Nuclear Physics

76 Nuclear Reactions

Learning Objectives

By the end of this section, you will be able to:

  • Describe and compare three types of nuclear radiation
  • Use nuclear symbols to describe changes that occur during nuclear reactions
  • Describe processes involved in the decay series of heavy elements

Early experiments revealed three types of nuclear “rays” or radiation: alpha (\alpha) rays, beta (\beta) rays, and gamma (\gamma) rays. These three types of radiation are differentiated by their ability to penetrate matter. Alpha radiation is barely able to pass through a thin sheet of paper. Beta radiation can penetrate aluminum to a depth of about 3 mm, and gamma radiation can penetrate lead to a depth of 2 or more centimeters ((Figure)).

A comparison of the penetration depths of alpha (\alpha), beta (\beta), and gamma (\gamma) radiation through various materials.

The figure shows from left to right: paper, metal, concrete and lead. Three types of radiation enter this setup from the left. Alpha radiation does not pass through paper. Beta radiation passes through paper but not through metal. Gamma radiation passes through paper, metal and concrete, but not through lead.

The electrical properties of these three types of radiation are investigated by passing them through a uniform magnetic field, as shown in (Figure). According to the magnetic force equation \stackrel{\to }{F}=q\stackrel{\to }{v}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\stackrel{\to }{B}, positively charged particles are deflected upward, negatively charged particles are deflected downward, and particles with no charge pass through the magnetic field undeflected. Eventually, \alpha rays were identified with helium nuclei \left({}^{4}\text{He}\right),\beta rays with electrons and positrons (positively charged electrons or antielectrons), and \gamma rays with high-energy photons. We discuss alpha, beta, and gamma radiation in detail in the remainder of this section.

The effect of a magnetic field on alpha (\alpha), beta (\beta), and gamma (\gamma) radiation. This figure is a schematic only. The relative paths of the particles depend on their masses and initial kinetic energies.

Figure shows a C-shaped material labeled lead. A small circle labeled radioactive source is shown in the hollow of the C-shape. Three rays radiate from this source towards the right. One curves upwards and is labeled alpha. One goes straight and is labeled gamma. The third curves downwards and is labeled beta minus. Magnetic field is shown as crosses. Two arrows originate from near the point where the rays emerge from the C-shape. The upwards pointing arrow is labeled F subscript alpha = q subscript alpha v B. The downwards pointing arrow is labeled F subscript beta = q subscript beta v B.

Alpha Decay

Heavy unstable nuclei emit \alpha radiation. In \alpha-particle decay (or alpha decay), the nucleus loses two protons and two neutrons, so the atomic number decreases by two, whereas its mass number decreases by four. Before the decay, the nucleus is called the parent nucleus. The nucleus or nuclei produced in the decay are referred to as the daughter nucleus or daughter nuclei. We represent an \alpha decay symbolically by

{}_{Z}^{A}\text{X}\to {}_{Z-2}^{A-4}\text{X}+{}_{2}^{4}\text{H}\text{e}

where {}_{Z}^{A}\text{X} is the parent nucleus, {}_{Z-2}^{A-4}\text{X} is the daughter nucleus, and {}_{2}^{4}\text{H}\text{e} is the \alpha particle. In \alpha decay, a nucleus of atomic number Z decays into a nucleus of atomic number Z-2 and atomic mass A-4. Interestingly, the dream of the ancient alchemists to turn other metals into gold is scientifically feasible through the alpha-decay process. The efforts of the alchemists failed because they relied on chemical interactions rather than nuclear interactions.

Watch alpha particles escape from a polonium nucleus, causing radioactive alpha decay. See how random decay times relate to the half-life. To try a simulation of alpha decay, visit alpha particles

An example of alpha decay is uranium-238:

{}_{\phantom{\rule{0.5em}{0ex}}92}^{238}\text{U}\to {}_{\phantom{\rule{0.5em}{0ex}}90}^{234}\text{X}+{}_{2}^{4}\text{H}\text{e}.

The atomic number has dropped from 92 to 90. The chemical element with Z=90 is thorium. Hence, Uranium-238 has decayed to Thorium-234 by the emission of an \alpha particle, written

{}_{\phantom{\rule{0.5em}{0ex}}92}^{238}\text{U}\to {}_{\phantom{\rule{0.5em}{0ex}}90}^{234}\text{T}\text{h}+{}_{2}^{4}\text{H}\text{e}.

Subsequently, {}_{\phantom{\rule{0.5em}{0ex}}90}^{234}\text{T}\text{h} decays by \beta emission with a half-life of 24 days. The energy released in this alpha decay takes the form of kinetic energies of the thorium and helium nuclei, although the kinetic energy of thorium is smaller than helium due to its heavier mass and smaller velocity.

Plutonium Alpha Decay Find the energy emitted in the \alpha decay of {}^{239}\text{Pu}.

Strategy The energy emitted in the \alpha decay of {}^{239}\text{Pu} can be found using the equation E=\left(\text{Δ}m\right){c}^{2}. We must first find \text{Δ}m, the difference in mass between the parent nucleus and the products of the decay.

Solution The decay equation is

{}^{239}\text{Pu}\to {}^{235}\text{U}+{}^{4}\text{He}.

Thus, the pertinent masses are those of {}^{239}\text{Pu}, {}^{235}\text{U}, and the \alpha particle or {}^{4}\text{He}, all of which are known. The initial mass was m\left({}^{239}\text{Pu}\right)=239.052157\phantom{\rule{0.2em}{0ex}}\text{u}. The final mass is the sum

\begin{array}{}\\ \\ \hfill m\left({}^{235}\text{U}\right)+m\left({}^{4}\text{He}\right)& =235.043924\phantom{\rule{0.2em}{0ex}}\text{u}+4.002602\phantom{\rule{0.2em}{0ex}}\text{u}\hfill \\ & =239.046526\phantom{\rule{0.2em}{0ex}}\text{u}.\hfill \end{array}

Thus,

\begin{array}{cc}\hfill \text{Δ}m& =m\left({}^{239}\text{Pu}\right)-\left[m\left({}^{235}\text{U}\right)+m\left({}^{4}\text{He}\right)\right]\hfill \\ & =239.052157\phantom{\rule{0.2em}{0ex}}\text{u}-239.046526\phantom{\rule{0.2em}{0ex}}\text{u}\hfill \\ & =0.0005631\phantom{\rule{0.2em}{0ex}}\text{u}.\hfill \end{array}

Now we can find E by entering \text{Δ}m into the equation:

E=\left(\text{Δ}m\right){c}^{2}=\left(0.005631\phantom{\rule{0.2em}{0ex}}\text{u}\right){c}^{2}.

We know 1\phantom{\rule{0.2em}{0ex}}\text{u}=931.5\phantom{\rule{0.2em}{0ex}}\text{MeV}\text{/}{c}^{2}, so we have

\begin{array}{cc}\hfill E& =\left(0.005631\right)\left(931.5\phantom{\rule{0.2em}{0ex}}\text{MeV}\text{/}{c}^{2}\right)\left({c}^{2}\right)\hfill \\ & =5.25\phantom{\rule{0.2em}{0ex}}\text{MeV}.\hfill \end{array}

Significance The energy released in this \alpha decay is in the MeV range, many times greater than chemical reaction energies. Most of this energy becomes kinetic energy of the \alpha particle (or {}^{4}\text{He} nucleus), which moves away at high speed. The energy carried away by the recoil of the {}^{235}\text{U} nucleus is much smaller due to its relatively large mass. The {}^{235}\text{U} nucleus can be left in an excited state to later emit photons (\gamma rays).

Beta Decay

In most \beta particle decays (or beta decay), either an electron ({\beta }^{-}) or positron ({\beta }^{+}) is emitted by a nucleus. A positron has the same mass as the electron, but its charge is \text{+}e. For this reason, a positron is sometimes called an antielectron. How does \beta decay occur? A possible explanation is the electron (positron) is confined to the nucleus prior to the decay and somehow escapes. To obtain a rough estimate of the escape energy, consider a simplified model of an electron trapped in a box (or in the terminology of quantum mechanics, a one-dimensional square well) that has the width of a typical nucleus ({10}^{-14}\text{m}). According to the Heisenberg uncertainty principle in Quantum Mechanics, the uncertainty of the momentum of the electron is:

\text{Δ}p>\frac{h}{\text{Δ}x}=\frac{6.6\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-34}\phantom{\rule{0.2em}{0ex}}{\text{m}}^{2}·\text{kg/s}}{{10}^{-14}\text{m}}=6.6\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-20}\text{kg}·\phantom{\rule{0.2em}{0ex}}\text{m/s}.

Taking this momentum value (an underestimate) to be the “true value,” the kinetic energy of the electron on escape is approximately

\frac{{\left(\text{Δ}p\right)}^{2}}{2{m}_{e}}=\frac{{\left(6.6\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-20}\phantom{\rule{0.2em}{0ex}}\text{kg}·\text{m/s}\right)}^{2}}{2\left(9.1\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-31}\text{kg}\right)}=2.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-9}\text{J}=12,400\phantom{\rule{0.2em}{0ex}}\text{MeV}.

Experimentally, the electrons emitted in {\beta }^{-} decay are found to have kinetic energies of the order of only a few MeV. We therefore conclude that the electron is somehow produced in the decay rather than escaping the nucleus. Particle production (annihilation) is described by theories that combine quantum mechanics and relativity, a subject of a more advanced course in physics.

Nuclear beta decay involves the conversion of one nucleon into another. For example, a neutron can decay to a proton by the emission of an electron ({\beta }^{-}) and a nearly massless particle called an antineutrino (\stackrel{-}{\nu }):

{}_{0}^{1}\text{n}\to {}_{1}^{1}\text{p}+{}_{-1}^{\phantom{\rule{0.7em}{0ex}}0}\text{e}\phantom{\rule{0.2em}{0ex}}\text{+}\phantom{\rule{0.2em}{0ex}}\stackrel{-}{v}.

The notation {}_{-1}^{\phantom{\rule{0.7em}{0ex}}0}\text{e} is used to designate the electron. Its mass number is 0 because it is not a nucleon, and its atomic number is -1 to signify that it has a charge of \text{−}e. The proton is represented by {}_{1}^{1}\text{p} because its mass number and atomic number are 1. When this occurs within an atomic nucleus, we have the following equation for beta decay:

{}_{Z}^{A}\text{X}\to {}_{Z+1}^{\phantom{\rule{1.5em}{0ex}}A}\text{X}+{}_{-1}^{\phantom{\rule{0.7em}{0ex}}0}\text{e}\phantom{\rule{0.2em}{0ex}}\text{+}\phantom{\rule{0.2em}{0ex}}\stackrel{-}{v}.

As discussed in another chapter, this process occurs due to the weak nuclear force.

Watch beta decay occur for a collection of nuclei or for an individual nucleus.

As an example, the isotope {}_{\phantom{\rule{0.5em}{0ex}}90}^{234}\text{T}\text{h} is unstable and decays by {\beta }^{-} emission with a half-life of 24 days. Its decay can be represented as

{}_{\phantom{\rule{0.5em}{0ex}}90}^{234}\text{T}\text{h}\to {}_{\phantom{\rule{0.5em}{0ex}}91}^{234}\text{X}+{}_{-1}^{\phantom{\rule{0.7em}{0ex}}0}\text{e}\phantom{\rule{0.2em}{0ex}}\text{+}\phantom{\rule{0.2em}{0ex}}\stackrel{-}{v}.

Since the chemical element with atomic number 91 is protactinium (Pa), we can write the {\beta }^{-} decay of thorium as

{}_{\phantom{\rule{0.5em}{0ex}}90}^{234}\text{T}\text{h}\to {}_{\phantom{\rule{0.5em}{0ex}}91}^{234}\text{P}\text{a}+{}_{-1}^{\phantom{\rule{0.7em}{0ex}}0}\text{e}\phantom{\rule{0.2em}{0ex}}\text{+}\phantom{\rule{0.2em}{0ex}}\stackrel{-}{v}.

The reverse process is also possible: A proton can decay to a neutron by the emission of a positron ({\beta }^{+}) and a nearly massless particle called a neutrino (v). This reaction is written as {}_{1}^{1}\text{p}\to {}_{0}^{1}\text{n}+{}_{\text{+}1}^{\phantom{\rule{0.7em}{0ex}}0}\text{e}\phantom{\rule{0.2em}{0ex}}\text{+}\phantom{\rule{0.2em}{0ex}}v.

The positron {}_{\text{+}1}^{\phantom{\rule{0.7em}{0ex}}0}\text{e} is emitted with the neutrino v, and the neutron remains in the nucleus. (Like {\beta }^{-} decay, the positron does not precede the decay but is produced in the decay.) For an isolated proton, this process is impossible because the neutron is heavier than the proton. However, this process is possible within the nucleus because the proton can receive energy from other nucleons for the transition. As an example, the isotope of aluminum {}_{13}^{26}\text{A}\text{l} decays by {\beta }^{+} emission with a half-life of 7.40\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{5}\text{y}. The decay is written as

{}_{13}^{26}\text{A}\text{l}\to {}_{12}^{26}\text{X}+{}_{\text{+}1}^{\phantom{\rule{0.7em}{0ex}}0}\text{e}\phantom{\rule{0.2em}{0ex}}\text{+}\phantom{\rule{0.2em}{0ex}}v.

The atomic number 12 corresponds to magnesium. Hence,

{}_{13}^{26}\text{A}\text{l}\to {}_{12}^{26}\text{M}\text{g}+{}_{\text{+}1}^{\phantom{\rule{0.7em}{0ex}}0}\text{e}\phantom{\rule{0.2em}{0ex}}\text{+}\phantom{\rule{0.2em}{0ex}}v.

As a nuclear reaction, positron emission can be written as

{}_{Z}^{A}\text{X}\to {}_{Z-1}^{\phantom{\rule{1.5em}{0ex}}A}\text{X}+{}_{\text{+}1}^{\phantom{\rule{0.7em}{0ex}}0}\text{e}\phantom{\rule{0.2em}{0ex}}\text{+}\phantom{\rule{0.2em}{0ex}}v.

The neutrino was not detected in the early experiments on \beta decay. However, the laws of energy and momentum seemed to require such a particle. Later, neutrinos were detected through their interactions with nuclei.

Bismuth Alpha and Beta Decay The {}_{\phantom{\rule{0.5em}{0ex}}83}^{211}\text{B}\text{i} nucleus undergoes both \alpha and {\beta }^{-} decay. For each case, what is the daughter nucleus?

Strategy We can use the processes described by (Figure) and (Figure), as well as the Periodic Table, to identify the resulting elements.

Solution The atomic number and the mass number for the \alpha particle are 2 and 4, respectively. Thus, when a bismuth-211 nucleus emits an \alpha particle, the daughter nucleus has an atomic number of 81 and a mass number of 207. The element with an atomic number of 81 is thallium, so the decay is given by

{}_{\phantom{\rule{0.5em}{0ex}}83}^{211}\text{B}\text{i}\to {}_{\phantom{\rule{0.5em}{0ex}}81}^{207}\text{T}\text{i}+{}_{2}^{4}\text{H}\text{e}.

In {\beta }^{-} decay, the atomic number increases by 1, while the mass number stays the same. The element with an atomic number of 84 is polonium, so the decay is given by

{}_{\phantom{\rule{0.5em}{0ex}}83}^{211}\text{B}\text{i}\to {}_{\phantom{\rule{0.5em}{0ex}}84}^{211}\text{P}\text{o}+{}_{-1}^{\phantom{\rule{0.7em}{0ex}}0}\text{e}\phantom{\rule{0.2em}{0ex}}\text{+}\phantom{\rule{0.2em}{0ex}}\stackrel{-}{v}.

Check Your Understanding In radioactive beta decay, does the atomic mass number, A, increase or decrease?

Neither; it stays the same.

Gamma Decay

A nucleus in an excited state can decay to a lower-level state by the emission of a “gamma-ray” photon, and this is known as gamma decay. This is analogous to de-excitation of an atomic electron. Gamma decay is represented symbolically by

{}_{Z}^{A}\text{X}*\to {}_{Z}^{A}\text{X}+\gamma

where the asterisk (*) on the nucleus indicates an excited state. In \gamma decay, neither the atomic number nor the mass number changes, so the type of nucleus does not change.

Radioactive Decay Series

Nuclei with Z>82 are unstable and decay naturally. Many of these nuclei have very short lifetimes, so they are not found in nature. Notable exceptions include {}_{\phantom{\rule{0.5em}{0ex}}90}^{232}\text{T}\text{h} (or Th-232) with a half-life of 1.39\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{10} years, and {}_{\phantom{\rule{0.5em}{0ex}}92}^{238}\text{U} (or U-238) with a half-life of 7.04\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{8} years. When a heavy nucleus decays to a lighter one, the lighter daughter nucleus can become the parent nucleus for the next decay, and so on. This process can produce a long series of nuclear decays called a decay series. The series ends with a stable nucleus.

To illustrate the concept of a decay series, consider the decay of Th-232 series ((Figure)). The neutron number, N, is plotted on the vertical y-axis, and the atomic number, Z, is plotted on the horizontal x-axis, so Th-232 is found at the coordinates \left(N,Z\right)=\left(142,90\right). Th-232 decays by \alpha emission with a half-life of 1.39\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{10} years. Alpha decay decreases the atomic number by 2 and the mass number by 4, so we have

{}_{\phantom{\rule{0.5em}{0ex}}90}^{232}\text{T}\text{h}\to {}_{\phantom{\rule{0.5em}{0ex}}88}^{228}\text{R}\text{a}+{}_{2}^{4}\text{H}\text{e}\phantom{\rule{0.2em}{0ex}}.

The neutron number for Radium-228 is 140, so it is found in the diagram at the coordinates \left(N,Z\right)=\left(140,88\right). Radium-228 is also unstable and decays by {\beta }^{-} emission with a half-life of 5.76 years to Actinum-228. The atomic number increases by 1, the mass number remains the same, and the neutron number decreases by 1. Notice that in the graph, \alpha emission appears as a line sloping downward to the left, with both N and Z decreasing by 2. Beta emission, on the other hand, appears as a line sloping downward to the right with N decreasing by 1, and Z increasing by 1. After several additional alpha and beta decays, the series ends with the stable nucleus Pb-208.

The relative frequency of different types of radioactive decays (alpha, beta, and gamma) depends on many factors, including the strength of the forces involved and the number of ways a given reaction can occur without violating the conservation of energy and momentum. How often a radioactive decay occurs often depends on a sensitive balance of the strong and electromagnetic forces. These forces are discussed in Particle Physics and Cosmology.

In the thorium {}_{\phantom{\rule{0.5em}{0ex}}90}^{232}\text{T}\text{h} decay series, alpha \left(\alpha \right) decays reduce the atomic number, as indicated by the red arrows. Beta ({\beta }^{-}) decays increase the atomic number, as indicated by the blue arrows. The series ends at the stable nucleus Pb-208.

A graph of neutron number N = A – Z versus atomic number Z is shown. Alpha decay is shown by red arrows pointing downward left, thus showing decrease in both N and Z. Beta decay is shown by blue arrows pointing downward right, indicating a decrease in N and increase in Z. The decay is shown as follows: Alpha decay from 232 Th to 228 Ra in 1.39 into 10 to the power 10 years. Beta decay from 228 Ra to 228 Ac in 5.76 years and from 228 Ac to 228 Th in 6.15 hours. Alpha decay from 228 Th to 224 Ra in 1.91 years, from 224 Ra to 220 Rn in 3.66 days, from 220 Rn to 216 Po in 55.6 seconds and from 216 Po to 212 Pb in 0.15 seconds. Beta decay from 212 Pb to 212 Bi in 10.6 hours and from 212 Bi to 212 Po in 60.6 minutes. Alpha decay from 212 Po to 208 Pb in 0.3 into 10 to the power minus 6 seconds.

As another example, consider the U-238 decay series shown in (Figure). After numerous alpha and beta decays, the series ends with the stable nucleus Pb-206. An example of a decay whose parent nucleus no longer exists naturally is shown in (Figure). It starts with Neptunium-237 and ends in the stable nucleus Bismuth-209. Neptunium is called a transuranic element because it lies beyond uranium in the periodic table. Uranium has the highest atomic number \left(Z=92\right) of any element found in nature. Elements with Z>92 can be produced only in the laboratory. They most probably also existed in nature at the time of the formation of Earth, but because of their relatively short lifetimes, they have completely decayed. There is nothing fundamentally different between naturally occurring and artificial elements.

In the Uranium-238 decay series, alpha (\alpha) decays reduce the atomic number, as indicated by the red arrows. Beta ({\beta }^{-}) decays increase the atomic number, as indicated by the blue arrows. The series ends at the stable nucleus Pb-206.

A graph of neutron number N = A – Z versus atomic number Z is shown. Alpha decay is shown by red arrows pointing downward left, thus showing decrease in both N and Z. Beta decay is shown by blue arrows pointing downward right, indicating a decrease in N and increase in Z. The decay is shown as follows: Alpha decay from 238 U to 234 Th in 4.46 into 10 to the power 9 years. Beta decay from 234 Th to 234 Pa in 24.1 days and from 234 Pa to 234 U in 6.66 hours. Alpha decay from 234 U to 230 Th in 2.48 into 10 to the power 5 years, from 230 Th to 226 Ra in 7.54 into 10 to the power 4 years, from 226 Ra to 222 Rn in 1600 years, from 222 Rn to 218 Po in 3.82 days, and from 218 Po to 214 Pb in 3.05 minutes. Beta decay from 214 Pb to 214 Bi in 26 minutes and from 214 Bi to 214 Po in 19.9 minutes. Alpha decay from 214 Bi to 210 Tl in 26 minutes and from 214 Po to 210 Pb in 1.64 into 10 to the power minus 4 seconds. Beta decay from 210 Tl to 210 Pb in 1.3 minutes, from 210 Pb to 210 Bi in 22.6 years and from 210 Bi to 210 Po in 5.01 days. Alpha decay from 210 Po to 206 Pb in 138 days.

Notice that for Bi (21), the decay may proceed through either alpha or beta decay.

In the Neptunium-237 decay series, alpha (\alpha) decays reduce the atomic number, as indicated by the red arrows. Beta ({\beta }^{-}) decays increase the atomic number, as indicated by the blue arrows. The series ends at the stable nucleus Bi-209.

A graph of neutron number N = A – Z versus atomic number Z is shown. Alpha decay is shown by red arrows pointing downward left, thus showing decrease in both N and Z. Beta decay is shown by blue arrows pointing downward right, indicating a decrease in N and increase in Z. The decay is shown as follows: Alpha decay from 237 Np to 233 Pa in 2.14 into 10 to the power 6 years. Beta decay from 233 Pa to 233 U in 27 days. Alpha decay from 233 U to 229 Th in 1.59 into 10 to the power 5 years and from 229 Th to 225 Ra in 7900 years. Beta decay from 225 Ra to 225 Ac in 14.8 days. Alpha decay from 225 Ac to 221 Fr in 10 days, from 221 Fr to 217 At in 4.8 minutes and from 217 At to 213 Bi in 0.032 seconds. Beta decay from 213 Bi to 213 Po in 45.6 minutes. Alpha decay from 213 Po to 209 Pb in 4 into 10 to the power minus 6 seconds. Beta decay from 209 Pb to 209 Bi in 3.25 hours.

Radioactivity in the Earth

According to geologists, if there were no heat source, Earth should have cooled to its present temperature in no more than 1 billion years. Yet, Earth is more than 4 billion years old. Why is Earth cooling so slowly? The answer is nuclear radioactivity, that is, high-energy particles produced in radioactive decays heat Earth from the inside ((Figure)).

Earth is heated by nuclear reactions (alpha, beta, and gamma decays). Without these reactions, Earth’s surface would be much cooler than it is now.

A sectioned figure of the earth showing different layers. A circular arrow, labeled convection, is shown near the core. Outward arrows from here are labeled conduction. Outward arrows outside the earth are labeled radiation. A section from within the earth is shown as a circle containing arrows for alpha, beta and gamma rays in all directions.

Candidate nuclei for this heating model are {}^{238}\text{U and}\phantom{\rule{0.2em}{0ex}}{}^{40}\text{K}, which possess half-lives similar to or longer than the age of Earth. The energy produced by these decays (per second per cubic meter) is small, but the energy cannot escape easily, so Earth’s core is very hot. Thermal energy in Earth’s core is transferred to Earth’s surface and away from it through the processes of convection, conduction, and radiation.

Summary

  • The three types of nuclear radiation are alpha (\alpha) rays, beta (\beta) rays, and gamma (\gamma) rays.
  • We represent \alpha decay symbolically by {}_{Z}^{A}\text{X}\to {}_{Z-2}^{A-4}\text{X}+{}_{2}^{4}\text{H}\text{e}. There are two types of \beta decay: either an electron ({\beta }^{-}) or a positron ({\beta }^{+}) is emitted by a nucleus. \gamma decay is represented symbolically by {}_{Z}^{A}\text{X}*\to {}_{Z}^{A}\text{X}+\gamma.
  • When a heavy nucleus decays to a lighter one, the lighter daughter nucleus can become the parent nucleus for the next decay, and so on, producing a decay series.

Conceptual Questions

What is the key difference and the key similarity between beta ({\beta }^{-}) decay and alpha decay?

What is the difference between \gamma rays and characteristic X-rays and visible light?

Gamma (γ) rays are produced by nuclear interactions and X-rays and light are produced by atomic interactions. Gamma rays are typically shorter wavelength than X-rays, and X-rays are shorter wavelength than light.

What characteristics of radioactivity show it to be nuclear in origin and not atomic?

Consider (Figure). If the magnetic field is replaced by an electric field pointed in toward the page, in which directions will the \alpha-, {\beta }^{+}-, and \gamma rays bend?

Assume a rectangular coordinate system with an xy-plane that corresponds to the plane of the paper. \alpha bends into the page (trajectory parabolic in the xz-plane); {\beta }^{+} bends into the page (trajectory parabolic in the xz-plane); and \gamma is unbent.

Why is Earth’s core molten?

Problems

{}^{249}\text{C}\text{f} undergoes alpha decay. (a) Write the reaction equation. (b) Find the energy released in the decay.

(a) Calculate the energy released in the \alpha decay of {}^{238}\text{U}. (b) What fraction of the mass of a single {}^{238}\text{U} is destroyed in the decay? The mass of {}^{234}\text{T}\text{h} is 234.043593 u. (c) 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?

a. 4.273 MeV; b. 1.927\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-5}; c. Since {}^{238}\text{U} is a slowly decaying substance, only a very small number of nuclei decay on human timescales; therefore, although those nuclei that decay lose a noticeable fraction of their mass, the change in the total mass of the sample is not detectable for a macroscopic sample.

The {\beta }^{-} particles emitted in the decay of {}^{3}\text{H} (tritium) interact with matter to create light in a glow-in-the-dark exit sign. At the time of manufacture, such a sign contains 15.0 Ci of {}^{3}\text{H}. (a) What is the mass of the tritium? (b) What is its activity 5.00 y after manufacture?

(a) Write the complete {\beta }^{-} decay equation for {}^{90}\text{S}\text{r}, a major waste product of nuclear reactors. (b) Find the energy released in the decay.

a. {}_{38}^{90}\text{S}{\text{r}}_{52}\to {}_{39}^{90}\text{Y}{}_{51}+{\beta }^{-1}+\stackrel{-}{{v}_{e}}; b. 0.546 MeV

Write a nuclear {\beta }^{-} decay reaction that produces the {}^{90}\text{Y} nucleus. (Hint: The parent nuclide is a major waste product of reactors and has chemistry similar to calcium, so that it is concentrated in bones if ingested.)

Write the complete decay equation in the complete {}_{Z}^{A}\text{X}{}_{N} notation for the beta ({\beta }^{-}) decay of {}^{3}\text{H} (tritium), a manufactured isotope of hydrogen used in some digital watch displays, and manufactured primarily for use in hydrogen bombs.

{}_{1}^{3}\text{H}{}_{2}\to {}_{2}^{3}\text{H}{\text{e}}_{1}+{\beta }^{-}+\stackrel{-}{{v}_{e}}

If a 1.50-cm-thick piece of lead can absorb 90.0\text{%} of the rays from a radioactive source, how many centimeters of lead are needed to absorb all but 0.100\text{%} of the rays?

An electron can interact with a nucleus through the beta-decay process:

{}_{Z}^{A}\text{X}\phantom{\rule{0.2em}{0ex}}\text{+}\phantom{\rule{0.2em}{0ex}}{e}^{-}\to \phantom{\rule{0.2em}{0ex}}Y+{v}_{e}.

(a) Write the complete reaction equation for electron capture by {}^{7}\text{Be}.

(b) Calculate the energy released.

a. {}_{4}^{7}\text{B}{\text{e}}_{3}+{e}^{-}\to {}_{3}^{7}\text{L}{\text{i}}_{4}+{v}_{e}; b. 0.862 MeV

(a) Write the complete reaction equation for electron capture by {}^{15}\text{O}.

(b) Calculate the energy released.

A rare decay mode has been observed in which {}^{222}\text{R}\text{a} emits a {}^{14}\text{C} nucleus. (a) The decay equation is {}^{222}\text{Ra}\to {}^{A}\text{X}+{}^{14}\text{C}. Identify the nuclide {}^{A}\text{X}. (b) Find the energy emitted in the decay. The mass of {}^{222}\text{Ra} is 222.015353 u.

a. \text{X}={}_{\phantom{\rule{0.5em}{0ex}}82}^{208}\text{P}{\text{b}}_{126}; b. 33.05 MeV

Glossary

alpha decay
radioactive nuclear decay associated with the emission of an alpha particle
alpha (α) rays
one of the types of rays emitted from the nucleus of an atom as alpha particles
antielectrons
another term for positrons
antineutrino
antiparticle of an electron’s neutrino in {\beta }^{-} decay
beta decay
radioactive nuclear decay associated with the emission of a beta particle
beta (\beta) rays
one of the types of rays emitted from the nucleus of an atom as beta particles
daughter nucleus
nucleus produced by the decay of a parent nucleus
decay series
series of nuclear decays ending in a stable nucleus
gamma decay
radioactive nuclear decay associated with the emission of gamma radiation
gamma (\gamma) rays
one of the types of rays emitted from the nucleus of an atom as gamma particles
neutrino
subatomic elementary particle which has no net electric charge
parent nucleus
original nucleus before decay
positron
electron with positive charge
transuranic element
element that lies beyond uranium in the periodic table

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University Physics Volume 3 by cnxuniphysics is licensed under a Creative Commons Attribution 4.0 International License, except where otherwise noted.

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