Learning Objectives

• To be able to track sucessive radioactive steps involves in the evolution of an unstable radioisotope into a stable isotope

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 (Figure $$\PageIndex{1}$$). 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 naturally occurring radioactive isotopes of the heaviest elements fall into chains of successive disintegrations, or decays, and all the species in one chain constitute a radioactive family, or radioactive decay series. Three of these series include most of the naturally radioactive elements of the periodic table. They are the uranium series, the actinide series, and the thorium series. The neptunium series is a fourth series, which is no longer significant on the earth because of the short half-lives of the species involved. Each series is characterized by a parent (first member) that has a long half-life and a series of daughter nuclides that ultimately lead to a stable end-product—that is, a nuclide on the band of stability. In all three series, the end-product is a stable isotope of lead. The neptunium series, previously thought to terminate with bismuth-209, terminates with thallium-205.

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

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

Exercise $$\PageIndex{1}$$: Thorium Series

In the thorium series, $$\ce{_{90}^{232}Th}$$ loses aotal of six α particles and four β particles in a 10-stage process. What isotope is finally produced in this series?

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