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19.2: Radioactive Series

  • Page ID
    49588
  • [ "article:topic", "ChemPrime", "radioactive series", "Decay Chain", "URANIUM-ACTINIUM SERIES", "Thorium Series", "authorname:chemprime", "showtoc:no" ]

    Naturally occurring uranium contains more than 99% 23892U, an isotope which decays to 23490Th by α emission, as shown in Equation (1) from Naturally Occurring Radioactivity. The product of this reaction is also radioactive, however, and undergoes β decay, as already shown in Equation (3) from that section. The 23491Pa produced in this second reaction also emits a β particle:

    \[{}_{\text{91}}^{\text{234}}\text{Pa }\to \text{ }{}_{\text{92}}^{\text{234}}\text{U + }{}_{-\text{1}}^{\text{0}}e\]

    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 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

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

    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 isotope23290Th and involves 10 successive stages, while the other starts with23592U 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 235 92U 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

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

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

    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 number 232 – 24 = 208 and a nuclear charge of 90 – 8 = 82. Since element 82 is Pb, we can write

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

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