Skip to main content

# 9.2: Radioactive Decay

Learning Objectives
• Define radioactive decay.
• Classify a radioactive decay as a combination or a decomposition reaction.
• Identify the indicator word associated with radioactive decay reactions.
• Define radiation.
• Name the four types of radiation that are generated during radioactive decay reactions.
• Write the Greek and nuclear symbols that are used to represent each of the four types of radiation.
• Define metastable.
• Define antimatter.
• Define daughter nucleus.
• Write a balanced nuclear equation that represents the radioactive decay of a radioisotope.

As stated in the previous section of this chapter, a radioisotope is defined as a nucleus that contains too many neutrons and, therefore, is highly unstable.  In order to generate a more stable daughter nucleus, a radioisotope must emit, or release, radiation through a process called radioactive decay.  The following paragraphs will present the process for developing a nuclear equation pattern that symbolically-represents this type of reaction and, subsequently, modifying the pattern that is established to reflect the identities of the specific nuclei that are involved in a particular transformation.

In order to indicate that an irreversible chemical change has occurred, a "forward," or left-to-right, reaction arrow is written in the nuclear equation pattern that is being developed, as shown below.  Additionally, since a nuclear equation is the symbolic representation of a nuclear reaction, the nuclear symbols, not the names, of the particles that are reacted and produced are incorporated into this pattern.  Based on the definition of radioactive decay that is provided above, this nuclear process can be classified as a decomposition reaction, and, therefore, the nuclear symbol for a single reactant, the unstable radioisotope, is written on the left side of the reaction arrow.  Two products, radiation and a daughter nucleus, should be represented on the right side of the arrow, and a plus sign, "+", must be used to separate their corresponding nuclear symbols, which can be written in either order.  Finally, in order to indicate that the equation pattern that is shown below should be applied to symbolically-represent a radioactive decay reaction, the word "decay" must be included in the verbal description of a specific decay reaction.

Four types of radiation, which can be generically-defined as small particles or bursts of energy that are emitted, or released, from an unstable radioisotope, can be produced during a radioactive decay reaction.  Because an alphabetical designation had already been applied to name "X-ray" radiation, the types of radiation that are produced during nuclear reactions were named after and, therefore, symbolized using, the first three letters of the Greek alphabet.  Additionally, corresponding a nuclear symbol for each type of radiation, which could be incorporated into the nuclear equation pattern that is shown above, was also derived from the description of the specific particles that are contained in each type of emission.

• The first type of radiation that was discovered, a helium nucleus, was designated as an alpha particle, and, therefore, is symbolized using the Greek letter "α."  Because the most common isotope of helium, He, has an atomic number of 2 and a mass number of 4, an alpha particle can also be represented using the nuclear symbol "$$\ce{^{4}_{2}He}$$."
• A high-energy electron was the second type of radiation that was detected during the investigations of radioactive decay reactions, and, consequently, this type of emission was called a beta particle.  This particle is represented using the Greek letter "β" and the nuclear symbol "$$\ce{^{0}_{–1}e}$$."  Because an electron is a subatomic particle, not an element, this type of radiation does not have a corresponding atomic number or an associated mass number.  Therefore, the nuclear symbol that is indicated above was derived from the subatomic particle symbolism for an electron, "e."
• Gamma radiation, which is symbolized using the third letter of the Greek alphabet, "$$\gamma$$," is defined as a burst of pure energy that is released from a nucleus that has achieved a high-energy, metastable state.  Since energy is not a physical substance, gamma radiation contains neither protons nor neutrons and, consequently, is represented using the nuclear symbol "$$\ce{^{0}_{0}\gamma}$$."
• The final type of radiation is the antimatter of and, therefore, is defined as a substance that has opposing properties and electrical charges relative to, a beta particle.  As a result, the name and associated symbolisms for this type of radiation were not assigned to correspond to a new Greek letter and, instead, were derived from the definition of a beta particle that is stated above.  Since a beta particle is defined as a high-energy electron, which is a negatively-charged subatomic particle, its antimatter must be positively-charged and, therefore, was called a positron.  Furthermore, in order to denote this positive charge, a plus sign, "+", is written as a superscript in the corresponding Greek symbolism, "β+."  Finally, the signs of the superscripts and subscripts in the nuclear symbol for beta radiation, "$$\ce{^{0}_{–1}e}$$," were negated in order to generate the nuclear symbol for a positron, "$$\ce{^{0}_{1}e}$$."

The remaining product that is synthesized during a radioactive decay reaction is referred to as a daughter nucleus, which, by definition, must be more stable than the radioisotope that was initially present.  In order to identify the element that is generated, the nuclear equation that is being developed must be "balanced."  Recall that the Law of Conservation of Matter is a fundamental principle that mandates that particles cannot be created or destroyed in the course of a chemical change.  Consequently, while the elemental symbols of substances must change during a nuclear reaction, the relative quantity of protons and neutrons that are involved in the reaction must be constant.  As stated previously, the atomic number of an element is defined as the number of protons that are contained in an atom of that element, and the mass number of a particular elemental isotope is calculated by adding the quantities of protons and neutrons that are present in that atom.  Therefore, since the atomic number and mass number of an isotope are written as the subscript and superscript, respectively, in the nuclear symbol that is used to represent that isotope, a nuclear equation is "balanced" if the sums of the atomic numbers and the mass numbers on the reactant side of the arrow are equal to the corresponding totals on the product side of the equation.  As a result, the atomic number and mass number of the daughter nucleus must be calculated by subtracting the atomic numbers and mass numbers, respectively, of the radiation that is generated from the atomic numbers and mass numbers, respectively, of the radioisotope that was initially present.  Furthermore, since the atomic number of an element is a unique value that directly corresponds to the identity of that element, the calculated subscript should be used to determine which elemental symbol represents the daughter nucleus that is produced in a particular radioactive decay reaction.  Finally, because the Law of Conservation of Matter is upheld by calculating the atomic number and mass number of the daughter nucleusbalancing coefficients are rarely incorporated into nuclear equations.

For example, write a balanced nuclear equation that represents the positron decay of Sr-80.

Because the word "decay" is included in the given statement, the nuclear equation pattern that corresponds to a radioactive decay should be applied to symbolically-represent this reaction.  Since a radioactive decay is a decomposition reaction, a single reactant should be written on the left side of the reaction arrow, and two products, separated by a plus sign, "+", should be represented on the right side of the equation, as shown below.

_____ $$\rightarrow$$ _____ _____

The reactant in a radioactive decay is, by definition, an unstable radioisotope.  Since the statement indicates that the decay of Sr-80 should be represented in the equation that is being developed, the radioisotope in the corresponding reaction is a Sr-80 nucleus.  Since nuclear symbols must be incorporated into a nuclear equation, the symbolism "$$\ce{^{80}_{38}Sr}$$" is written on the left side of this pattern, as shown below.  The subscript in this symbolism, 38, is determined by locating "Sr" on the periodic table and recording the atomic number that corresponds to this element.

$$\ce{^{80}_{38}Sr}$$ $$\rightarrow$$ _____ _____

One of the products of a radioactive decay reaction is, by definition, classified as radiation.  Because the given statement indicates that Sr-80 undergoes positron decay, the nuclear symbol for a positron, "$$\ce{^{0}_{1}e}$$," is written on the right side of this equation, as shown below.

$$\ce{^{80}_{38}Sr}$$ $$\rightarrow$$ $$\ce{^{0}_{1}e}$$ _____

Finally, in order to identify the daughter nucleus that is generated, the nuclear equation that is written above must be "balanced."  Recall that the Law of Conservation of Matter mandates that particles cannot be created or destroyed in the course of a chemical change.  Consequently, while the elemental symbols of substances must change during a nuclear reaction, the relative quantity of protons and neutrons that are involved in the reaction must be constant.  As stated previously, the atomic number of an element is defined as the number of protons contained in an atom of that element, and the mass number of a particular elemental isotope is calculated by adding the quantities of protons and neutrons that are present in that atom.  Therefore, since the atomic number and mass number of an isotope are written as the subscript and superscript, respectively, in the nuclear symbol that is used to represent that isotope, a nuclear equation is "balanced" if the sums of the atomic numbers and the mass numbers on the reactant side of the arrow are equal to the corresponding totals on the product side of the equation.  The atomic number of the daughter nucleus37, is calculated by subtracting the atomic number of the radiation that is generated, 1, from the atomic number of the radioisotope that was initially present, 38.  Furthermore, since the atomic number of an element is a unique value that directly corresponds to the identity of that element, this calculated subscript indicates that the daughter nucleus that is produced in this radioactive decay reaction, rubidium, is symbolized as "Rb."  The mass number of the daughter nucleus80, is calculated by subtracting the mass number of the radiation0, from the mass number of the radioisotope, 80.  Finally, since the Law of Conservation of Matter is upheld by calculating the atomic number and mass number of the daughter nucleusbalancing coefficients are not incorporated into the nuclear equation that is shown below.  Because the following reaction equation is balanced, this nuclear equation is the chemically-correct representation of the positron decay of Sr-80.

$$\ce{^{80}_{38}Sr}$$ $$\rightarrow$$ $$\ce{^{0}_{1}e}$$ $$\ce{^{80}_{37}Rb}$$

Exercise $$\PageIndex{1}$$

Write a balanced nuclear equation that represents the beta decay of tin-126.

Answer
Because the word "decay" is included in the given statement, the nuclear equation pattern that corresponds to a radioactive decay should be applied to symbolically-represent this reaction.  Since a radioactive decay is a decomposition reaction, a single reactant should be written on the left side of the reaction arrow, and two products, separated by a plus sign, "+", should be represented on the right side of the equation.  Furthermore, the reactant in a radioactive decay is, by definition, an unstable radioisotope.  Because the given statement indicates that the decay of tin-126 should be represented in the equation that is being developed, the radioisotope in the corresponding reaction is a tin-126 nucleus.  Since nuclear symbols must be incorporated into a nuclear equation, the symbolism "$$\ce{^{126}_{50}Sn}$$" is written on the left side of this pattern.  One of the products of a radioactive decay reaction is, by definition, classified as radiation.  Because the given statement indicates that tin-126 undergoes beta decay, the nuclear symbol for a beta particle, "$$\ce{^{0}_{–1}e}$$," is written on the right side of this equation, as shown below.

$$\ce{^{126}_{50}Sn}$$ $$\rightarrow$$ $$\ce{^{0}_{–1}e}$$ _____

Finally, in order to identify the daughter nucleus that is generated, the nuclear equation that is written above must be "balanced" by equating the sums of the atomic number subscripts and the mass number superscripts on the reactant side of the arrow to the corresponding totals on the product side of the equation.  The atomic number of the daughter nucleus51, is calculated by subtracting the atomic number of the radiation that is generated, 1, from, or, equivalently, adding 1 to, the atomic number of the radioisotope that was initially present, 50.  Furthermore, since the atomic number of an element directly corresponds to the identity of that element, this calculated subscript indicates that the daughter nucleus that is produced in this radioactive decay reaction, antimony, is symbolized as "Sb."  The mass number of the daughter nucleus126, is calculated by subtracting the mass number of the radiation0, from the mass number of the radioisotope, 126.  Finally, since the Law of Conservation of Matter is upheld by calculating the atomic number and mass number of the daughter nucleus, balancing coefficients are not incorporated into the nuclear equation that is shown below.  Because the following reaction equation is balanced, this nuclear equation is the chemically-correct representation of the beta decay of tin-126.

$$\ce{^{126}_{50}Sn}$$ $$\rightarrow$$ $$\ce{^{0}_{–1}e}$$ $$\ce{^{126}_{51}Sb}$$

Exercise $$\PageIndex{2}$$

Write a balanced nuclear equation that represents the reaction that generates lead-206 through the alpha decay of an unstable radioisotope.

Answer
Because the word "decay" is included in the given statement, the nuclear equation pattern that corresponds to a radioactive decay should be applied to symbolically-represent this reaction.  Since a radioactive decay is a decomposition reaction, a single reactant should be written on the left side of the reaction arrow, and two products, separated by a plus sign, "+", should be represented on the right side of the equation.  Because the given statement indicates that the generation of lead-206 should be represented in the equation that is being developed, the daughter nucleus that is produced in the corresponding reaction is a lead-206 nucleus.  Since nuclear symbols must be incorporated into a nuclear equation, the symbolism "$$\ce{^{206}_{82}Pb}$$" is written on the right side of this pattern.  The remaining product of a radioactive decay reaction is, by definition, classified as radiation.  Because the given statement indicates that an unstable radioisotope undergoes alpha decay, the nuclear symbol for an alpha particle, "$$\ce{^{4}_{2}He}$$," is written on the product side of this equation, as shown below.

_____ $$\rightarrow$$ $$\ce{^{4}_{2}He}$$ $$\ce{^{206}_{82}Pb}$$

Finally, in order to identify the unstable radioisotope, the nuclear equation that is written above must be "balanced" by equating the sums of the atomic number subscripts and the mass number superscripts on the reactant side of the arrow to the corresponding totals on the product side of the equation.  The atomic number of the unstable radioisotope, 84, is calculated by adding the atomic number of the radiation that is generated, 2, to the atomic number of the daughter nucleus that is produced, 82.  Furthermore, since the atomic number of an element directly corresponds to the identity of that element, this calculated subscript indicates that the unstable radioisotope that decomposes in this radioactive decay reaction, polonium, is symbolized as "Po."  The mass number of the radioisotope, 210, is calculated by adding the mass number of the radiation4, to the mass number of the daughter nucleus, 206.  Finally, since the Law of Conservation of Matter is upheld by calculating the atomic number and mass number of the radioisotope, balancing coefficients are not incorporated into the nuclear equation that is shown below.  Because the following reaction equation is balanced, this nuclear equation is the chemically-correct representation of the reaction that generates lead-206 through the alpha decay of an unstable radioisotope.

$$\ce{^{210}_{84}Po}$$ $$\rightarrow$$ $$\ce{^{4}_{2}He}$$ $$\ce{^{206}_{82}Pb}$$

Exercise $$\PageIndex{3}$$

Write a balanced nuclear equation that represents the gamma decay of metastable Pa-234.

Answer
Because the word "decay" is included in the given statement, the nuclear equation pattern that corresponds to a radioactive decay should be applied to symbolically-represent this reaction.  Since a radioactive decay is a decomposition reaction, a single reactant should be written on the left side of the reaction arrow, and two products, separated by a plus sign, "+", should be represented on the right side of the equation.  Furthermore, the reactant in a radioactive decay is, by definition, an unstable radioisotope.  Because the given statement indicates that the decay of metastable Pa-234 should be represented in the equation that is being developed, the radioisotope in the corresponding reaction is a metastable Pa-234 nucleus.  Since nuclear symbols must be incorporated into a nuclear equation, the symbolism "$$\ce{^{234m}_{91}Pa}$$" is written on the left side of this pattern, as shown below.  The high energy, metastable state of this radioisotope is represented by the lower-case "m" that is included in this notation.  One of the products of a radioactive decay reaction is, by definition, classified as radiation.  Because the given statement indicates that metastable Pa-234 undergoes gamma decay, the nuclear symbol for a gamma ray, "$$\ce{^{0}_{0}\gamma}$$," is written on the right side of this equation, as shown below.

$$\ce{^{234m}_{91}Pa}$$ $$\rightarrow$$ $$\ce{^{0}_{0}\gamma}$$ _____

Finally, in order to identify the daughter nucleus that is generated, the nuclear equation that is written above must be "balanced" by equating the sums of the atomic number subscripts and the mass number superscripts on the reactant side of the arrow to the corresponding totals on the product side of the equation.  The atomic number of the daughter nucleus91, is calculated by subtracting the atomic number of the radiation that is generated, 0, from the atomic number of the radioisotope that was initially present, 91.  Furthermore, since the atomic number of an element directly corresponds to the identity of that element, this calculated subscript indicates that the daughter nucleus that is produced in this radioactive decay reaction, protactinium, is symbolized as "Pa."  The mass number of the daughter nucleus234, is calculated by subtracting the mass number of the radiation0, from the mass number of the radioisotope, 234.  Since, as stated above, a metastable nucleus has achieved a high-energy state, and gamma radiation is released as a burst of pure energy, the resultant daughter nucleus is a lower-energy, more stable version of the unstable radioisotope that was initially present.  Finally, since the Law of Conservation of Matter is upheld by calculating the atomic number and mass number of the daughter nucleus, balancing coefficients are not incorporated into the nuclear equation that is shown below.  Because the following reaction equation is balanced, this nuclear equation is the chemically-correct representation of the gamma decay of metastable Pa-234.

$$\ce{^{234m}_{91}Pa}$$ $$\rightarrow$$ $$\ce{^{0}_{0}\gamma}$$ $$\ce{^{234}_{91}Pa}$$

• Was this article helpful?