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12.1 Introduction to Nuclear Chemistry

  • Page ID
    218440
  • Nuclear chemistry is the study of reactions that involve changes in nuclear structure. The sections on atoms, molecules, and ions introduced the basic idea of nuclear structure; that the nucleus of an atom is composed of protons and neutrons (with the exception of \(\ce{^1_1H}\)). Recall that the number of protons in the nucleus is called the atomic number (\(Z\)) of the element, and the sum of the number of protons and the number of neutrons is the mass number (\(A\)). Atoms with the same atomic number but different mass numbers are isotopes of the same element. When referring to a single type of nucleus, we often use the term nuclide and identify it by the notation:

    \[ \large \ce{^{A}_{Z}X} \label{Eq1a}\]

    where

    • \(X\) is the symbol for the element,
    • \(A\) is the mass number, and
    • \(Z\) is the atomic number.

    Often a nuclide is referenced by the name of the element followed by a hyphen and the mass number. For example, \(\ce{^{14}_6C}\) is called “carbon-14.”

    Protons and neutrons, collectively called nucleons, are packed together tightly in a nucleus. With a radius of about 10−15 meters, a nucleus is quite small compared to the radius of the entire atom, which is about 10−10 meters. Nuclei are extremely dense compared to bulk matter, averaging \(1.8 \times 10^{14}\) grams per cubic centimeter. For example, water has a density of 1 gram per cubic centimeter, and iridium, one of the densest elements known, has a density of 22.6 g/cm3. If the earth’s density were equal to the average nuclear density, the earth’s radius would be only about 200 meters (earth’s actual radius is approximately \(6.4 \times 10^6\) meters, 30,000 times larger).

    Changes of nuclei that result in changes in their atomic numbers, mass numbers, or energy states are nuclear reactions. To describe a nuclear reaction, we use an equation that identifies the nuclides involved in the reaction, their mass numbers and atomic numbers, and the other particles involved in the reaction.

    Nuclear Equations

    A balanced chemical reaction equation reflects the fact that during a chemical reaction, bonds break and form, and atoms are rearranged, but the total numbers of atoms of each element are conserved and do not change. A balanced nuclear reaction equation indicates that there is a rearrangement during a nuclear reaction, but subatomic particles are rearranged rather than atoms. Nuclear reactions also follow conservation laws, and they are balanced in two ways:

    1. The sum of the mass numbers of the reactants equals the sum of the mass numbers of the products.
    2. The sum of the charges of the reactants equals the sum of the charges of the products.

    If the atomic number and the mass number of all but one of the particles in a nuclear reaction are known, we can identify the particle by balancing the reaction. For instance, we could determine that \(\ce{^{17}_8O}\) is a product of the nuclear reaction of \(\ce{^{14}_7N}\) and \(\ce{^4_2He}\) if we knew that a proton, \(\ce{^1_1H}\), was one of the two products. Example \(\PageIndex{1}\) shows how we can identify a nuclide by balancing the nuclear reaction.

    Example \(\PageIndex{1}\): Balancing Equations for Nuclear Reactions

    The reaction of an α particle with magnesium-25 \( (\ce{^{25}_{12}Mg})\) produces a proton and a nuclide of another element. Identify the new nuclide produced.

    Solution

    The nuclear reaction can be written as:

    \[\ce{^{25}_{12}Mg + ^4_2He \rightarrow ^1_1H + ^{A}_{Z}X}\]

    where

    • \(\ce A\) is the mass number and
    • \(\ce Z\) is the atomic number of the new nuclide, \(\ce X\).

    Because the sum of the mass numbers of the reactants must equal the sum of the mass numbers of the products:

    \[\mathrm{25+4=A+1}\]

    so

    \[ \mathrm{A=28}\]

    Similarly, the charges must balance, so:

    \[\mathrm{12+2=Z+1}\]

    so

    \[\mathrm{Z=13}\]

    Check the periodic table: The element with nuclear charge = +13 is aluminum. Thus, the product is \(\ce{^{28}_{13}Al}\).

    Exercise \(\PageIndex{1}\)

    The nuclide \(\ce{^{125}_{53}I}\) combines with an electron and produces a new nucleus and no other massive particles. What is the equation for this reaction?

    Answer

    \[\ce{^{125}_{53}I + ^0_{−1}e \rightarrow ^{125}_{52}Te} \nonumber\]

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