.693 }\times \text{ 15}\text{.0 min}}{\text{25}\text{.0 min}} = – 0.4158

Thus


\frac{n}{n_{\text{0}}} = e– 0.4158 = 0.6598

or             n = 0.6598n0 = 0.6598 × 3.65 amol = 2.41 amol



Equation (2) describes how the amount of a radioactive isotope decreases with time, but similar formulas can also be written for the massA measure of the force required to impart unit acceleration to an object; mass is proportional to chemical amount, which represents the quantity of matter in an object. m and also for the rate of disintegrationThe event of a nuclear transformation caused by radioactive decay or nuclear bombardment. r. This is because both the mass and the rate are proportional to the amount of isotope. Thus the rate at which an isotope decays is given by


ln\frac{r}{r_{\text{0}}} = \frac{-\text{0}\text{.693}}{t_{\text{1/2}}}t      (4)


where r0 is the initial rateThe rate of a reaction measured close enough to the beginning of a reaction that the reactant concentations have not yet changed appreciably. at time zero.

The following example uses equation 4 to determine the half life of the disintegration of 9038Sr using decay rates taken at two time points.


EXAMPLE 2 A sample of 9038Sr has a decay rate of 1.682×106 disintegrations per minute. A year later, the rate of decay has decreased to 1.641×106 disintigrations per minute. What is the half life of 9038Sr?


Solution We have r0 = 1.682×106 min–1 g–1, while r = 1.641×106 min–1 g–1, and t = 1yr. Substituting into Eq. (4), we then have


ln\frac{r}{r_{\text{0}}} = \frac{-\text{0}\text{.693}}{t_{\text{1/2}}}t


ln\frac{1.641 \times  10^6}{1.682 \times  10^6} = \frac{-\text{0}\text{.693}}{t_{\text{1/2}}}1 yr


– 0.0247 = \frac{-\text{0}\text{.693}}{t_{\text{1/2}}}1 yr


t1/2 = \frac{-\text{0}\text{.693}}{-0.0247}1 yr=28.1 years


While rates of decay may be used as in example 2, if masses of either the radioactive substanceA material that is either an element or that has a fixed ratio of elements in its chemical formula. or the productA substance produced by a chemical reaction. formed from decay are given, they must first be converted into moles. Then equation 3 may be used to find any remaining unknown quantities. Example three asks for the age of the sample given masses of the radioactive isotope, product, and half life:


EXAMPLE 3 21084Po decays to 20682Pb by emitting an alpha particleA species consisting of two protons and two neutrons and formed by radioactive decay or nuclear bombardment; identical in composition to the nucleus of a helium-4 atom.. Analysis of a sample originally containing only 21084Po is found to have 0.387 μg of 21084Po and 0.294 μg of 20682Pb. Given a half life of 138 days, how long ago was the original sample produced?


Solution

Amount of 210Po = nPo = \frac{\text{0}\text{.387 }\mu \text{g}}{\text{210 g mol}^{-\text{1}}} = 1.84 × 10–3 μmol
Amount of 206Pb = nPb = \frac{\text{0}\text{.294 }\mu \text{g}}{\text{206 g mol}^{-\text{1}}} = 1.43 × 10–3 μmol


Since each mol of 206Pb was originally a mole of 210Po, the original amount of 210Po, n0, is given by

n0 = (1.84 + 1.43) × 10–3 μmol = 3.27 × 10–3 μmol


Substituting into Eq. (3), we have


ln\frac{n}{n_{\text{0}}} = \frac{-\text{0}\text{.693}}{t_{\text{1/2}}}t
ln\frac{\text{1}\text{.84 }\times \text{ 10}^{-\text{3}}\text{ }\mu \text{mol}}{\text{3}\text{.27 }\times \text{ 10}^{-\text{3}}\text{ }\mu \text{mol}} = \frac{-\text{0}\text{.693}}{138\text{ days}} t


– 0.575 = – 5.02 × 10–3 days–1 t

or                    t = 115 days