Radioactivity, Nuclear Physics and its Medical Applications

118 Nuclear Decay and Conservation Laws

Learning Objectives

  • Describe the process of nuclear decay and explain why some nuclei are unstable.
  • Apply the conservation laws that govern nuclear decay.
  • Distinguish between parent nuclei and daughter nuclei.
  • Calculate the energy released during radioactive decay using the mass–energy relationship.

Radioactive decay has provided an extraordinary window into the structure of matter at the smallest scales. The study of radioactive materials gave scientists the first experimental evidence that mass and energy are fundamentally related, and it led to the discovery of two of nature's four fundamental forces: the strong nuclear force and the weak nuclear force. As physicists investigated radioactive decay, they also uncovered entirely new particles and established conservation laws that remain central to modern nuclear and particle physics.

Some atomic nuclei are stable and can exist indefinitely. Others are radioactive, meaning they spontaneously transform into more stable nuclei while emitting particles or electromagnetic radiation. The original unstable nucleus is called the parent nucleus, and the nucleus produced after the decay is called the daughter nucleus.

Some radioactive nuclei reach stability in a single decay. For example, [latex]{}^{60}\mathrm{Co}[/latex] decays directly into stable [latex]{}^{60}\mathrm{Ni}[/latex]. Many radionuclides, however, decay through a sequence of intermediate radioactive nuclei known as a decay series. In these chains, each daughter nucleus is itself unstable and continues to decay until a stable isotope is finally produced.

The decay series that begins with [latex]{}^{238}\mathrm{U}[/latex] is especially important because it produces several historically and medically significant radioactive isotopes, including [latex]{}^{226}\mathrm{Ra}[/latex] and [latex]{}^{210}\mathrm{Po}[/latex], both first isolated by Marie and Pierre Curie. The series also produces [latex]{}^{222}\mathrm{Rn}[/latex], a naturally occurring radioactive gas. Because radon is a noble gas, it can escape from rocks and soil that contain even small amounts of uranium and accumulate inside buildings. When inhaled, radon and its radioactive decay products expose lung tissue to ionizing radiation, making radon a significant environmental health concern. The uranium-238 decay series ultimately ends with the stable isotope [latex]{}^{206}\mathrm{Pb}[/latex].

Healthcare Connection

Radon is the second leading cause of lung cancer after cigarette smoking and the leading cause among people who have never smoked. Because it is colorless, odorless, and chemically inert, elevated indoor radon concentrations cannot be detected without specialized testing. Many public health agencies recommend routine radon testing in homes, schools, and healthcare facilities located in regions with uranium-rich bedrock.

Decay series of uranium-238 showing successive alpha and beta decays, the half-life of each radionuclide, and the final stable lead-206 nucleus.
Figure 118.1. The decay series of [latex]{}^{238}\mathrm{U}[/latex], the most abundant naturally occurring uranium isotope. Each radionuclide undergoes either alpha or beta decay, and the half-life of every step is indicated. Several members of the series, including radium, radon, and polonium, occur naturally in uranium-containing rocks and soils. The series terminates with stable [latex]{}^{206}\mathrm{Pb}[/latex].

Notice that every daughter nucleus produced by [latex]\alpha[/latex] decay has two fewer protons and two fewer neutrons than its parent. This is expected because an alpha particle is simply the nucleus of a helium-4 atom, containing two protons and two neutrons. In contrast, daughter nuclei produced by [latex]\beta[/latex] decay have one fewer neutron and one additional proton than the parent nucleus. As we will see later in this section, beta decay involves a different physical process than alpha decay. Gamma decay is not represented in Figure 118.1 because it does not change one nuclide into another; instead, it allows an excited nucleus to release excess energy while remaining the same element and isotope.

Alpha Decay

In alpha decay, an unstable nucleus emits an alpha particle, which is the nucleus of a helium-4 atom, consisting of two protons and two neutrons. As a result, the daughter nucleus contains two fewer protons and two fewer neutrons than the parent nucleus (Figure 118.2).

One example of alpha decay is the decay of [latex]{}^{238}\mathrm{U}[/latex], shown previously in Figure 118.1. Another common example is the alpha decay of plutonium-239:

[latex]{}^{238}\mathrm{U} \rightarrow {}^{234}\mathrm{Th} + {}^{4}\mathrm{He}[/latex]

and

[latex]{}^{239}\mathrm{Pu} \rightarrow {}^{235}\mathrm{U} + {}^{4}\mathrm{He}.[/latex]
Illustration of alpha decay showing a parent nucleus emitting a helium-4 nucleus to form a daughter nucleus with two fewer protons and two fewer neutrons.
Figure 118.2. During alpha decay, the parent nucleus emits an alpha particle (a helium-4 nucleus). The daughter nucleus therefore has two fewer protons and two fewer neutrons than the parent. Alpha decay occurs spontaneously only when the combined mass of the daughter nucleus and alpha particle is less than that of the original nucleus.

These examples illustrate an important feature of alpha decay. Because the emitted alpha particle contains two protons, the daughter element always has an atomic number that is two less than that of the parent. For example, uranium has atomic number [latex]Z=92[/latex], while thorium has [latex]Z=90[/latex]. Likewise, plutonium has [latex]Z=94[/latex], and its daughter, uranium, has [latex]Z=92[/latex].

In general, the alpha-decay equation is written as

[latex]{}_{Z}^{A}\mathrm{X} \rightarrow {}_{Z-2}^{A-4}\mathrm{Y} + {}_{2}^{4}\mathrm{He} \qquad (\alpha\ \text{decay}).[/latex]

Here, [latex]A[/latex] is the mass number and [latex]Z[/latex] is the atomic number of the parent nucleus. The daughter nucleus has a mass number four units smaller and an atomic number two units smaller because the emitted alpha particle removes four nucleons—two protons and two neutrons—from the nucleus.

For example, suppose you are told that [latex]{}^{239}\mathrm{Pu}[/latex] undergoes alpha decay. The daughter nucleus must have an atomic number two less than plutonium, making it uranium. Since four nucleons are emitted, the mass number decreases from 239 to 235. The complete decay equation is therefore

[latex]{}^{239}\mathrm{Pu} \rightarrow {}^{235}\mathrm{U} + {}^{4}\mathrm{He}.[/latex]

Conservation Laws in Alpha Decay

Like all nuclear reactions, alpha decay obeys several fundamental conservation laws. The total electric charge is conserved because the charge carried away by the alpha particle exactly accounts for the decrease in the charge of the daughter nucleus. Likewise, the total number of nucleons remains constant before and after the decay.

Linear momentum is also conserved. If the parent nucleus is initially at rest, its total momentum is zero. After the decay, the alpha particle and the daughter nucleus recoil in opposite directions with equal and opposite momenta. Because the daughter nucleus is much more massive than the alpha particle, the alpha particle carries away most of the kinetic energy released in the decay.

Angular momentum is conserved as well, although verifying this requires considering the intrinsic spins and orbital angular momenta of the particles involved.

Finally, mass-energy is conserved. The energy released during the decay comes from a small decrease in the total mass of the system, according to Einstein's mass-energy relationship,

[latex]E=\Delta mc^2.[/latex]

Here, [latex]E[/latex] is the nuclear reaction energy, and [latex]\Delta m[/latex] is the difference between the total mass of the reactants and the total mass of the products. When the products have a smaller combined mass than the parent nucleus, [latex]\Delta m[/latex] is positive, and the reaction releases energy. Such reactions are called exothermic. If the products have a greater combined mass than the reactants, [latex]\Delta m[/latex] is negative, and energy must be supplied for the reaction to occur. Consequently, alpha decay can occur spontaneously only when the combined mass of the daughter nucleus and alpha particle is less than the mass of the parent nucleus.

Example 118.1: Calculating the Energy Released in Alpha Decay

Problem: Determine the energy released when [latex]{}^{239}\mathrm{Pu}[/latex] undergoes alpha decay.

Strategy

The energy released in a nuclear decay is determined from the mass difference between the parent nucleus and the decay products using Einstein's mass–energy relationship,

[latex]E=\Delta mc^2.[/latex]

Our first step is therefore to calculate [latex]\Delta m[/latex], the difference between the total initial mass and the total final mass. The required atomic masses can be obtained from Appendix A.

Solution

The alpha-decay equation for plutonium-239 is

[latex]{}^{239}\mathrm{Pu} \rightarrow {}^{235}\mathrm{U} + {}^{4}\mathrm{He}.[/latex]

From Appendix A, the relevant atomic masses are

  • [latex]m({}^{239}\mathrm{Pu})=239.052157\ \mathrm{u}[/latex]
  • [latex]m({}^{235}\mathrm{U})=235.043924\ \mathrm{u}[/latex]
  • [latex]m({}^{4}\mathrm{He})=4.002602\ \mathrm{u}[/latex]

The combined mass of the decay products is therefore

[latex]m({}^{235}\mathrm{U})+m({}^{4}\mathrm{He}) = 235.043924\ \mathrm{u} + 4.002602\ \mathrm{u} = 239.046526\ \mathrm{u}.[/latex]

The mass difference is

[latex]\Delta m = m({}^{239}\mathrm{Pu}) - \left[ m({}^{235}\mathrm{U}) + m({}^{4}\mathrm{He}) \right][/latex]
[latex]\Delta m = 239.052157\ \mathrm{u} - 239.046526\ \mathrm{u} = 0.005631\ \mathrm{u}.[/latex]

Using the conversion

[latex]1\ \mathrm{u} = 931.5\ \mathrm{MeV}/c^2,[/latex]

the energy released is

[latex]E = (\Delta m)c^2 = (0.005631\ \mathrm{u})(931.5\ \mathrm{MeV}/c^2)(c^2) = 5.25\ \mathrm{MeV}.[/latex]

Discussion

This decay releases [latex]5.25\ \mathrm{MeV}[/latex] of energy, which is roughly one million times greater than the energy involved in a typical chemical reaction. This enormous energy difference is one of the defining characteristics of nuclear processes.

Most of the released energy becomes kinetic energy of the emitted alpha particle because it is much lighter than the recoiling [latex]{}^{235}\mathrm{U}[/latex] nucleus. A small fraction appears as kinetic energy of the uranium nucleus, ensuring that linear momentum is conserved. In some decays, the daughter nucleus is left in an excited state and subsequently emits one or more gamma rays before reaching its lowest-energy state.

This reaction occurs spontaneously because the total mass of the products is smaller than the mass of the original plutonium nucleus. The "missing" mass has been converted into energy according to [latex]E=\Delta mc^2[/latex]. We will examine the physical origin of this mass difference in more detail in the chapter on Binding Energy.

Finally, note that the masses listed in Appendix A are the masses of neutral atoms, not bare nuclei. Since the same total number of electrons (94) exists before and after the decay, the electron masses cancel when calculating [latex]\Delta m[/latex], allowing atomic masses to be used directly.

Beta Decay

Unlike alpha decay, beta decay can occur in several different ways. In this chapter, we will examine three forms of beta decay: beta-minus ([latex]\beta^-[/latex]) decay, beta-plus ([latex]\beta^+[/latex]) decay, and electron capture. We begin with the most common type, beta-minus decay, which was the first form of beta decay to be discovered.

In [latex]\beta^-[/latex] decay, the nucleus emits an electron. Unlike the electrons that occupy atomic orbitals, this electron is created during the nuclear decay process itself. An example is the decay of cobalt-60:

[latex]{}^{60}\mathrm{Co} \rightarrow {}^{60}\mathrm{Ni} + \beta^- + \text{electron antineutrino}.[/latex]

The discovery of beta decay presented physicists with an unexpected puzzle. Careful measurements showed that the emitted electron did not carry away all of the energy expected from the decay. In addition, the measured momentum and angular momentum of the products did not satisfy the known conservation laws. Either these fundamental conservation laws were failing, or another particle was escaping detection.

In 1930, Wolfgang Pauli proposed that an additional neutral particle was emitted during beta decay. This particle, later named the neutrino by Enrico Fermi, carries away the "missing" energy, momentum, and angular momentum. Fermi subsequently developed the first successful theory of beta decay and showed that the process is governed by the weak nuclear force, one of the four fundamental forces of nature.

Neutrinos are extraordinarily difficult to detect. They have no electric charge, interact only through the weak nuclear force (and gravity), and possess extremely small masses. As a result, billions of neutrinos pass through your body every second with almost no interaction. Although they are elusive, neutrinos play an essential role in nuclear physics because they ensure that the conservation laws remain valid in beta decay.

The experimental confirmation of Fermi's theory came largely through the pioneering work of physicist Chien-Shiung Wu. By designing exceptionally precise experiments, Wu verified the principles of beta decay and demonstrated the fundamental role of the weak interaction, making lasting contributions to modern nuclear physics.

Portraits of Enrico Fermi and Chien-Shiung Wu, two physicists whose work established the modern understanding of beta decay.
Figure 118.3. (a) Enrico Fermi developed the first successful theory of beta decay and identified the weak nuclear force as the interaction responsible for the process. The unit of length called the fermi (fm), the element fermium, Fermilab, and the class of particles known as fermions are all named in his honor. (b) Chien-Shiung Wu performed the landmark experiments that confirmed Fermi's theory of beta decay and established many of the fundamental properties of the weak interaction. (Credits: U.S. Department of Energy; Smithsonian Institution.)

Electron Family Number

The discovery of the neutrino also revealed a new conservation law. Elementary particles belong to different particle families, and electrons and their associated neutrinos belong to the electron family. In every nuclear reaction, the total electron family number must remain constant.

Before a [latex]\beta^-[/latex] decay occurs, no electron-family particles are created or destroyed, so the total electron family number is zero. After the decay, an electron is produced, which is assigned an electron family number of [latex]+1[/latex]. To preserve the total, the decay must also produce an electron antineutrino, written [latex]\overline{\nu}_e[/latex], which has an electron family number of [latex]-1[/latex]. The sum therefore remains zero both before and after the decay.

This conservation law is analogous to conservation of electric charge. Just as positive and negative charges must balance, particles and their corresponding antiparticles must also be created in combinations that preserve the total electron family number.

If a nuclide undergoes [latex]\beta^-[/latex] decay, the general decay equation is

[latex]{}_{Z}^{A}\mathrm{X} \rightarrow {}_{Z+1}^{A}\mathrm{Y} + \beta^- + \overline{\nu}_e \qquad (\beta^-\ \text{decay}).[/latex]

Notice that the daughter nucleus has the same mass number but an atomic number that is one unit larger than that of the parent. This occurs because one neutron inside the nucleus is converted into a proton while simultaneously producing an electron and an electron antineutrino.

For example, cobalt has an atomic number of [latex]Z=27[/latex]. After [latex]\beta^-[/latex] decay, the daughter nucleus must therefore have [latex]Z=28[/latex], corresponding to nickel. Outside the nucleus, free neutrons undergo exactly the same transformation:

[latex]n \rightarrow p + \beta^- + \overline{\nu}_e.[/latex]
Diagram illustrating beta-minus decay in which a parent nucleus emits an electron and an electron antineutrino, producing a daughter nucleus with one additional proton.
Figure 118.4. During [latex]\beta^-[/latex] decay, a neutron is converted into a proton while an electron and an electron antineutrino are emitted. The daughter nucleus therefore contains one more proton and one fewer neutron than the parent nucleus. Because neutrinos interact only through the weak nuclear force, they are extremely difficult to detect directly.

Conservation Laws in Beta Decay

Beta-minus decay obeys all of the fundamental conservation laws. Electric charge is conserved because the daughter nucleus gains one unit of positive charge while the emitted electron carries one unit of negative charge. For example, cobalt-60 has a total nuclear charge of [latex]+27[/latex]. After decay, the nickel nucleus has charge [latex]+28[/latex], and the emitted electron has charge [latex]-1[/latex], leaving the total unchanged.

Linear momentum and angular momentum are also conserved. The emitted electron and antineutrino share most of the kinetic energy released in the decay, while the much heavier daughter nucleus recoils with comparatively little kinetic energy.

Another important conservation law states that the total number of nucleons ([latex]A[/latex]) remains constant. Although a neutron changes into a proton, the total number of nucleons is unchanged. Consequently, the mass number remains the same throughout beta decay, even though the numbers of protons and neutrons change individually.

As with alpha decay, the energy released during beta decay can be calculated from the difference in mass between the parent nucleus and the decay products using Einstein's relation, [latex]E=\Delta mc^2[/latex]. The next example illustrates this calculation.

Example 118.2: Calculating the Energy Released in Beta-Minus Decay

Problem: Determine the energy released when [latex]{}^{60}\mathrm{Co}[/latex] undergoes [latex]\beta^-[/latex] decay.

Strategy

As in the previous example, we determine the energy released by first calculating the mass difference between the parent nucleus and the decay products. The energy is then obtained using Einstein's mass–energy relationship,

[latex]E=\Delta mc^2.[/latex]

Because the tabulated values in Appendix A are atomic masses (which include electrons), an important simplification occurs. The daughter atom contains one more orbital electron than the parent atom, exactly matching the electron emitted during beta decay. Consequently, the electron masses cancel, and the mass difference is simply

[latex]\Delta m = m({}^{60}\mathrm{Co}) - m({}^{60}\mathrm{Ni}).[/latex]

Solution

The beta-minus decay equation is

[latex]{}^{60}_{27}\mathrm{Co} \rightarrow {}^{60}_{28}\mathrm{Ni} + \beta^- + \overline{\nu}_e.[/latex]

Using the atomic masses from Appendix A,

  • [latex]m({}^{60}\mathrm{Co})=59.933820\ \mathrm{u}[/latex]
  • [latex]m({}^{60}\mathrm{Ni})=59.930789\ \mathrm{u}[/latex]

The mass difference is therefore

[latex]\Delta m = 59.933820\ \mathrm{u} - 59.930789\ \mathrm{u} = 0.003031\ \mathrm{u}.[/latex]

The energy released is

[latex]E = (\Delta m)c^2 = (0.003031\ \mathrm{u})c^2.[/latex]

Using

[latex]1\ \mathrm{u} = 931.5\ \mathrm{MeV}/c^2,[/latex]

we obtain

[latex]E = (0.003031)(931.5\ \mathrm{MeV}/c^2)(c^2) = 2.82\ \mathrm{MeV}.[/latex]

Discussion

The beta decay of [latex]{}^{60}\mathrm{Co}[/latex] releases [latex]2.82\ \mathrm{MeV}[/latex] of energy, again demonstrating that nuclear reactions release far more energy than chemical reactions.

Unlike alpha decay, the released energy is shared among several particles. Most of the kinetic energy is carried by the emitted electron and the electron antineutrino, while the much heavier daughter nucleus receives only a small recoil energy. In many decays, the daughter nucleus [latex]{}^{60}\mathrm{Ni}[/latex] is also produced in an excited state and subsequently emits one or more gamma rays as it returns to its ground state.

Finally, remember that the emitted electron is not one of the atom's orbital electrons. It is created during the nuclear decay itself as a neutron inside the nucleus is transformed into a proton.

Beta-Plus Decay (Positron Emission)

The second form of beta decay is beta-plus ([latex]\beta^+[/latex]) decay, also known as positron emission. Instead of emitting an electron, the nucleus emits a positron, which is the electron's antimatter counterpart (Figure 118.5).

Diagram of beta-plus decay showing a parent nucleus emitting a positron and producing a daughter nucleus with one fewer proton.
Figure 118.5. During [latex]\beta^+[/latex] decay, a proton is converted into a neutron while a positron and an electron neutrino are emitted. The positron eventually encounters an electron, and the pair annihilates, producing two gamma rays that travel in opposite directions.

A positron has exactly the same mass and spin as an electron but carries a positive electric charge. It is often represented by the symbol [latex]e^+[/latex], although the symbol [latex]\beta^+[/latex] is commonly used when it is produced in nuclear decay.

When a positron encounters an ordinary electron, the two particles annihilate each other. Their entire mass is converted into electromagnetic energy, typically producing two gamma-ray photons that travel in opposite directions:

[latex]e^+ + e^- \rightarrow \gamma + \gamma.[/latex]

Healthcare Connection: PET Imaging

Positron emission is the physical principle behind positron emission tomography (PET), one of the most important medical imaging techniques. Radioactive isotopes such as fluorine-18 emit positrons inside the body. Each positron quickly annihilates with a nearby electron, producing two gamma rays that travel in opposite directions. PET scanners detect these paired gamma rays simultaneously, allowing physicians to reconstruct detailed three-dimensional images of metabolic activity. PET imaging plays a major role in cancer diagnosis, neurology, and cardiology.

The general equation for beta-plus decay is

[latex]{}_{Z}^{A}\mathrm{X} \rightarrow {}_{Z-1}^{A}\mathrm{Y} + \beta^+ + \nu_e \qquad (\beta^+\ \text{decay}).[/latex]

In this process, one proton inside the nucleus is transformed into a neutron. Consequently, the daughter nucleus has one fewer proton, one additional neutron, and the same mass number as the parent nucleus. To conserve electron family number, an electron neutrino, [latex]\nu_e[/latex], is emitted together with the positron.

For example, sodium-22 undergoes beta-plus decay according to

[latex]{}^{22}_{11}\mathrm{Na} \rightarrow {}^{22}_{10}\mathrm{Ne} + \beta^+ + \nu_e.[/latex]

When calculating the energy released during beta-plus decay using atomic masses, an additional correction is required. One electron mass is missing from the daughter atom, and another electron mass appears as the emitted positron. Therefore,

[latex]\Delta m = m(\text{parent}) - \left[ m(\text{daughter}) + 2m_e \right].[/latex]

Electron Capture

The third type of beta decay is electron capture (EC). Instead of emitting a positron, the nucleus captures one of the atom's inner-shell electrons. This process has the same overall effect as beta-plus decay: a proton is converted into a neutron while an electron neutrino is emitted.

[latex]{}_{Z}^{A}\mathrm{X} + e^- \rightarrow {}_{Z-1}^{A}\mathrm{Y} + \nu_e \qquad (\text{electron capture}).[/latex]

Electron capture can occur only when the combined mass of the daughter nucleus and emitted neutrino is less than the combined mass of the parent atom and captured electron. Any nuclide capable of beta-plus decay can also undergo electron capture, and many radionuclides exhibit both decay modes.

Both electron capture and beta-plus decay obey the same conservation laws for charge, nucleon number, energy, momentum, angular momentum, and electron family number.

The Region of Stability

All forms of beta decay occur because the parent nucleus lies outside the region of stability on the chart of nuclides.

  • Nuclei containing too many neutrons tend to undergo [latex]\beta^-[/latex] decay, converting a neutron into a proton and moving toward a more stable neutron-to-proton ratio.
  • Nuclei containing too many protons tend to undergo [latex]\beta^+[/latex] decay or electron capture, converting a proton into a neutron and likewise moving toward greater stability.

In all cases, beta decay helps unstable nuclei evolve toward a more stable balance of protons and neutrons.

Gamma Decay

Gamma decay is the simplest major form of nuclear decay. It occurs when a nucleus in an excited energy state releases excess energy by emitting one or more high-energy photons called gamma rays.

Just as electrons in atoms can occupy excited energy levels, protons and neutrons within a nucleus can also exist in higher-energy states. When the nucleons move to lower-energy states, the nucleus releases the energy difference as electromagnetic radiation. Because nuclear energy-level differences are much larger than atomic energy-level differences, gamma-ray photons generally have much higher energies than visible light or X-rays produced by atomic transitions.

Nuclear excited states typically have very short lifetimes, often about [latex]10^{-14}\ \mathrm{s}[/latex]. This rapid transition reflects the strength of the forces acting within the nucleus.

The general equation for gamma decay is

[latex]{}_{Z}^{A}\mathrm{X}_{N}^{*} \rightarrow {}_{Z}^{A}\mathrm{X}_{N} + \gamma_1 + \gamma_2 + \cdots \qquad (\gamma\ \text{decay}),[/latex]

where the asterisk indicates that the nucleus is in an excited state. Depending on how the nucleus reaches its lowest-energy state, it may emit one gamma ray or a sequence of gamma rays.

Unlike alpha and beta decay, gamma decay does not change the number of protons or neutrons in the nucleus. The atomic number [latex]Z[/latex], neutron number [latex]N[/latex], and mass number [latex]A[/latex] all remain unchanged. Only the energy state of the nucleus changes.

Gamma emission commonly follows another nuclear decay that leaves the daughter nucleus in an excited state. For example, when [latex]{}^{60}\mathrm{Co}[/latex] undergoes beta-minus decay, the resulting nickel-60 nucleus is often produced in an excited state, written [latex]{}^{60}\mathrm{Ni}^{*}[/latex]. The excited nucleus then emits two gamma-ray photons:

[latex]{}^{60}\mathrm{Ni}^{*} \rightarrow {}^{60}\mathrm{Ni} + \gamma_1 + \gamma_2.[/latex]

Healthcare Connection: Gamma Rays in Cancer Treatment

The gamma rays associated with cobalt-60 have been used in radiation therapy to treat cancer. Although the photons are emitted by excited nickel-60 nuclei formed after cobalt-60 decays, they are commonly called cobalt-60 gamma rays.

Because gamma rays are highly penetrating, carefully directed beams can deliver energy to tumors located inside the body. The radiation damages the DNA of cancer cells, reducing their ability to divide and grow. Treatment systems are designed to concentrate the dose in the tumor while limiting exposure to surrounding healthy tissue.

Gamma decay obeys the same fundamental conservation laws as other nuclear processes. Energy, electric charge, linear momentum, angular momentum, nucleon number, and electron family number are all conserved. The emitted photon carries energy and momentum away from the nucleus, while the nucleus experiences a small recoil.

Because gamma decay does not transform one nuclide into another, it is usually not shown prominently in decay-series diagrams such as Figure 118.1.

Other forms of nuclear decay also exist, but they occur less frequently than alpha, beta, and gamma decay. The most important of these is spontaneous fission, in which a heavy nucleus divides into two smaller nuclei. Fission and its applications in nuclear energy are discussed in the next chapter.

Section Summary

  • Radioactive decay transforms an unstable parent nucleus into one or more products. The resulting nucleus is called the daughter nucleus. The three major types of radioactive decay are alpha, beta, and gamma decay.
  • In alpha decay, the nucleus emits a helium-4 nucleus containing two protons and two neutrons:
    [latex]{}_{Z}^{A}\mathrm{X}_{N} \rightarrow {}_{Z-2}^{A-4}\mathrm{Y}_{N-2} + {}_{2}^{4}\mathrm{He}_{2}.[/latex]

    The daughter nucleus therefore has two fewer protons, two fewer neutrons, and four fewer nucleons than the parent.

  • The energy released in nuclear decay is related to the decrease in mass by
    [latex]E=\Delta mc^2.[/latex]

    A decay can occur spontaneously when the products have less total mass than the parent nucleus.

  • In beta-minus decay, a neutron is converted into a proton while an electron and an electron antineutrino are emitted:
    [latex]{}_{Z}^{A}\mathrm{X}_{N} \rightarrow {}_{Z+1}^{A}\mathrm{Y}_{N-1} + \beta^- + \overline{\nu}_e.[/latex]
  • In beta-plus decay, a proton is converted into a neutron while a positron and an electron neutrino are emitted:
    [latex]{}_{Z}^{A}\mathrm{X}_{N} \rightarrow {}_{Z-1}^{A}\mathrm{Y}_{N+1} + \beta^+ + \nu_e.[/latex]
  • In electron capture, the nucleus absorbs an inner-shell electron, converting a proton into a neutron and emitting an electron neutrino:
    [latex]{}_{Z}^{A}\mathrm{X}_{N} + e^- \rightarrow {}_{Z-1}^{A}\mathrm{Y}_{N+1} + \nu_e.[/latex]
  • The symbol [latex]\beta^-[/latex] represents an electron emitted in nuclear decay, while [latex]\beta^+[/latex] represents a positron. The symbol [latex]\nu_e[/latex] represents an electron neutrino, and [latex]\overline{\nu}_e[/latex] represents an electron antineutrino.
  • Beta decay demonstrates conservation of electron family number. It also preserves the total number of nucleons, even though the individual numbers of protons and neutrons may change.
  • In gamma decay, an excited nucleus releases energy as one or more high-energy photons:
    [latex]{}_{Z}^{A}\mathrm{X}_{N}^{*} \rightarrow {}_{Z}^{A}\mathrm{X}_{N} + \gamma_1 + \gamma_2 + \cdots.[/latex]

    Gamma decay does not change the identity of the nuclide because its numbers of protons and neutrons remain unchanged.

Conceptual Questions

  1. Star Trek fans have often heard the term antimatter drive. Describe how a magnetic field could be used to trap antimatter produced in nuclear decay and later allow it to combine with ordinary matter to release energy. In your answer, identify the type of antimatter involved, explain why it must be stored in a vacuum, and discuss what fraction of the matter is converted into energy.
  2. Which conservation law requires an electron neutrino to be produced during electron capture? Remember that the captured electron no longer exists after the reaction.
  3. Neutrinos have been shown experimentally to possess an extremely small mass. During Supernova 1987A, located about 100,000 light-years from Earth, neutrinos and visible light from the explosion were detected at nearly the same time. Explain how comparing the arrival times of the neutrinos and the light can be used to place limits on the mass of neutrinos.
  4. What characteristic is shared by all three types of beta decay that distinguishes them from alpha decay?

Problems & Exercises

In the following eight problems, write the complete decay equation for the given nuclide using the complete [latex]{}_{Z}^{A}X_{N}[/latex] notation. Refer to the periodic table for values of [latex]Z[/latex].

  1. [latex]\beta^-[/latex] decay of [latex]{}^{3}\mathrm{H}[/latex] (tritium), a manufactured isotope of hydrogen used in some digital watch displays, and manufactured primarily for use in hydrogen bombs.
  2. [latex]\beta^-[/latex] decay of [latex]{}^{40}\mathrm{K}[/latex], a naturally occurring rare isotope of potassium responsible for some of our exposure to background radiation.
  3. [latex]\beta^+[/latex] decay of [latex]{}^{50}\mathrm{Mn}[/latex].
  4. [latex]\beta^+[/latex] decay of [latex]{}^{52}\mathrm{Fe}[/latex].
  5. Electron capture by [latex]{}^{7}\mathrm{Be}[/latex].
  6. Electron capture by [latex]{}^{106}\mathrm{In}[/latex].
  7. [latex]\alpha[/latex] decay of [latex]{}^{210}\mathrm{Po}[/latex], the isotope of polonium in the decay series of [latex]{}^{238}\mathrm{U}[/latex] that was discovered by the Curies. A favorite isotope in physics laboratories because it has a short half-life and decays to a stable nuclide.
  8. [latex]\alpha[/latex] decay of [latex]{}^{226}\mathrm{Ra}[/latex], another isotope in the decay series of [latex]{}^{238}\mathrm{U}[/latex], first recognized as a new element by the Curies. It poses special problems because its daughter is a radioactive noble gas.

In the following four problems, identify the parent nuclide and write the complete decay equation using the [latex]{}_{Z}^{A}X_{N}[/latex] notation. Refer to the periodic table for values of [latex]Z[/latex].

  1. [latex]\beta^-[/latex] decay producing [latex]{}^{137}\mathrm{Ba}[/latex]. The parent nuclide is a major waste product of nuclear reactors and has chemistry similar to potassium and sodium, resulting in its concentration in your cells if ingested.
  2. [latex]\beta^-[/latex] decay producing [latex]{}^{90}\mathrm{Y}[/latex]. The parent nuclide is a major waste product of nuclear reactors and has chemistry similar to calcium, so that it is concentrated in bones if ingested ([latex]{}^{90}\mathrm{Y}[/latex] is also radioactive).
  3. [latex]\alpha[/latex] decay producing [latex]{}^{228}\mathrm{Ra}[/latex]. The parent nuclide is nearly 100% of the natural element and is found in gas lantern mantles and in metal alloys used in jet engines ([latex]{}^{228}\mathrm{Ra}[/latex] is also radioactive).
  4. [latex]\alpha[/latex] decay producing [latex]{}^{208}\mathrm{Pb}[/latex]. The parent nuclide is in the decay series produced by [latex]{}^{232}\mathrm{Th}[/latex], the only naturally occurring isotope of thorium.
  5. When an electron and positron annihilate, both their masses are destroyed, creating two equal-energy photons to conserve momentum.
    1. Confirm that the annihilation equation [latex]e^{+}+e^{-}\rightarrow\gamma+\gamma[/latex] conserves charge, electron family number, and total number of nucleons. Identify the values of each before and after the annihilation.
    2. Find the energy of each [latex]\gamma[/latex] ray, assuming the electron and positron are initially nearly at rest.
    3. Explain why the two [latex]\gamma[/latex] rays travel in exactly opposite directions if the center of mass of the electron–positron system is initially at rest.
  6. Confirm that charge, electron family number, and the total number of nucleons are all conserved by the rule for [latex]\alpha[/latex] decay given by
    [latex]{}_{Z}^{A}X_{N}\rightarrow{}_{Z-2}^{A-4}Y_{N-2}+{}_{2}^{4}\mathrm{He}_{2}.[/latex]

    Identify the values of each before and after the decay.

  7. Confirm that charge, electron family number, and the total number of nucleons are all conserved by the rule for [latex]\beta^-[/latex] decay given by
    [latex]{}_{Z}^{A}X_{N}\rightarrow{}_{Z+1}^{A}Y_{N-1}+\beta^-+\overline{\nu}_{e}.[/latex]

    Identify the values of each before and after the decay.

  8. Confirm that charge, electron family number, and the total number of nucleons are all conserved by the rule for [latex]\beta^+[/latex] decay given by
    [latex]{}_{Z}^{A}X_{N}\rightarrow{}_{Z-1}^{A}Y_{N+1}+\beta^++\nu_{e}.[/latex]

    Identify the values of each before and after the decay.

  9. Confirm that charge, electron family number, and the total number of nucleons are all conserved by the rule for electron capture given by
    [latex]{}_{Z}^{A}X_{N}+e^{-}\rightarrow{}_{Z-1}^{A}Y_{N+1}+\nu_{e}.[/latex]

    Identify the values of each before and after the capture.

  10. A rare decay mode has been observed in which [latex]{}^{222}\mathrm{Ra}[/latex] emits a [latex]{}^{14}\mathrm{C}[/latex] nucleus.
    1. The decay equation is [latex]{}^{222}\mathrm{Ra}\rightarrow{}^{A}X+{}^{14}\mathrm{C}[/latex]. Identify the nuclide [latex]{}^{A}X[/latex].
    2. Find the energy emitted in the decay. The mass of [latex]{}^{222}\mathrm{Ra}[/latex] is 222.015353 u.
    1. Write the complete [latex]\alpha[/latex] decay equation for [latex]{}^{226}\mathrm{Ra}[/latex].
    2. Find the energy released in the decay.
    1. Write the complete [latex]\alpha[/latex] decay equation for [latex]{}^{249}\mathrm{Cf}[/latex].
    2. Find the energy released in the decay.
    1. Write the complete [latex]\beta^-[/latex] decay equation for the neutron.
    2. Find the energy released in the decay.
    1. Write the complete [latex]\beta^-[/latex] decay equation for [latex]{}^{90}\mathrm{Sr}[/latex], a major waste product of nuclear reactors.
    2. Find the energy released in the decay.
  11. Calculate the energy released in the [latex]\beta^+[/latex] decay of [latex]{}^{22}\mathrm{Na}[/latex], the equation for which is given in the text. The masses of [latex]{}^{22}\mathrm{Na}[/latex] and [latex]{}^{22}\mathrm{Ne}[/latex] are 21.994434 u and 21.991383 u, respectively.
    1. Write the complete [latex]\beta^+[/latex] decay equation for [latex]{}^{11}\mathrm{C}[/latex].
    2. Calculate the energy released in the decay. The masses of [latex]{}^{11}\mathrm{C}[/latex] and [latex]{}^{11}\mathrm{B}[/latex] are 11.011433 u and 11.009305 u, respectively.
    1. Calculate the energy released in the [latex]\alpha[/latex] decay of [latex]{}^{238}\mathrm{U}[/latex].
    2. What fraction of the mass of a single [latex]{}^{238}\mathrm{U}[/latex] is destroyed in the decay? The mass of [latex]{}^{234}\mathrm{Th}[/latex] is 234.043593 u.
    3. Although the fractional mass loss is large for a single nucleus, it is difficult to observe for an entire macroscopic sample of uranium. Why is this?
    1. Write the complete reaction equation for electron capture by [latex]{}^{7}\mathrm{Be}[/latex].
    2. Calculate the energy released.
    1. Write the complete reaction equation for electron capture by [latex]{}^{15}\mathrm{O}[/latex].
    2. Calculate the energy released.

Glossary

alpha decay
A type of radioactive decay in which a nucleus emits an alpha particle ([latex]{}^{4}\mathrm{He}[/latex] nucleus).
antielectron
Another name for a positron, the antimatter counterpart of the electron.
antimatter
Matter composed of antiparticles, which have the same mass as ordinary particles but opposite electric charge and certain other quantum properties.
beta decay
A class of radioactive decays in which a nucleus emits an electron, a positron, or captures an orbital electron.
daughter nucleus
The nucleus produced when a radioactive parent nucleus undergoes nuclear decay.
decay
The spontaneous transformation of an unstable atomic nucleus into a more stable configuration through the emission of particles or electromagnetic radiation.
decay equation
A nuclear equation showing the parent nucleus, the emitted particles, and the daughter nucleus produced during radioactive decay.
decay series
A sequence of radioactive decays in which one unstable daughter nucleus decays into another until a stable nucleus is produced.
electron capture
A form of beta decay in which a proton-rich nucleus captures an inner-shell electron, converting a proton into a neutron while emitting an electron neutrino.
electron capture equation
The nuclear equation representing an electron-capture reaction.
electron neutrino ([latex]\nu_e[/latex])
An electrically neutral elementary particle emitted during beta-plus decay and electron capture.
electron antineutrino ([latex]\overline{\nu}_e[/latex])
The antimatter counterpart of the electron neutrino, emitted during beta-minus decay.
gamma decay
A nuclear transition in which an excited nucleus emits one or more high-energy gamma-ray photons without changing its atomic or mass number.
neutrino
An electrically neutral elementary particle that interacts only through the weak nuclear force (and gravity) and has an extremely small mass.
nuclear reaction energy
The energy released or absorbed during a nuclear reaction as a result of a change in mass.
parent nucleus
The original unstable nucleus that undergoes radioactive decay.
positron
The positively charged antimatter counterpart of the electron, emitted during beta-plus decay.
positron decay
Another name for beta-plus decay, in which a proton is converted into a neutron while a positron and an electron neutrino are emitted.
definition

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Introductory Physics for the Health and Life Sciences II Copyright © 2012 by OSCRiceUniversity is licensed under a Creative Commons Attribution 4.0 International License, except where otherwise noted.