19.3: Radioactive Series

Naturally occurring uranium contains more than 99% \(\ce{_{92}^{238}U}\) that decays to \(\ce{_{90}^{234}Th}\) by \(α\) emission:

\[\ce{_{92}^{238}U -> _{90}^{234}Th + _{2}^{4}He} \nonumber \]

The product of this reaction is also radioactive, however, and undergoes \(β\) decay:

\[\ce{ _{90}^{234}Th -> _{91}^{234}Pa + _{-1}^{0}e} \nonumber \]

The \(\ce{_{91}^{234}Pa}\) produced in this second reaction also emits a \(β\) particle:

\[\ce{ _{91}^{234}Pa -> _{92}^{234}U + _{-1}^{0}e} \nonumber \]

These three reactions are only the first of 14 steps. After emission of eight \(α\) particles and six \(β\) particles, the isotope \(\ce{ _{82}^{206}Pb}\) is produced. It has a stable nucleus 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

\[{}_{\text{92}}^{\text{238}}\text{U }\to \text{ }{}_{\text{82}}^{\text{206}}\text{Pb + 8}{}_{\text{2}}^{\text{4}}\text{He + 6}{}_{-\text{1}}^{\text{0}}e \nonumber \]

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 \(\ce{ _{90}^{232}Th}\) and involves 10 successive stages, while the other starts with \(\ce{_{92}^{235}U}\) and involves 11 stages. Each of the three series produces a different stable isotope of lead.

Example \(\PageIndex{1}\): Uranium-Actinium Series

The first four stages in the uranium-actinium series involve the emission of an α particle from a \(\ce{_{92}^{235}U}\) 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.

Solution

The emission of an a particle lowers the atomic number by 2 (from 92 to 90). Since element 90 is thorium, we have 

\[\ce{ _{92}^{235}U -> _{90}^{231}Th + _{2}^{4}He} \nonumber \]

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

\[\ce{_{90}^{231}Th -> _{91}^{231}Pa + _{-1}^{0}e} \nonumber \]

The next two stages follow similarly:

\[\ce{_{91}^{231}Pa -> _{89}^{227}Ac + _{2}^{4}He} \nonumber \]

and

\[\ce{_{89}^{227}Ac -> _{90}^{227}Th + _{-1}^{0}e} \nonumber \]

Example \(\PageIndex{2}\): Thorium Series

In the thorium series, \(\ce{_{90}^{232}Th}\) 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:

\[\ce{6 _{2}^{4}He + 4 _{-1}^{0}e} \nonumber \]

involves the total loss of 24 nucleons and 6 × 2 – 4 = 8 positive charges from the \(\ce{_{90}^{232}Th}\) nucleus. The eventual result will be an isotope of mass number 232 – 24 = 208 and a nuclear charge of 90 – 8 = 82. Since element 82 is \(\ce{Pb}\), we can write

\[\ce{ _{90}^{232}Th -> _{82}^{208}Pb + 6 _{2}^{4}He + 4 _{-1}^{0}e} \nonumber \]