Submitted by ChemPRIME Staff on Thu, 12/16/2010 - 15:56

Naturally occurring uranium contains more than 99% 23892U, an isotopeOne of two or more samples of an element whose atoms differ in the number of neutrons found in the nucleus. which decays to 23490Th by α emission, as shown in Eq. (1) from Naturally Occurring Radioactivity. The productA substance produced by a chemical reaction. of this reaction is also radioactive, however, and undergoes β decay, as already shown in Eq. (3) from that section. The 23491Pa produced in this second reaction also emits a β particle:

${}_{\text{91}}^{\text{234}}\text{Pa }\to \text{ }{}_{\text{92}}^{\text{234}}\text{U + }{}_{-\text{1}}^{\text{0}}e$      (1)

These three reactions are only the first of 14 steps. After emission of eight α particles and six β particles, the isotope 20682Pb is produced. It has a stable nucleusThe collection of protons and neutrons at the center of an atom that contains nearly all of the atoms's mass. which does not disintegrate further. The complete process may be written as follows:

$\text{ }{}_{\text{92}}^{\text{238}}\text{U}\xrightarrow{\alpha }{}_{\text{90}}^{\text{234}}\text{Th}\xrightarrow{\beta }{}_{\text{91}}^{\text{234}}\text{Pa}\xrightarrow{\beta }{}_{\text{92}}^{\text{234}}\text{U}\xrightarrow{\alpha }{}_{\text{90}}^{\text{230}}\text{Th}\xrightarrow{\alpha }{}_{\text{88}}^{\text{226}}\text{Ra}\xrightarrow{\alpha }{}_{\text{88}}^{\text{222}}\text{Rn}$

$\downarrow ^{\alpha }$      (2a)

${}_{\text{82}}^{\text{206}}\text{Pb}\xleftarrow{\alpha }{}_{\text{84}}^{\text{210}}\text{Po}\xleftarrow{\beta }{}_{\text{83}}^{\text{210}}\text{Bi}\xleftarrow{\alpha }{}_{\text{82}}^{\text{210}}\text{Pb}\xleftarrow{\alpha }{}_{\text{84}}^{\text{214}}\text{Po}\xleftarrow{\beta }{}_{\text{83}}^{\text{214}}\text{Bi}\xleftarrow{\beta }{}_{\text{82}}^{\text{214}}\text{Pb}\xleftarrow{\alpha }{}_{\text{84}}^{\text{218}}\text{Po }$

While the net reaction is

${}_{\text{92}}^{\text{238}}\text{U }\to \text{ }{}_{\text{82}}^{\text{206}}\text{Pb + 8}{}_{\text{2}}^{\text{4}}\text{He + 6}{}_{-\text{1}}^{\text{0}}e$      (2b)

Such a series of successive nuclear reactions is called a radioactive series. Two other radioactive series similar to the one just described occur in nature. One of these starts with the isotope

23290Th and involves 10 successive stages, while the other starts with

23592U and involves 11 stages. Each of the three series produces a different stable isotope of lead.

EXAMPLE 1 The first four stages in the uranium-actinium series involve the emission of an α particle from a

23592U nucleus, followed successively by the emission of a β particle, a second α particle, and then a second β particle. Write out equations to describe all four nuclear reactions.

SolutionA mixture of one or more substances dissolved in a solvent to give a homogeneous mixture. The emission of an a particle lowers the atomic numberThe number of protons in the nucleus of an atom; used to define the position of an element in the periodic table; represented by the letter Z. by 2 (from 92 to 90). Since elementA substance containing only one kind of atom and that therefore cannot be broken down into component substances by chemical means. 90 is thorium, we have

${}_{\text{92}}^{\text{235}}\text{U }\to \text{ }{}_{\text{90}}^{\text{231}}\text{Th + }{}_{\text{2}}^{\text{4}}\text{He}$

The emission of a β particle now increases the atomic number by 1 to give an isotope of element 91, protactinium:

${}_{\text{90}}^{\text{231}}\text{Th }\to \text{ }{}_{\text{91}}^{\text{231}}\text{Pa + }{}_{-\text{1}}^{\text{0}}e$

The next two stages follow similarly:

${}_{\text{91}}^{\text{231}}\text{Pa }\to \text{ }{}_{\text{89}}^{\text{227}}\text{Ac + }{}_{\text{2}}^{\text{4}}\text{He}$     and     ${}_{\text{89}}^{\text{227}}\text{Ac }\to \text{ }{}_{\text{90}}^{\text{227}}\text{Th + }{}_{-\text{1}}^{\text{0}}e$

EXAMPLE 2 In the thorium series,

23290Th loses a total of six α particles and four β particles in a 10-stage process. What isotope is finally produced in this series?

Solution The loss of six α particles and four β particles:

$\text{6}{}_{\text{2}}^{\text{4}}\text{He + 4}{}_{-\text{1}}^{\text{0}}e$

involves the total loss of 24 nucleons and 6 × 2 – 4 = 8 positive charges from the 23290Th nucleus. The eventual result will be an isotope of mass numberThe sum of the numbers of protons and neutrons in an atom; these two kinds of particles contain almost all of the mass of an atom. 232 – 24 = 208 and a nuclear charge of 90 – 8 = 82. Since element 82 is Pb, we can write

${}_{\text{90}}^{\text{232}}\text{Th }\to \text{ }{}_{\text{82}}^{\text{208}}\text{Pb + 6}{}_{\text{2}}^{\text{4}}\text{He + 4}{}_{-\text{1}}^{\text{0}}e$