# 21.3: Nuclear Transmutations

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##### Learning Objectives
• To understand how nuclear transmutation reactions lead to the formation of the elements in stars and how they can be used to synthesize transuranium elements.

The relative abundances of the elements in the known universe vary by more than 12 orders of magnitude. For the most part, these differences in abundance cannot be explained by differences in nuclear stability. Although the 56Fe nucleus is the most stable nucleus known, the most abundant element in the known universe is not iron, but hydrogen (1H), which accounts for about 90% of all atoms. In fact, 1H is the raw material from which all other elements are formed. In this section, we explain why 1H and 2He together account for at least 99% of all the atoms in the known universe. We also describe the nuclear reactions that take place in stars, which transform one nucleus into another and create all the naturally occurring elements.

## Relative Abundances of the Elements on Earth and in the Known Universe

The relative abundances of the elements in the known universe and on Earth relative to silicon are shown in Figure $$\PageIndex{1}$$. The data are estimates based on the characteristic emission spectra of the elements in stars, the absorption spectra of matter in clouds of interstellar dust, and the approximate composition of Earth as measured by geologists. The data in Figure $$\PageIndex{1}$$ illustrate two important points. First, except for hydrogen, the most abundant elements have even atomic numbers. Not only is this consistent with the known trends in nuclear stability, but it also suggests that heavier elements are formed by combining helium nuclei (Z = 2). Second, the relative abundances of the elements in the known universe and on Earth are often very different, as indicated by the data in Table $$\PageIndex{1}$$ for some common elements.

Some of these differences are easily explained. For example, nonmetals such as H, He, C, N, O, Ne, and Kr are much less abundant relative to silicon on Earth than they are in the rest of the universe. These elements are either noble gases (He, Ne, and Kr) or elements that form volatile hydrides, such as NH3, CH4, and H2O. Because Earth’s gravity is not strong enough to hold such light substances in the atmosphere, these elements have been slowly diffusing into outer space ever since our planet was formed. Argon is an exception; it is relatively abundant on Earth compared with the other noble gases because it is continuously produced in rocks by the radioactive decay of isotopes such as 40K. In contrast, many metals, such as Al, Na, Fe, Ca, Mg, K, and Ti, are relatively abundant on Earth because they form nonvolatile compounds, such as oxides, that cannot escape into space. Other metals, however, are much less abundant on Earth than in the universe; some examples are Ru and Ir. This section explains some of the reasons for the great differences in abundances of the metallic elements.

Table $$\PageIndex{1}$$: Relative Abundances of Elements on Earth and in the Known Universe
Terrestrial/Universal Element Abundance Ratio
H 0.0020
He 2.4 × 10−8
C 0.36
N 0.02
O 46
Ne 1.9 × 10−6
Na 1200
Mg 48
Al 1600
Si 390
S 0.84
K 5000
Ca 710
Ti 2200
Fe 57

All the elements originally present on Earth (and on other planets) were synthesized from hydrogen and helium nuclei in the interiors of stars that have long since exploded and disappeared. Six of the most abundant elements in the universe (C, O, Ne, Mg, Si, and Fe) have nuclei that are integral multiples of the helium-4 nucleus, which suggests that helium-4 is the primary building block for heavier nuclei.

## Synthesis of the Elements in Stars

Elements are synthesized in discrete stages during the lifetime of a star, and some steps occur only in the most massive stars known (Figure $$\PageIndex{2}$$). Initially, all stars are formed by the aggregation of interstellar “dust,” which is mostly hydrogen. As the cloud of dust slowly contracts due to gravitational attraction, its density eventually reaches about 100 g/cm3, and the temperature increases to about 1.5 × 107 K, forming a dense plasma of ionized hydrogen nuclei. At this point, self-sustaining nuclear reactions begin, and the star “ignites,” creating a yellow star like our sun.

In the first stage of its life, the star is powered by a series of nuclear fusion reactions that convert hydrogen to helium:

$_1^1\textrm H+\,_1^1\textrm H\rightarrow\,_1^2\textrm H+\,_{+1}^0\beta \\_1^2\textrm H+\,_1^1\textrm H\rightarrow\,_2^3\textrm{He}+\,_{0}^0\gamma \\_2^3\textrm{He}+\,_2^3\textrm{He}\rightarrow\,_2^4\textrm{He}+2_{1}^1\textrm H\label{Eq1}$

The overall reaction is the conversion of four hydrogen nuclei to a helium-4 nucleus, which is accompanied by the release of two positrons, two $$\gamma$$ rays, and a great deal of energy:

$4_1^1\textrm H\rightarrow\,_2^4\textrm{He}+2_{+1}^0\beta+2_0^0\gamma\label{Eq2}$

These reactions are responsible for most of the enormous amount of energy that is released as sunlight and solar heat. It takes several billion years, depending on the size of the star, to convert about 10% of the hydrogen to helium.

Once large amounts of helium-4 have been formed, they become concentrated in the core of the star, which slowly becomes denser and hotter. At a temperature of about 2 × 108 K, the helium-4 nuclei begin to fuse, producing beryllium-8:

$2_2^4\textrm{He}\rightarrow\,_4^8\textrm{Be}\label{Eq3}$

Although beryllium-8 has both an even mass number and an even atomic number, its also has a low neutron-to-proton ratio (and other factors beyond the scope of this text) that makes it unstable; it decomposes in only about 10−16 s. Nonetheless, this is long enough for it to react with a third helium-4 nucleus to form carbon-12, which is very stable. Sequential reactions of carbon-12 with helium-4 produce the elements with even numbers of protons and neutrons up to magnesium-24:

$_4^8\textrm{Be}\xrightarrow{_2^4\textrm{He}}\,_6^{12}\textrm C\xrightarrow{_2^4\textrm{He}}\,_8^{16}\textrm O\xrightarrow{_2^4\textrm{He}}\,_{10}^{20}\textrm{Ne}\xrightarrow{_2^4\textrm{He}}\,_{12}^{24}\textrm{Mg}\label{Eq4}$

So much energy is released by these reactions that it causes the surrounding mass of hydrogen to expand, producing a red giant that is about 100 times larger than the original yellow star.

As the star expands, heavier nuclei accumulate in its core, which contracts further to a density of about 50,000 g/cm3, so the core becomes even hotter. At a temperature of about 7 × 108 K, carbon and oxygen nuclei undergo nuclear fusion reactions to produce sodium and silicon nuclei:

$_6^{12}\textrm C+\,_6^{12}\textrm C\rightarrow \,_{11}^{23}\textrm{Na}+\,_1^1\textrm H\label{Eq5}$

$_6^{12}\textrm C+\,_8^{16}\textrm O\rightarrow \,_{14}^{28}\textrm{Si}+\,_0^0\gamma\label{Eq6}$

At these temperatures, carbon-12 reacts with helium-4 to initiate a series of reactions that produce more oxygen-16, neon-20, magnesium-24, and silicon-28, as well as heavier nuclides such as sulfur-32, argon-36, and calcium-40:

$_6^{12}\textrm C\xrightarrow{_2^4\textrm{He}}\,_8^{16}\textrm O\xrightarrow{_2^4\textrm{He}}\,_{10}^{20}\textrm{Ne}\xrightarrow{_2^4\textrm{He}}\,_{12}^{24}\textrm{Mg}\xrightarrow{_2^4\textrm{He}}\,_{14}^{28}\textrm{Si}\xrightarrow{_2^4\textrm{He}}\,_{16}^{32}\textrm S\xrightarrow{_2^4\textrm{He}}\,_{18}^{36}\textrm{Ar}\xrightarrow{_2^4\textrm{He}}\,_{20}^{40}\textrm{Ca}\label{Eq7}$

The energy released by these reactions causes a further expansion of the star to form a red supergiant, and the core temperature increases steadily. At a temperature of about 3 × 109 K, the nuclei that have been formed exchange protons and neutrons freely. This equilibration process forms heavier elements up to iron-56 and nickel-58, which have the most stable nuclei known.

## The Formation of Heavier Elements in Supernovas

None of the processes described so far produces nuclei with Z > 28. All naturally occurring elements heavier than nickel are formed in the rare but spectacular cataclysmic explosions called supernovas (Figure $$\PageIndex{2}$$). When the fuel in the core of a very massive star has been consumed, its gravity causes it to collapse in about 1 s. As the core is compressed, the iron and nickel nuclei within it disintegrate to protons and neutrons, and many of the protons capture electrons to form neutrons. The resulting neutron star is a dark object that is so dense that atoms no longer exist. Simultaneously, the energy released by the collapse of the core causes the supernova to explode in what is arguably the single most violent event in the universe. The force of the explosion blows most of the star’s matter into space, creating a gigantic and rapidly expanding dust cloud, or nebula (Figure $$\PageIndex{3}$$). During the extraordinarily short duration of this event, the concentration of neutrons is so great that multiple neutron-capture events occur, leading to the production of the heaviest elements and many of the less stable nuclides. Under these conditions, for example, an iron-56 nucleus can absorb as many as 64 neutrons, briefly forming an extraordinarily unstable iron isotope that can then undergo multiple rapid β-decay processes to produce tin-120:

$_{26}^{56}\textrm{Fe}+64_0^1\textrm n\rightarrow \,_{26}^{120}\textrm{Fe}\rightarrow\,_{50}^{120}\textrm{Sn}+24_{-1}^0\beta\label{Eq8}$

Although a supernova occurs only every few hundred years in a galaxy such as the Milky Way, these rare explosions provide the only conditions under which elements heavier than nickel can be formed. The force of the explosions distributes these elements throughout the galaxy surrounding the supernova, and eventually they are captured in the dust that condenses to form new stars. Based on its elemental composition, our sun is thought to be a second- or third-generation star. It contains a considerable amount of cosmic debris from the explosion of supernovas in the remote past.

##### Example $$\PageIndex{1}$$: Carbon Burning Stars

The reaction of two carbon-12 nuclei in a carbon-burning star can produce elements other than sodium. Write a balanced nuclear equation for the formation of

1. magnesium-24.
2. neon-20 from two carbon-12 nuclei.

Given: reactant and product nuclides

Strategy:

Use conservation of mass and charge to determine the type of nuclear reaction that will convert the reactant to the indicated product. Write the balanced nuclear equation for the reaction.

Solution

1. A magnesium-24 nucleus (Z = 12, A = 24) has the same nucleons as two carbon-12 nuclei (Z = 6, A = 12). The reaction is therefore a fusion of two carbon-12 nuclei, and no other particles are produced: $$_{6}^{12}\textrm C+\,_{6}^{12}\textrm C\rightarrow\,_{12}^{24}\textrm{Mg}$$.
2. The neon-20 product has Z = 10 and A = 20. The conservation of mass requires that the other product have A = (2 × 12) − 20 = 4; because of conservation of charge, it must have Z = (2 × 6) − 10 = 2. These are the characteristics of an α particle. The reaction is therefore $$_{6}^{12}\textrm C+\,_{6}^{12}\textrm C\rightarrow\,_{10}^{20}\textrm{Ne}+\,_2^4\alpha$$.
##### Exercise $$\PageIndex{1}$$

How many neutrons must an iron-56 nucleus absorb during a supernova explosion to produce an arsenic-75 nucleus? Write a balanced nuclear equation for the reaction.

19 neutrons; $$_{26}^{56}\textrm{Fe}+19_0^1\textrm n \rightarrow \,_{26}^{75}\textrm{Fe}\rightarrow \,_{33}^{75}\textrm{As}+7_{-1}^0\beta$$