10.4 Nuclear Reactions

Early experiments revealed three types of nuclear “rays” or radiation: alpha ( $α$ ) rays, beta ( $β$ ) rays, and gamma ( $γ$ ) 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 10.11).

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. — **Figure 10.11** A comparison of the penetration depths of alpha ( $α$ ), beta ( $β$ ), and gamma ( $γ$ ) radiation through various materials.

The electrical properties of these three types of radiation are investigated by passing them through a uniform magnetic field, as shown in Figure 10.12. According to the magnetic force equation $\vec{F} = q \vec{v} \times \vec{B},$ positively charged particles are deflected upward, negatively charged particles are deflected downward, and particles with no charge pass through the magnetic field undeflected. Eventually, $α$ rays were identified with helium nuclei $open parenthesis to the power 4 He close parenthesis, β$ rays with electrons and positrons (positively charged electrons or antielectrons), and $γ$ rays with high-energy photons. We discuss alpha, beta, and gamma radiation in detail in the remainder of this section.

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. — **Figure 10.12** The effect of a magnetic field on alpha ( $α$ ), beta ( $β$ ), and gamma ( $γ$ ) radiation. This figure is a schematic only. The relative paths of the particles depend on their masses and initial kinetic energies.

Alpha Decay

Heavy unstable nuclei emit $α$ radiation. In $α$ -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 $α$ decay symbolically by

$_{Z}^{A} X \to_{Z - 2}^{A - 4} X +_{2}^{4} H e$

(10.21)

where $_{Z}^{A} X$ is the parent nucleus, $_{Z - 2}^{A - 4} X$ is the daughter nucleus, and $2 4 H e$ is the $α$ particle. In $α$ decay, a nucleus of atomic number Z decays into a nucleus of atomic number $Z minus 2$ and atomic mass $A minus 4.$ 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.

An example of alpha decay is uranium-238:

$92 238 U → 90 234 X plus 2 4 H e.$

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

$92 238 U → 90 234 T h plus 2 4 H e.$

Subsequently, $90 234 T h$ decays by $β$ 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.

Example 1 Plutonium Alpha Decay

Find the energy emitted in the $α$ decay of $to the power 239 Pu$ .

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

Solution The decay equation is

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

Thus, the pertinent masses are those of $to the power 239 Pu$ , $to the power 235 U$ , and the $α$ particle or $to the power 4 He$ , all of which are known. The initial mass was $m (^{239} Pu) = 239.052157 u .$ The final mass is the sum

$\begin{matrix} m (^{235} U) + m (^{4} He) & = 235.043924 u + 4.002602 u \\ = 239.046526 u . \end{matrix}$

Thus,

$\begin{matrix} Δ m & = m (^{239} Pu) - [m (^{235} U) + m (^{4} He)] \\ = 239.052157 u - 239.046526 u \\ = 0.0005631 u . \end{matrix}$

Now we can find E by entering $Δ m$ into the equation:

$E = (Δ m) c^{2} = (0.005631 u) c^{2} .$

We know $1 u equals 931.5 MeV / c squared,$ so we have

$\begin{matrix} E & = (0.005631) (931.5 MeV / c^{2}) (c^{2}) \\ = 5.25 MeV . \end{matrix}$

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

Beta Decay

In most $β$ particle decays (or beta decay), either an electron ( $β^{-}$ ) or positron ( $β^{+}$ ) is emitted by a nucleus. A positron has the same mass as the electron, but its charge is $+ e$ . For this reason, a positron is sometimes called an antielectron. How does $β$ 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 ( $10 to the power −14 m$ ). According to the Heisenberg uncertainty principle in Quantum Mechanics, the uncertainty of the momentum of the electron is:

$Δ p > \frac{h}{Δ x} = \frac{6.6 \times 10^{−34} m^{2} \cdot kg/s}{10^{−14} m} = 6.6 \times 10^{−20} kg \cdot m/s .$

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

$\frac{{(Δ p)}^{2}}{2 m_{e}} = \frac{{(6.6 \times 10^{−20} kg \cdot m/s)}^{2}}{2 (9.1 \times 10^{−31} kg)} = 2.0 \times 10^{−9} J = 12,400 MeV .$

Experimentally, the electrons emitted in $β^{-}$ 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 ( $β^{-}$ ) and a nearly massless particle called an antineutrino ( $\bar{ν}$ ):

$_{0}^{1} n \to_{1}^{1} p +_{−1}^{0} e + \bar{v} .$

The notation $−1 0 e$ 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 $− e$ . The proton is represented by $1 1 p$ because its mass number and atomic number are 1. When this occurs within an atomic nucleus, we have the following equation for beta decay:

$_{Z}^{A} X \to_{Z + 1}^{A} X +_{−1}^{0} e + \bar{v} .$

(10.22)

Enrico Fermi proposed a theory of beta decay in 1934, but his work was initially rejected. Other physicists' experiments to prove it were unsuccessful, casting further doubt on the theory. Chinese-born physicist Chien-Shiung Wu, who had developed a number of processes critical to the Manhattan Project and related research, identified a number of flaws in the earlier experimental methods and materials. She designed a new method, verified Fermi's theory, and later went on to establish the core principles of beta decay. As discussed in another chapter, this process occurs due to the weak nuclear force.

As an example, the isotope $90 234 T h$ is unstable and decays by $β^{-}$ emission with a half-life of 24 days. Its decay can be represented as

$_{90}^{234} T h \to_{91}^{234} X +_{−1}^{0} e + \bar{v} .$

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

$_{90}^{234} T h \to_{91}^{234} P a +_{−1}^{0} e + \bar{v} .$

The reverse process is also possible: A proton can decay to a neutron by the emission of a positron ( $β^{+}$ ) and a nearly massless particle called a neutrino (v). This reaction is written as $1 1 p → 0 1 n plus + 1 0 e + v.$

The positron $+ 1 0 e$ is emitted with the neutrino v, and the neutron remains in the nucleus. (Like $β^{-}$ 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 $13 26 A l$ decays by $β^{+}$ emission with a half-life of $7.40 \times 10^{5} y .$ The decay is written as

$13 26 A l → 12 26 X plus + 1 0 e + v.$

The atomic number 12 corresponds to magnesium. Hence,

$13 26 A l → 12 26 M g plus + 1 0 e + v.$

As a nuclear reaction, positron emission can be written as

$_{Z}^{A} X \to_{Z - 1}^{A} X +_{+ 1}^{0} e + v .$

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

Example 2 Bismuth Alpha and Beta Decay

The $83 211 B i$ nucleus undergoes both $α$ and $β^{-}$ decay. For each case, what is the daughter nucleus?

Strategy We can use the processes described by Equation 10.21 and Equation 10.22, as well as the Periodic Table, to identify the resulting elements.

Solution The atomic number and the mass number for the $α$ particle are 2 and 4, respectively. Thus, when a bismuth-211 nucleus emits an $α$ 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

$83 211 B i → 81 207 T i plus 2 4 H e.$

In $β^{-}$ 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

$_{83}^{211} B i \to_{84}^{211} P o +_{−1}^{0} e + \bar{v} .$

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

$Z A X * → Z A X plus γ$

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

Radioactive Decay Series

Nuclei with $Z greater than 82$ are unstable and decay naturally. Many of these nuclei have very short lifetimes, so they are not found in nature. Notable exceptions include $90 232 T h$ (or Th-232) with a half-life of $1.39 \times 10^{10}$ years, and $92 238 U$ (or U-238) with a half-life of $7.04 \times 10^{8}$ 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 10.13). 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 $open parenthesis N, Z close parenthesis equals open parenthesis 142, 90 close parenthesis.$ Th-232 decays by $α$ emission with a half-life of $1.39 \times 10^{10}$ years. Alpha decay decreases the atomic number by 2 and the mass number by 4, so we have

$90 232 T h → 88 228 R a plus 2 4 H e.$

The neutron number for Radium-228 is 140, so it is found in the diagram at the coordinates $open parenthesis N, Z close parenthesis equals open parenthesis 140, 88 close parenthesis.$ Radium-228 is also unstable and decays by $β^{-}$ 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, $α$ 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.

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. — **Figure 10.13** In the thorium $90 232 T h$ decay series, alpha $(α)$ decays reduce the atomic number, as indicated by the red arrows. Beta ( $β^{-}$ ) decays increase the atomic number, as indicated by the blue arrows. The series ends at the stable nucleus Pb-208.

As another example, consider the U-238 decay series shown in Figure 10.14. 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 10.15. It starts with Neptunium-237, which decays to Bismuth-209, long thought to be stable, but which decays with a half-life of 2.01 × 10¹⁹ years (much longer than the age of the universe) to the stable nucleus Thallium-205. Neptunium is called a transuranic element because it lies beyond uranium in the periodic table. Uranium has the highest atomic number $open parenthesis Z equals 92 close parenthesis$ of any element found in nature. Elements with $Z greater than 92$ 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.

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

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 × 10⁹ years. Yet, Earth is more than 4 × 10⁹ 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 10.16).

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. — **Figure 10.16** Earth is heated by nuclear reactions (alpha, beta, and gamma decays). Without these reactions, Earth’s core and mantle would be much cooler than it is now.

Candidate nuclei for this heating model are $^{238} U and^{40} K$ , 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 ( $α$ ) rays, beta ( $β$ ) rays, and gamma ( $γ$ ) rays.
We represent $α$ decay symbolically by $_{Z}^{A} X \to_{Z - 2}^{A - 4} X +_{2}^{4} H e$ . There are two types of $β$ decay: either an electron ( $β^{-}$ ) or a positron ( $β^{+}$ ) is emitted by a nucleus. $γ$ decay is represented symbolically by $Z A X * → Z A X plus γ$ .
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 ( $β^{-}$ ) decay and alpha decay?

What is the difference between $γ$ 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 10.12. If the magnetic field is replaced by an electric field pointed in toward the page, in which directions will the $α$ -, $β^{+}$ -, and $γ$ rays bend?

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

Why is Earth’s core molten?

Problems

$to the power 249 C f$ undergoes alpha decay. (a) Write the reaction equation. (b) Find the energy released in the decay.

(a) Calculate the energy released in the $α$ decay of $to the power 238 U$ . (b) What fraction of the mass of a single $to the power 238 U$ is destroyed in the decay? The mass of $to the power 234 T h$ 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. $1.927 \times 10^{−5}$ ; c. Since $to the power 238 U$ 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 $β^{-}$ particles emitted in the decay of $^{3} H$ (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 $^{3} H$ . (a) What is the mass of the tritium? (b) What is its activity 5.00 y after manufacture?

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

a. $_{38}^{90} S r_{52} \to_{39}^{90} Y_{51} + β^{−1} + \bar{v_{e}}$ ; b. 0.546 MeV

Write a nuclear $β^{-}$ decay reaction that produces the $to the power 90 Y$ 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 $_{Z}^{A} X_{N}$ notation for the beta ( $β^{-}$ ) decay of $^{3} H$ (tritium), a manufactured isotope of hydrogen used in some digital watch displays, and manufactured primarily for use in hydrogen bombs.

$_{1}^{3} H_{2} \to_{2}^{3} H e_{1} + β^{-} + \bar{v_{e}}$

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

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

$_{Z}^{A} X + e^{-} \to Y + v_{e}$ .

(a) Write the complete reaction equation for electron capture by $to the power 7 Be$ .

(b) Calculate the energy released.

a. $_{4}^{7} B e_{3} + e^{-} \to_{3}^{7} L i_{4} + v_{e}$ ; b. 0.862 MeV

(a) Write the complete reaction equation for electron capture by $to the power 15 O.$

(b) Calculate the energy released.

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

a. $X equals 82 208 P b sub 126$ ; b. 33.05 MeV