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Chapter 24.2: Nuclear Reactions

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    28634

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    Learning Objectives

    • To know the different kinds of radioactive decay.
    • To balance a nuclear reaction.

    The two general kinds of nuclear reactions are nuclear decay reactions and nuclear transmutation reactions. In a nuclear decay reactionA nuclear reaction that occurs when an unstable nucleus emits radiation and is transformed into the nucleus of one or more other elements., also called radioactive decay, an unstable nucleus emits radiation and is transformed into the nucleus of one or more other elements. The resulting daughter nuclei have a lower mass and are lower in energy (more stable) than the parent nucleus that decayed. In contrast, in a nuclear transmutation reactionA nuclear reaction in which a nucleus reacts with a subatomic particle or another nuleus to give a product nucleus that is more massive than the starting material., a nucleus reacts with a subatomic particle or another nucleus to form a product nucleus that is more massive than the starting material. As we shall see, nuclear decay reactions occur spontaneously under all conditions, but nuclear transmutation reactions occur only under very special conditions, such as the collision of a beam of highly energetic particles with a target nucleus or in the interior of stars. We begin this section by considering the different classes of radioactive nuclei, along with their characteristic nuclear decay reactions and the radiation they emit.

    Note the Pattern

    Nuclear decay reactions occur spontaneously under all conditions, whereas nuclear transmutation reactions are induced.

    Classes of Radioactive Nuclei

    The three general classes of radioactive nuclei are characterized by a different decay process or set of processes:

    1. Neutron-rich nuclei. The nuclei on the upper left side of the band of stable nuclei in Figure 24.1.2  have a neutron-to-proton ratio that is too high to give a stable nucleus. These nuclei decay by a process that converts a neutron to a proton, thereby decreasing the neutron-to-proton ratio.
    2. Neutron-poor nuclei. Nuclei on the lower right side of the band of stable nuclei have a neutron-to-proton ratio that is too low to give a stable nucleus. These nuclei decay by processes that have the net effect of converting a proton to a neutron, thereby increasing the neutron-to-proton ratio.
    3. Heavy nuclei. With very few exceptions, heavy nuclei (those with A ≥ 200) are intrinsically unstable regardless of the neutron-to-proton ratio, and all nuclei with Z > 83 are unstable. This is presumably due to the cumulative effects of electrostatic repulsions between the large number of positively charged protons, which cannot be totally overcome by the strong nuclear force, regardless of the number of neutrons present. Such nuclei tend to decay by emitting an α particle (a helium nucleus, \( _{2}^{4}He\) ), which decreases the number of protons and neutrons in the original nucleus by 2. Because the neutron-to-proton ratio in an α particle is 1, the net result of alpha emission is an increase in the neutron-to-proton ratio.

    Note the Pattern

    Nuclear decay reactions always produce daughter nuclei that have a more favorable neutron-to- proton ratio and hence are more stable than the parent nucleus.

    Nuclear Decay Reactions

    Just as we use the number and type of atoms present to balance a chemical equation, we can use the number and type of nucleons present to write a balanced nuclear equation for a nuclear decay reaction. This procedure also allows us to predict the identity of either the parent or the daughter nucleus if the identity of only one is known. Regardless of the mode of decay, the total number of nucleons is conserved in all nuclear reactions.

    To describe nuclear decay reactions, chemists have extended the \( _{Z}^{A}X \) notation for nuclides to include radioactive emissions. Table 24.2.1 " lists the name and symbol for each type of emitted radiation. We introduced the most common of these, α and β particles and γ rays, in Chapter 1 and Chapter 14. The most notable addition is the positron, a particle that has the same mass as an electron but a positive charge rather than a negative charge.

    Table 24.2.1 Nuclear Decay Emissions and Their Symbols

    Identity Symbol Charge Mass (amu)
    helium nucleus \( _{2}^{4}\alpha \) +2 4.001506
    electron \( _{-1}^{0}\beta \; or \; \beta ^{- } \) −1 0.000549
    photon \( _{0}^{0}\gamma \)
    neutron \( _{0}^{1}n \) 0 1.008665
    proton \( _{1}^{1}p \) +1 1.007276
    positron \( _{+1}^{0}\beta \; or \; \beta ^{+} \) +1 0.000549

    Like the notation used to indicate isotopes, the upper left superscript in the symbol for a particle gives the mass number, which is the total number of protons and neutrons. For a proton or a neutron, A = 1. Because neither an electron nor a positron contains protons or neutrons, its mass number is 0. The numbers should not be taken literally, however, as meaning that these particles have zero mass; ejection of a beta particle (an electron) simply has a negligible effect on the mass of a nucleus.

    Similarly, the lower left subscript gives the charge of the particle. Because protons carry a positive charge, Z = +1 for a proton. In contrast, a neutron contains no protons and is electrically neutral, so Z = 0. In the case of an electron, Z = −1, and for a positron, Z = +1. Because γ rays are high-energy photons, both A and Z are 0. In some cases, two different symbols are used for particles that are identical but produced in different ways. For example, the symbol \( _{-1}^{0}e \) which is usually simplified to e, represents a free electron or an electron associated with an atom, whereas the symbol \( _{-1}^{0}\beta \) which is often simplified to β, denotes an electron that originates from within the nucleus, which is a β particle. Similarly, \( _{2}^{4}He^{2+} \)  refers to the nucleus of a helium atom, and \( _{2}^{4}\alpha \) denotes an identical particle that has been ejected from a heavier nucleus.

    There are six fundamentally different kinds of nuclear decay reactions, and each releases a different kind of particle or energy. The essential features of each reaction are shown in Figure 24.2.1. The most common are alpha and beta decay and gamma emission, but the others are essential to an understanding of nuclear decay reactions.

    21052210add8d84594c96ca623d39804.jpg

    Figure 24.2.1 Common Modes of Nuclear Decay

    Alpha Decay

    Many nuclei with mass numbers greater than 200 undergo alpha (α) decayA nuclear decay reaction that results in the emission of a helium-4 nucleus as an α particle., which results in the emission of a helium-4 nucleus as an alpha (α) particleA helium nucleus: \( _{2}^{4}\alpha \) The general reaction is as follows:

    \( \underset{parent}{_{Z}^{A}X} \rightarrow \underset{daughter}{_{Z-2}^{A-4}X} + \underset{alpha \; particle}{_{2}^{4}\alpha} \tag{24.2.1} \)

    The daughter nuclide contains two fewer protons and two fewer neutrons than the parent. Thus α-particle emission produces a daughter nucleus with a mass number A − 4 and a nuclear charge Z − 2 compared to the parent nucleus. Radium-226, for example, undergoes alpha decay to form radon-222:

    \( _{88}^{226}Ra \rightarrow _{86}^{222}Rn + _{2}^{4}\alpha  \tag{24.2.2} \)

    Because nucleons are conserved in this and all other nuclear reactions, the sum of the mass numbers of the products, 222 + 4 = 226, equals the mass number of the parent. Similarly, the sum of the atomic numbers of the products, 86 + 2 = 88, equals the atomic number of the parent. Thus the nuclear equation is balanced.

    Note the Pattern

    Just as the total number of atoms is conserved in a chemical reaction, the total number of nucleons is conserved in a nuclear reaction.

    Beta Decay

    Nuclei that contain too many neutrons often undergo beta (β) decayA nuclear decay reaction in which a neutron is converted to a proton and a high-energy electron that is ejected from the nucleus as a β particle., in which a neutron is converted to a proton and a high-energy electron that is ejected from the nucleus as a β particle:

    \( \underset{unstable neutron}{_{0}^{1}n} \rightarrow \underset{retained proton}{_{1}^{1}p} + \underset{emitted \; \beta \; particle}{_{-1}^{0}\beta} \tag{24.2.3} \)

    The general reaction for beta decay is therefore

    \( \underset{parent}{_{Z}^{A}X} \rightarrow \underset{daughter}{_{Z+1}^{A}X'} + \underset{beta}{_{-1}^{0}\beta} \tag{24.2.4} \)

    Although beta decay does not change the mass number of the nucleus, it does result in an increase of +1 in the atomic number because of the addition of a proton in the daughter nucleus. Thus beta decay decreases the neutron-to-proton ratio, moving the nucleus toward the band of stable nuclei. For example, carbon-14 undergoes beta decay to form nitrogen-14:

    \( _{6}^{14}C \rightarrow _{7}^{14}N + _{-1}^{\;\;0}\beta \tag{24.2.5} \)

    Once again, the number of nucleons is conserved, and the charges are balanced. The parent and the daughter nuclei have the same mass number, 14, and the sum of the atomic numbers of the products is 6, which is the same as the atomic number of the carbon-14 parent.

    Positron Emission

    Because a positron has the same mass as an electron but opposite charge, positron emissionA nuclear decay reaction in which a proton is transformed into a neutron, and a high-energy positron is emitted. is the opposite of beta decay. Thus positron emission is characteristic of neutron-poor nuclei, which decay by transforming a proton to a neutron and emitting a high-energy positron:

    \( _{1}^{1}p \rightarrow _{0}^{1}n + _{+1}^{\;\;0}\beta \tag{24.2.6} \)

    The general reaction for positron emission is therefore

    \( \underset{parent}{_{Z}^{A}X} \rightarrow \underset{daughter}{_{Z-1}^{A}X'} + \underset{positron}{_{+1}^{0}\beta} \tag{24.2.7} \)

    Like beta decay, positron emission does not change the mass number of the nucleus. In this case, however, the atomic number of the daughter nucleus is lower by 1 than that of the parent. Thus the neutron-to-proton ratio has increased, again moving the nucleus closer to the band of stable nuclei. For example, carbon-11 undergoes positron emission to form boron-11:

    \( _{6}^{11}C \rightarrow _{5}^{11}N + _{+1}^{\;\;0}\beta \tag{24.2.8} \)

    Nucleons are conserved, and the charges balance. The mass number, 11, does not change, and the sum of the atomic numbers of the products is 6, the same as the atomic number of the parent carbon-11 nuclide.

    Electron Capture

    A neutron-poor nucleus can decay by either positron emission or electron capture (EC)A nuclear decay reaction in which an electron in an inner shell reacts with a proton to produce a neutron., in which an electron in an inner shell reacts with a proton to produce a neutron:

    \( _{1}^{1}p + _{\; \; 0}^{-1}e \rightarrow \;\;_{0}^{1}n \tag{24.2.9} \)

    When a second electron moves from an outer shell to take the place of the lower-energy electron that was absorbed by the nucleus, an x-ray is emitted. The overall reaction for electron capture is thus

    \( \underset{parent}{_{Z}^{A}X} + \underset{electron}{_{-1}^{\;\;0}e} \rightarrow \underset{daughter}{_{Z-1}^{A}X'} + x-ray  \tag{24.2.10} \)

    Electron capture does not change the mass number of the nucleus because both the proton that is lost and the neutron that is formed have a mass number of 1. As with positron emission, however, the atomic number of the daughter nucleus is lower by 1 than that of the parent. Once again, the neutron-to-proton ratio has increased, moving the nucleus toward the band of stable nuclei. For example, iron-55 decays by electron capture to form manganese-55, which is often written as follows:

    \( _{26}^{55}Fe \overset{EC}{\rightarrow} \;\;_{25}^{55}Mn + x-ray \tag{24.2.11} \)

    The atomic numbers of the parent and daughter nuclides differ in Equation 20.11, although the mass numbers are the same. To write a balanced nuclear equation for this reaction, we must explicitly include the captured electron in the equation:

    \( _{26}^{55}Fe _{-1}^{\;\;0}e \;\rightarrow  \;\;_{25}^{55}Mn + x-ray \tag{24.2.12} \)

    Both positron emission and electron capture are usually observed for nuclides with low neutron-to-proton ratios, but the decay rates for the two processes can be very different.

    Gamma Emission

    Many nuclear decay reactions produce daughter nuclei that are in a nuclear excited state, which is similar to an atom in which an electron has been excited to a higher-energy orbital to give an electronic excited state. Just as an electron in an electronic excited state emits energy in the form of a photon when it returns to the ground state, a nucleus in an excited state releases energy in the form of a photon when it returns to the ground state. (For more information on electron ground states, see Chapter 2 ) These high-energy photons are γ rays. Gamma (γ) emissionA nuclear decay reaction that results when a nucleus in an excited state releases energy in the form of a high-energy photon (a γ ray) when it returns to the ground state. can occur virtually instantaneously, as it does in the alpha decay of uranium-238 to thorium-234, where the asterisk denotes an excited state:

    \( _{92}^{238}U \rightarrow   \;\;\underset{excited \;state}{_{90}^{234}Th^{*}} + _{2}^{4}\alpha \overset{relaxation}{\longrightarrow} _{90}^{234}Th + _{0}^{0}\gamma\tag{24.2.13} \) 

    If we disregard the decay event that created the excited nucleus, then

    \( _{90}^{234}Th^{*} \rightarrow \; _{90}^{234}Th + _{0}^{0}\gamma\tag{24.2.14} \) 

    or more generally,

    \( _{Z}^{A}X^{*} \rightarrow \; _{Z}^{A}X + _{0}^{0}\gamma\tag{24.2.15} \)

    Gamma emission can also occur after a significant delay. For example, technetium-99m has a half-life of about 6 hours before emitting a γ ray to form technetium-99. (The m is for metastable, which is explained in Section 14.5.)

    Because γ rays are energy, their emission does not affect either the mass number or the atomic number of the daughter nuclide. Gamma-ray emission is therefore the only kind of radiation that does not necessarily involve the conversion of one element to another, although it is almost always observed in conjunction with some other nuclear decay reaction.

    Spontaneous Fission

    Only very massive nuclei with high neutron-to-proton ratios can undergo spontaneous fissionA nuclear decay reaction in which the nucleus breaks into two pieces with different atomic numbers and atomic masses., in which the nucleus breaks into two pieces that have different atomic numbers and atomic masses. This process is most important for the transactinide elements, with Z ≥ 104. Spontaneous fission is invariably accompanied by the release of large amounts of energy, and it is usually accompanied by the emission of several neutrons as well. An example is the spontaneous fission of \(_{98}^{254}Cf \) which gives a distribution of fission products; one possible set of products is shown in the following equation:

    \( _{98}^{254}Cf \rightarrow \; _{46}^{118}Pd + _{52}^{132}Te + 4_{0}^{1}n \tag{24.2.16} \) 

    Once again, the number of nucleons is conserved. Thus the sum of the mass numbers of the products (118 + 132 + 4 = 254) equals the mass number of the reactant. Similarly, the sum of the atomic numbers of the products [46 + 52 + (4 × 0) = 98] is the same as the atomic number of the parent nuclide.

    Example 24.2.1

    Write a balanced nuclear equation to describe each reaction.

    1. the beta decay of \( _{16}^{35}S \)
    2. the decay of \( _{80}^{201}Hg \)  by electron capture
    3. the decay of  \( _{15}^{30}P \) by positron emission 

    Given: radioactive nuclide and mode of decay

    Asked for: balanced nuclear equation

    Strategy:

    A Identify the reactants and the products from the information given.

    B Use the values of A and Z to identify any missing components needed to balance the equation.

    Solution:

    A We know the identities of the reactant and one of the products (a β particle). We can therefore begin by writing an equation that shows the reactant and one of the products and indicates the unknown product as  \( _{Z}^{A}X \)

    \( _{16}^{35}S \rightarrow \; _{Z}^{A}X + _{\;\;0}^{-1}\beta \) 

    B Because both protons and neutrons must be conserved in a nuclear reaction, the unknown product must have a mass number of A = 35 − 0 = 35 and an atomic number of Z = 16 − (−1) = 17. The element with Z = 17 is chlorine, so the balanced nuclear equation is as follows:

    \( _{16}^{35}S \rightarrow \; _{16}^{35}Cl + _{\;\;0}^{-1}\beta \) 

    A We know the identities of both reactants: \( _{80}^{201}Hg \).  The reaction is as follows:

    \( _{80}^{201}Hg + _{\;\;0}^{-1}e \rightarrow \; _{Z}^{A}X  \)

    B Both protons and neutrons are conserved, so the mass number of the product must be A = 201 + 0 = 201, and the atomic number of the product must be Z = 80 + (−1) = 79, which corresponds to the element gold. The balanced nuclear equation is thus

    \( _{80}^{201}Hg + _{\;\;0}^{-1}e \rightarrow \; _{79}^{201}Au  \)

    1. A As in part (a), we are given the identities of the reactant and one of the products—in this case, a positron. The unbalanced nuclear equation is therefore

      \( _{15}^{30}P +\rightarrow \; _{Z}^{A}X +_{+1}^{\;\;0}\beta \)

      B The mass number of the second product is A = 30 − 0 = 30, and its atomic number is Z = 15 − 1 = 14, which corresponds to silicon. The balanced nuclear equation for the reaction is as follows:

      \( _{15}^{30}P +\rightarrow \; _{14}^{30}Si + _{+1}^{\;\;0}\beta  \)

    Exercise

    Write a balanced nuclear equation to describe each reaction.

    1. \( _{6}^{11}C \) by positron emission
    2. the beta decay of molybdenum-99
    3. the emission of an α particle followed by gamma emission from
    4. \( _{74}^{185}C \)

    Answer:

    1. \( _{6}^{11}C \rightarrow \; _{5}^{11}B + _{+1}^{\;\;0}\beta  \)
    2. \( _{42}^{99}C \rightarrow \; _{43}^{90m}Tc + _{-1}^{\;\;0}\beta  \) 
    3. \( _{74}^{185}W \rightarrow \; _{72}^{181}Hf + _{0}^{0}\gamma \beta + _{2}^{4}\alpha \) 

    Example 24.2.2

    Predict the kind of nuclear change each unstable nuclide undergoes when it decays.

    1. \( _{22}^{45}Ti \)
    2. \( _{94}^{292}Pu \)
    3. \( _{5}^{12}B \)
    4. \( _{100}^{256}Fm \)

    Given: nuclide

    Asked for: type of nuclear decay

    Strategy:

    Based on the neutron-to-proton ratio and the value of Z, predict the type of nuclear decay reaction that will produce a more stable nuclide.

    Solution:

    1. This nuclide has a neutron-to-proton ratio of only 1.05, which is much less than the requirement for stability for an element with an atomic number in this range. Nuclei that have low neutron-to-proton ratios decay by converting a proton to a neutron. The two possibilities are positron emission, which converts a proton to a neutron and a positron, and electron capture, which converts a proton and a core electron to a neutron. In this case, both are observed, with positron emission occurring about 86% of the time and electron capture about 14% of the time.
    2. Nuclei with Z > 83 are too heavy to be stable and usually undergo alpha decay, which decreases both the mass number and the atomic number. Thus \( _{94}^{292}Pu \) is expected to decay by alpha emission.
    3. This nuclide has a neutron-to-proton ratio of 1.4, which is very high for a light element. Nuclei with high neutron-to-proton ratios decay by converting a neutron to a proton and an electron. The electron is emitted as a β particle, and the proton remains in the nucleus, causing an increase in the atomic number with no change in the mass number. We therefore predict that \( _{5}^{12}B \) will undergo beta decay.
    4. This is a massive nuclide, with an atomic number of 100 and a mass number much greater than 200. Nuclides with A ≥ 200 tend to decay by alpha emission, and even heavier nuclei tend to undergo spontaneous fission. We therefore predict that \( _{100}^{256}Fm \) will decay by either or both of these two processes. In fact, it decays by both spontaneous fission and alpha emission, in a 97:3 ratio.

    Exercise

    Predict the kind of nuclear change each unstable nuclide undergoes when it decays.

    1. \( _{14}^{32}Si \)
    2. \( _{21}^{43}Sc \)
    3. \( _{91}^{231}Pa \)

    Answer:

    1. beta decay
    2. positron emission or electron capture
    3. alpha decay

    Radioactive Decay Series

    The nuclei of all elements with atomic numbers greater than 83 are unstable. Thus all isotopes of all elements beyond bismuth in the periodic table are radioactive. Because alpha decay decreases Z by only 2, and positron emission or electron capture decreases Z by only 1, it is impossible for any nuclide with Z > 85 to decay to a stable daughter nuclide in a single step, except via nuclear fission. Consequently, radioactive isotopes with Z > 85 usually decay to a daughter nucleus that is radiaoctive, which in turn decays to a second radioactive daughter nucleus, and so forth, until a stable nucleus finally results. This series of sequential alpha- and beta-decay reactions is called a radioactive decay seriesA series of sequential alpha- and beta-decay reactions that occur until a stable nucleus is finally obtained.. The most common is the uranium-238 decay series, which produces lead-206 in a series of 14 sequential alpha- and beta-decay reactions (Figure 24.2.2). Although a radioactive decay series can be written for almost any isotope with Z > 85, only two others occur naturally: the decay of uranium-235 to lead-207 (in 11 steps) and thorium-232 to lead-208 (in 10 steps). A fourth series, the decay of neptunium-237 to bismuth-209 in 11 steps, is known to have occurred on the primitive Earth. With a half-life of “only” 2.14 million years, all the neptunium-237 present when Earth was formed decayed long ago, and today all the neptunium on Earth is synthetic.

     

    60346894cc83fb1ed443a5e618c94bd5.jpg

    Figure 24.2.2 A Radioactive Decay Series Three naturally occurring radioactive decay series are known to occur currently: the uranium-238 decay series, the decay of uranium-235 to lead-207, and the decay of thorium-232 to lead-208.

    Due to these radioactive decay series, small amounts of very unstable isotopes are found in ores that contain uranium or thorium. These rare, unstable isotopes should have decayed long ago to stable nuclei with a lower atomic number, and they would no longer be found on Earth. Because they are generated continuously by the decay of uranium or thorium, however, their amounts have reached a steady state, in which their rate of formation is equal to their rate of decay. In some cases, the abundance of the daughter isotopes can be used to date a material or identify its origin, as described in Chapter 14.

    Induced Nuclear Reactions

    The discovery of radioactivity in the late 19th century showed that some nuclei spontaneously transform into nuclei with a different number of protons, thereby producing a different element. When scientists realized that these naturally occurring radioactive isotopes decayed by emitting subatomic particles, they realized that—in principle—it should be possible to carry out the reverse reaction, converting a stable nucleus to another more massive nucleus by bombarding it with subatomic particles in a nuclear transmutation reaction.

    The first successful nuclear transmutation reaction was carried out in 1919 by Ernest Rutherford, who showed that α particles emitted by radium could react with nitrogen nuclei to form oxygen nuclei. As shown in the following equation, a proton is emitted in the process:

    \( _{2}^{4}\alpha  + _{7}^{14}N \rightarrow _{8}^{17}O +\;   _{1}^{1}p \tag{24.2.17} \)

    Rutherford’s nuclear transmutation experiments led to the discovery of the neutron. He found that bombarding the nucleus of a light target element with an α particle usually converted the target nucleus to a product that had an atomic number higher by 1 and a mass number higher by 3 than the target nucleus. Such behavior is consistent with the emission of a proton after reaction with the α particle. Very light targets such as Li, Be, and B reacted differently, however, emitting a new kind of highly penetrating radiation rather than a proton. Because neither a magnetic field nor an electrical field could deflect these high-energy particles, Rutherford concluded that they were electrically neutral. (For more information on high-energy particles, see Chapter 1.) Other observations suggested that the mass of the neutral particle was similar to the mass of the proton. In 1932, James Chadwick (Nobel Prize in Physics, 1935), who was a student of Rutherford’s at the time, named these neutral particles neutrons and proposed that they were fundamental building blocks of the atom. The reaction that Chadwick initially used to explain the production of neutrons was as follows:

    \( _{2}^{4}\alpha  + _{4}^{9}Be \rightarrow _{6}^{12}C +\;   _{0}^{1}n \tag{24.2.18} \)

    Because α particles and atomic nuclei are both positively charged, electrostatic forces cause them to repel each other. Only α particles with very high kinetic energy can overcome this repulsion and collide with a nucleus (Figure 24.2.3). Neutrons have no electrical charge, however, so they are not repelled by the nucleus. Hence bombardment with neutrons is a much easier way to prepare new isotopes of the lighter elements. In fact, carbon-14 is formed naturally in the atmosphere by bombarding nitrogen-14 with neutrons generated by cosmic rays:

    \( _{0}^{1}n  + _{7}^{14}N \rightarrow _{6}^{14}C +\;   _{1}^{1}p \tag{24.2.19} \)

     

    3a6a199f5a8d7c46c559e04d47c81294.jpg

    Figure 24.2.3 A Nuclear Transmutation Reaction Bombarding a target of one element with high-energy nuclei or subatomic particles can create new elements. Electrostatic repulsions normally prevent a positively charged particle from colliding and reacting with a positively charged nucleus. If the positively charged particle is moving at a very high speed, however, its kinetic energy may be great enough to overcome the electrostatic repulsions, and it may collide with the target nucleus. Such collisions can result in a nuclear transmutation reaction.

    Example 24.2.3

    In 1933, Frédéric Joliot and Iréne Joliot-Curie (daughter of Marie and Pierre Curie) prepared the first artificial radioactive isotope by bombarding aluminum-27 with α particles. For each 27Al that reacted, one neutron was released. Identify the product nuclide and write a balanced nuclear equation for this transmutation reaction.

    Given: reactants in a nuclear transmutation reaction

    Asked for: product nuclide and balanced nuclear equation

    Strategy:

    A Based on the reactants and one product, identify the other product of the reaction. Use conservation of mass and charge to determine the values of Z and A of the product nuclide and thus its identity.

    B Write the balanced nuclear equation for the reaction.

    Solution:

    A Bombarding an element with α particles usually produces an element with an atomic number that is 2 greater than the atomic number of the target nucleus. Thus we expect that aluminum (Z = 13) will be converted to phosphorus (Z = 15). With one neutron released, conservation of mass requires that the mass number of the other product be 3 greater than the mass number of the target. In this case, the mass number of the target is 27, so the mass number of the product will be 30. The second product is therefore phosphorus-30, \( _{15}^{30}P \)

    B The balanced nuclear equation for the reaction is as follows:

    \( _{2}^{4}\alpha + _{13}^{27}Al \rightarrow _{15}^{30}P +\;   _{0}^{1}n  \)

    Exercise

    Because all isotopes of technetium are radioactive and have short half-lives, it does not exist in nature. Technetium can, however, be prepared by nuclear transmutation reactions. For example, bombarding a molybdenum-96 target with deuterium nuclei \( _{1}^{2}H \) produces technetium-97. Identify the other product of the reaction and write a balanced nuclear equation for this transmutation reaction.

    Answer: neutron, \( _{0}^{1}n ;  _{42}^{96}Mo + _{1}^{2}H \rightarrow _{43}^{97}Tc +\;   _{0}^{1}n  \)

    We noted earlier in this section that very heavy nuclides, corresponding to Z ≥ 104, tend to decay by spontaneous fission. Nuclides with slightly lower values of Z, such as the isotopes of uranium (Z = 92) and plutonium (Z = 94), do not undergo spontaneous fission at any significant rate. Some isotopes of these elements, however, such as \(  _{92}^{235}U \) and \(  _{94}^{239}Pu​ \) undergo induced nuclear fission when they are bombarded with relatively low-energy neutrons, as shown in the following equation for uranium-235 and in Figure 24.2.4:

    \( _{0}^{1}n +  _{92}^{235}U  \rightarrow _{92}^{236}U \rightarrow _{56}^{141}Ba + \;_{36}^{92}Kr +\;   3_{0}^{1}n  \tag{24.2.20} \)

     

    76a0a494f334a195bab06106babc1e90.jpg

    Figure 24.2.4 Neutron-Induced Nuclear FissionCollision of a relatively slow-moving neutron with a fissile nucleus can split it into two smaller nuclei with the same or different masses. Neutrons are also released in the process, along with a great deal of energy.

    Any isotope that can undergo a nuclear fission reaction when bombarded with neutrons is called a fissile isotope.

    During nuclear fission, the nucleus usually divides asymmetrically rather than into two equal parts, as shown in Figure 24.2.4. Moreover, every fission event of a given nuclide does not give the same products; more than 50 different fission modes have been identified for uranium-235, for example. Consequently, nuclear fission of a fissile nuclide can never be described by a single equation. Instead, as shown in Figure 24.2.5 , a distribution of many pairs of fission products with different yields is obtained, but the mass ratio of each pair of fission products produced by a single fission event is always roughly 3:2.

     

    ca0d2e2d5a1a9a6f34ace942efcc354f.jpg

    Figure 24.2.5 Mass Distribution of Nuclear Fission Products of 235UNuclear fission usually produces a range of products with different masses and yields, although the mass ratio of each pair of fission products from a fission event is approximately 3:2. As shown in this plot, more than 50 different fission products are known for 235U.

    Synthesis of Transuranium Elements

    Uranium (Z = 92) is the heaviest naturally occurring element. Consequently, all the elements with Z > 92, the transuranium elementsAn artificial element that has been prepared by bombarding suitable target nuclei with smaller particles., are artificial and have been prepared by bombarding suitable target nuclei with smaller particles. The first of the transuranium elements to be prepared was neptunium (Z = 93), which was synthesized in 1940 by bombarding a 238U target with neutrons. As shown in Equation 24.2.21, this reaction occurs in two steps. Initially, a neutron combines with a 238U nucleus to form 239U, which is unstable and undergoes beta decay to produce 239Np:

    \( _{0}^{1}n +  _{92}^{238}U  \rightarrow _{92}^{239}U \rightarrow _{93}^{239}Np + _{-1}^{\;\;0}\beta \tag{24.2.21} \)

    Subsequent beta decay of 239Np produces the second transuranium element, plutonium (Z = 94):

    \(  _{93}^{239}Np  \rightarrow _{94}^{239}Pu  + _{-1}^{\;\;0}\beta \tag{24.2.22} \)

    Bombarding the target with more massive nuclei creates elements that have atomic numbers significantly greater than that of the target nucleus (Table 20.2.2). Such techniques have resulted in the creation of the superheavy elements 114 and 116, both of which lie in or near the “island of stability” discussed in Section 24.1 .

    Table 20.2.2 Some Reactions Used to Synthesize Transuranium Elements

    \(  _{94}^{239}Pu + _{2}^{4}\alpha   \rightarrow _{96}^{242}Cm  + _{0}^{1}n \)
    \(  _{94}^{239}Pu + _{2}^{4}\alpha   \rightarrow _{95}^{241}Am  + _{1}^{1}p+ _{0}^{1}n \)
    \(  _{96}^{242}Cm + _{2}^{4}\alpha   \rightarrow _{97}^{243}Bk  + _{1}^{1}p+ 2_{0}^{1}n \)
    \(  _{99}^{253}Es + _{2}^{4}\alpha   \rightarrow _{101}^{256}Md  + _{0}^{1}n \)
    \(  _{92}^{238}U + _{6}^{12}C   \rightarrow _{98}^{246}Cf  + 4_{0}^{1}n \)
    \(  _{98}^{252}Cf + _{5}^{10}B   \rightarrow _{103}^{256}Lr  + 6_{0}^{1}n \)

    A device called a particle accelerator is used to accelerate positively charged particles to the speeds needed to overcome the electrostatic repulsions between them and the target nuclei by using electrical and magnetic fields. Operationally, the simplest particle accelerator is the linear accelerator (Figure 24.2.6), in which a beam of particles is injected at one end of a long evacuated tube. Rapid alternation of the polarity of the electrodes along the tube causes the particles to be alternately accelerated toward a region of opposite charge and repelled by a region with the same charge, resulting in a tremendous acceleration as the particle travels down the tube. A modern linear accelerator such as the Stanford Linear Accelerator (SLAC) at Stanford University is about 2 miles long.

     

    eb701a3e8ec9a55cbdccc63ffef735bf.jpg

    Figure 24.2.6 A Linear Particle Accelerator (a) An aerial view of the SLAC, the longest linear particle accelerator in the world; the overall length of the tunnel is 2 miles. (b) Rapidly reversing the polarity of the electrodes in the tube causes the charged particles to be alternately attracted as they enter one section of the tube and repelled as they leave that section. As a result, the particles are continuously accelerated along the length of the tube.

    To achieve the same outcome in less space, a particle accelerator called a cyclotron forces the charged particles to travel in a circular path rather than a linear one. The particles are injected into the center of a ring and accelerated by rapidly alternating the polarity of two large D-shaped electrodes above and below the ring, which accelerates the particles outward along a spiral path toward the target.

    The length of a linear accelerator and the size of the D-shaped electrodes in a cyclotron severely limit the kinetic energy that particles can attain in these devices. These limitations can be overcome by using a synchrotron, a hybrid of the two designs. A synchrotron contains an evacuated tube similar to that of a linear accelerator, but the tube is circular and can be more than a mile in diameter (Figure 24.2.7). Charged particles are accelerated around the circle by a series of magnets whose polarities rapidly alternate.

     

    bb21eee49de9b25c6e5c06a48951787b.jpg

    Figure 24.2.7 A SynchrotronAn aerial photograph of what is currently the world’s most powerful particle accelerator, the Tevatron at the Fermi National Accelerator Laboratory (Fermilab) in Illinois. The large tube characteristic of a synchrotron is 4 miles in circumference, contains 1000 superconducting magnets cooled by liquid helium, and can accelerate a beam of protons to almost the speed of light, giving them an energy greater than 1 TeV (teraelectronvolt = 103 GeV = 1012 eV) for collisions with other particles.

    Summary

    In nuclear decay reactions (or radioactive decay), the parent nucleus is converted to a more stable daughter nucleus. Nuclei with too many neutrons decay by converting a neutron to a proton, whereas nuclei with too few neutrons decay by converting a proton to a neutron. Very heavy nuclei (with A ≥ 200 and Z > 83) are unstable and tend to decay by emitting an α particle. When an unstable nuclide undergoes radioactive decay, the total number of nucleons is conserved, as is the total positive charge. Six different kinds of nuclear decay reactions are known. Alpha decay results in the emission of an α particle, <math display="inline" xml:id="av_1.0-20_m109"><semantics><mrow><mmultiscripts/></mrow></semantics></math><mi>α</mi><mprescripts/><mn>2</mn><mn>4</mn><mtext>,</mtext> and produces a daughter nucleus with a mass number that is lower by 4 and an atomic number that is lower by 2 than the parent nucleus. Beta decay converts a neutron to a proton and emits a high-energy electron, producing a daughter nucleus with the same mass number as the parent and an atomic number that is higher by 1. Positron emission is the opposite of beta decay and converts a proton to a neutron plus a positron. Positron emission does not change the mass number of the nucleus, but the atomic number of the daughter nucleus is lower by 1 than the parent. In electron capture (EC), an electron in an inner shell reacts with a proton to produce a neutron, with emission of an x-ray. The mass number does not change, but the atomic number of the daughter is lower by 1 than the parent. In gamma emission, a daughter nucleus in a nuclear excited state undergoes a transition to a lower-energy state by emitting a γ ray. Very heavy nuclei with high neutron-to-proton ratios can undergo spontaneous fission, in which the nucleus breaks into two pieces that can have different atomic numbers and atomic masses with the release of neutrons. Many very heavy nuclei decay via a radioactive decay series—a succession of some combination of alpha- and beta-decay reactions. In nuclear transmutation reactions, a target nucleus is bombarded with energetic subatomic particles to give a product nucleus that is more massive than the original. All transuranium elements—elements with Z > 92—are artificial and must be prepared by nuclear transmutation reactions. These reactions are carried out in particle accelerators such as linear accelerators, cyclotrons, and synchrotrons.

    Key Takeaway

    • Nuclear decay reactions occur spontaneously under all conditions and produce more stable daughter nuclei, whereas nuclear transmutation reactions are induced and form a product nucleus that is more massive than the starting material.

    Key Equations

    alpha decay

    Equation 24.2.1:  \( _{Z}^{A}X \rightarrow _{Z-2}^{A-4}X'  + _{2}^{4}\alpha \)

    beta decay

    Equation 24.2.4: \( _{Z}^{A}X \rightarrow _{Z+1}^{A}X'  + _{-1}^{\;\;0}\beta \)

    positron emission

    Equation 24.2.7: \( _{Z}^{A}X \rightarrow _{Z-1}^{A}X'  + _{+1}^{\;\;0}\beta \)

    electron capture

    Equation 24.2.100: \( _{Z}^{A}X + _{-1}^{\;\;0}e \rightarrow _{Z-1}^{A}X'  + x-ray \)

    gamma emission

    Equation 24.2.15: \( _{Z}^{A}X^{*} \rightarrow _{Z}^{A}X'  + _{0}^{0}\gamma \)

    Conceptual Problems

    1. Describe the six classifications of nuclear decay reactions. What is the most common mode of decay for elements that have heavy nuclei? Why?

    2. Complete the following table for these five nuclear reactions.

        Alpha Decay Beta Decay Gamma Emission Positron Emission Electron Capture
      identity of particle or radiation helium-4 nucleus        
      mass number of parent − mass number of daughter 4        
      atomic number of parent − atomic number of daughter   −1      
      effect on neutron-to-proton ratio   decreases      
    3. What is the most common decay process for elements in row 5 of the periodic table that contain too few neutrons for the number of protons present? Why?

    4. Explain the difference between the symbols e and β. What is the difference in meaning between the symbols \( _{2}^{4}He \) and \( _{2}^{4}\alpha \)

    5. What is a mass number? Which particles have a mass number of zero?

    6. What are the key differences between the equations written for chemical reactions and for nuclear reactions? How are they similar?

    7. Can all the kinds of nuclear decay reactions discussed be characterized by the general equation: parent → daughter + particle? Explain your answer.

    8. Which types of nuclear decay reactions conserve both mass number and atomic number? In which do the parent and daughter nuclei have the same mass number but different atomic numbers? Which do not convert one element to another?

    9. Describe a radioactive decay series. How many series occur naturally? Of these, which one no longer occurs in nature? Why?

    10. Only nine naturally occurring elements have an atomic number greater than 83, and all of them are radioactive. Except for some isotopes of uranium that have a very long half-life, the half-lives of the heavy elements are so short that these elements should have been completely converted to lighter, more stable elements long ago. Why are these elements still present in nature?

    11. Why are neutrons preferred to protons when preparing new isotopes of the lighter elements?

    12. Why are particle accelerators and cyclotrons needed to create the transuranium elements?

    Answers

    1.  
    2.  
    3. Both positron decay and electron capture increase the neutron-to-proton ratio; electron capture is more common for heavier elements such those of row 5.

    4.  
    5. The mass number is the sum of the numbers of protons and neutrons present. Particles with a mass number of zero include β particles (electrons) and positrons; gamma rays and x-rays also have a mass number of zero.

    6.  
    7.  
    8.  
    9.  
    10.  
    11. Unlike protons, neutrons have no charge, which minimizes the electrostatic barrier to colliding and reacting with a positively charged nucleus.

    12.  

    Numerical Problems

    1. What type of particle is emitted in each nuclear reaction?

      1. 238Pu → 234U
      2. 32Si → 32P
      3. 18F → 18O
      4. 206Tl → 206Pb
    2. What type of particle is emitted in each nuclear reaction?

      1. 230Th → 226Ra
      2. 224Rn → 224Fr
      3. 210Bi → 206Tl
      4. 36Cl → 36S
    3. Predict the mode of decay and write a balanced nuclear reaction for each isotope.

      1. 235U
      2. 254Es
      3. 36S
      4. 99Mo
    4. Predict the mode of decay and write a balanced nuclear reaction for each isotope.

      1. 13N
      2. 231Pa
      3. 7Be
      4. 77Ge
    5. Balance each nuclear reaction.

      1. 208Po → α + Pb
      2. 226Ra → α + Rn
      3. 228Th → Ra + α + γ
      4. 231Pa → Ac + α + γ
      5. Ho → 166Er + β + γ
      6. Ac → 226Th + β + γ
    6. Complete each nuclear reaction.

      1. \( _{84}^{210}Po \rightarrow ^{206}Pb \)
      2. \( _{85}^{217}At \rightarrow Bi + \alpha  \)
      3. \( Ra \rightarrow _{86}^{220}Rn + \alpha   \)
      4. \( ^{208}Tl \rightarrow _{82}Pb + \beta^{-}  \)
      5. \( Np \rightarrow ^{239}Pu + \beta^{-}  \)
      6. \( Fe \rightarrow ^{52}Mn + \beta^{+} + \gamma  \)
    7. Write a balanced nuclear equation for each reaction.

      1. β decay of 87Rb
      2. β+ decay of 20Mg
      3. α decay of 268Mt
    8. Write a balanced nuclear equation for each reaction.

      1. β decay of 45K
      2. β+ decay of 41Sc
      3. α decay of 146Sm
    9. The decay products of several isotopes are listed here. Identify the type of radiation emitted and write a balanced nuclear equation for each.

      1. 218Po → 214Pb
      2. 32Si → 32P
      3. an excited state of an iron-57 nucleus decaying to its ground state
      4. conversion of thallium-204 to lead-204
    10. The decay products of several isotopes are listed here. Identify the type of radiation emitted and write a balanced nuclear equation for each.

      1. 218Po → 218At
      2. 216Po → 212Pb
      3. bismuth-211 converted to thallium-207
      4. americium-242 converted to rhodium-107 with the emission of four neutrons
    11. Predict the most likely mode of decay and write a balanced nuclear reaction for each isotope.

      1. 238U
      2. 208Po
      3. 40S
      4. molybdenum-93m
    12. Predict the most likely mode of decay and write a balanced nuclear reaction for each isotope.

      1. 16N
      2. 224Th
      3. 118In
      4. 64Ge
    13. For each nuclear reaction, identify the type(s) of decay and write a balanced nuclear equation.

      1. 216Po → ? + At
      2. ? → α + 231Pa
      3. 228Th → ? + α + γ
      4. 231Pa → ? + β + γ
    14. For each nuclear reaction, identify the type(s) of decay and write a balanced nuclear equation.

      1. 212Po → 208Pb + ?
      2. 192Ir → Pt + ?
      3. 241Am → 57Fe + 184? + ?
      4. Ge → 77Ge + ?
    15. Identify the parent isotope and write a balanced nuclear reaction for each process.

      1. Lead-205 is formed via an alpha emission.
      2. Titanium-46 is formed via beta and gamma emission.
      3. Argon-36 is formed via a beta decay and a gamma emission.
    16. Identify the parent isotope and write a balanced nuclear reaction for each process.

      1. Iodine-130 is formed by ejecting an electron and a gamma ray from a nucleus.
      2. Uranium-240 is formed by alpha decay.
      3. Curium-247 is formed by releasing a helium dication and a gamma ray.
    17. Write a balanced nuclear equation for each process.

      1. Bromine undergoes a decay and produces a gas with an atomic mass of 80 amu.
      2. An element emits two neutrons while decaying into two metals, each of which can be extracted and converted to chlorides with the formula MCl2. The masses of the two salts are 162.9 and 210.9 g/mol, respectively.
    18. Write a balanced nuclear equation for each process.

      1. An unknown element emits γ rays plus particles that are readily blocked by paper. The sample also contains a substantial quantity of tin-104.
      2. An unstable element undergoes two different decay reactions: beta decay to produce a material with a mass of 222 amu and alpha decay to astatine.
    19. Bombarding 249Cf with 12C produced a transuranium element with a mass of 257 amu, plus several neutral subatomic particles. Identify the element and write a nuclear reaction for this transmutation.

    20. One transuranium element, 253Es, is prepared by bombarding 238U with 15 neutrons. What is the other product of this reaction? Write a balanced transmutation reaction for this conversion.

    21. Complete this radioactive decay series:

      \( _{88}^{223}Rn \overset{\alpha }{\rightarrow} Rn \overset{\alpha }{\rightarrow} Po \overset{\alpha }{\rightarrow} Pb \overset{\beta }{\rightarrow} Bi \overset{\alpha }{\rightarrow} Tl \overset{\beta  }{\rightarrow} Pb \)

    22. Complete each nuclear fission reaction.

      1. \( _{92}^{235}U + _{0}^{1}n \rightarrow _{36}^{90}Kr + ? +2_{0}^{1}n  \)
      2. \( ? + _{0}^{1}n \rightarrow ^{140}Cs + ^{96}Y + 4_{0}^{1}n  \)
    23. Complete each nuclear fission reaction.

      1. \( _{92}^{235}U + _{0}^{1}n \rightarrow _{57}^{145}La + ? +3_{0}^{1}n \)
      2. \( ? + _{0}^{1}n \rightarrow _{42}^{95}Mo + _{57}^{145}La + 2_{0}^{1}n + 7_{-1}^{\;\;0}\beta \)
    24. A stable nuclide absorbs a neutron, emits an electron, and then splits into two α particles. Identify the nuclide.

    25. Using 18O, how would you synthesize an element with atomic number 106 from 249Cf? Write a balanced nuclear equation for the reaction.

    26. Using 10B and 252Cf, how would you synthesize an element with atomic number 103? Write a balanced nuclear equation for the reaction.

    Answers

    1.  
    2.  
      1. α decay; \( _{92}^{235}U \rightarrow _{2}^{4}\alpha + _{90}^{231}Th \)
      2. α decay; \( _{99}^{254}Es \rightarrow _{2}^{4}\alpha + _{97}^{250}Bk \)
      3. β decay; \( _{16}^{36}S \rightarrow _{-1}^{\;\;0}\beta + _{17}^{36}Cl \)
      4. β decay; \( _{42}^{99}Mo \rightarrow _{-1}^{\;\;0}\beta + _{43}^{99m}Tc \)
    3.  
      1. \( _{84}^{208}Po \rightarrow _{2}^{4}\alpha + _{82}^{204}Pb \)
      2. \( _{88}^{226}Ra \rightarrow _{2}^{4}\alpha + _{86}^{222}Rn \)
      3. \( _{90}^{228}Th \rightarrow _{2}^{4}\alpha + \gamma + _{88}^{224}Ra \)
      4. \( _{91}^{231}Pa \rightarrow _{2}^{4}\alpha + \gamma + _{89}^{227}Ac \)
      5. \( _{67}^{166}Ho \rightarrow _{-1}^{\;\;0}\beta  + \gamma + _{68}^{166}Er \)
      6. \( _{89}^{226}Ac \rightarrow _{-1}^{\;\;0}\beta  + \gamma + _{90}^{226}Th \)
    4.  
      1. \( _{37}^{87}Rb \rightarrow _{-1}^{\;\;0}\beta + _{38}^{87}Sr \)
      2. \( _{12}^{20}Mg \rightarrow _{+1}^{0}\beta  + _{11}^{20}Na \)
      3. \( _{109}^{268}Mt \rightarrow _{2}^{4}\alpha + _{107}^{264}Bh \)
    5.  
      1. α particle; \( _{84}^{218}Po \rightarrow _{2}^{4}\alpha + _{82}^{214}Pb \)
      2. β particle; \( _{14}^{32}Si \rightarrow _{-1}^{\;\;0}\beta + _{15}^{32}P \)
      3. γ ray; \( _{26}^{57m}Fe \rightarrow \gamma + _{26}^{57}Fe \)
      4. β particle; \( _{81}^{204}Tl \rightarrow _{-1}^{\;\;0}\beta + _{82}^{204}Pb \)
    6.  
      1. α emission; \( _{92}^{238}U \rightarrow _{2}^{4}\alpha + _{90}^{234}Th \)
      2. α emission; \( _{84}^{208}Po \rightarrow _{2}^{4}\alpha + _{82}^{204}Pb \)
      3. β emission; \( _{16}^{40}S \rightarrow _{-1}^{\;\;0}\beta + _{17}^{40}Cl \)
      4. γ emission; \( _{42}^{93m}Mo \rightarrow \gamma + _{42}^{93}Mo \)
    7.  
      1. β decay; \( _{84}^{216}Po \rightarrow _{-1}^{\;\;0}\beta +  _{85}^{216}At  \)
      2. α decay; \( _{93}^{235}Np \rightarrow _{2}^{4}\alpha + _{91}^{231}Pa \)
      3. α decay; γ emission;\( _{90}^{228}Th \rightarrow _{2}^{4}\alpha + \gamma + _{88}^{224}Ra \)
      4. β decay, γ emission; \( _{91}^{231}Pa \rightarrow _{-1}^{\;\;0}\beta + \gamma +  _{92}^{231}U  \)
    8.  
    9.  
    10.  
      1. \( _{35}^{80}Br  \rightarrow _{-1}^{\;\;0}\beta  + _{36}^{80}Kr \)
      2. \( _{94}^{234}Pu  \rightarrow 2_{0}^{1}n + _{56}^{140}Ba + _{38}^{92}Sr \)>
    11.  
    12. \( _{98}^{249}Cf +  _{6}^{12}C \rightarrow 4_{0}^{1}n + _{104}^{257}Rf \)
    13.  
    14.  
    15.  
      1. \( _{92}^{235}U + _{0}^{1}n  \rightarrow 3_{0}^{1}n + _{57}^{145}La + _{35}^{88}Br \)
      2. \( _{92}^{235}U + _{0}^{1}n  \rightarrow 2_{0}^{1}n + 7_{-1}^{\;\;0}\beta  + _{43}^{95}Mo + _{57}^{139}La \)
    16.  
    17.  
    18.  

    Contributors

    • Anonymous

    Modified by Joshua B. Halpern