Learning Objectives To write nuclear equations to represent nuclear reactions. To identify a missing particle when provided with the rest of a nuclear equation. Alpha Emission When a radioactive atom emits an alpha particle, the original atom’s atomic number decreases by two (because of the loss of two protons), and its mass number decreases by four (because of the loss of four nuclear particles). We can represent the emission of an alpha particle with a nuclear equation—for example, the alpha-particle emission of uranium-235 is as follows: \[\ce{^{235}_{92}U \rightarrow \,_2^4He + \, _{90}^{231}Th} \label{Eq2}\] Like mathematical equations, nuclear equations relate things that are equal. However, nuclear equations use an arrow instead of an equal sign to show the passage of time. Substances written on the left are present at the beginning of the reaction and substances written on the right are present at the end. How do we know that a product of the reaction is \(\ce{^{231}_{90}Th}\)? We use the law of conservation of matter, which says that matter cannot be created or destroyed. This means we must have the same number of protons and neutrons on both sides of the equation. If our uranium nucleus loses 2 protons, there are 90 protons remaining, identifying the element as thorium. Moreover, if we lose 4 nuclear particles of the original 235, there are 231 remaining. Thus, we use subtraction to identify the isotope of the thorium atom—in this case, \(\ce{^{231}_{90}Th}\). Chemists often use the names parent isotope and daughter isotope to represent the original atom and the product other than the alpha particle. In the previous example, \(\ce{^{235}_{92}U}\) is the parent isotope, and \(\ce{^{231}_{90}Th}\) is the daughter isotope. When one element changes into another in this manner, it undergoes radioactive decay . Example \(\PageIndex{1}\): Radon-222 Write the nuclear equation that represents the radioactive decay of radon-222 by alpha particle emission and identify the daughter isotope. Solution Radon has an atomic number of 86, so the parent isotope is represented as \(\ce{^{222}_{86}Rn}\). We represent the alpha particle as \(\ce{^{4}_{2}He}\) and use subtraction (222 − 4 = 218 and 86 − 2 = 84) to identify the daughter isotope as an isotope of polonium, \(\mathrm{^{218}_{84}Po}\): \(\ce{_{86}^{222}Rn\rightarrow \, _2^4He + \, _{84}^{218}Po}\) Exercise \(\PageIndex{1}\): Polonium-209 Write the nuclear equation that represents the radioactive decay of polonium-209 by alpha particle emission and identify the daughter isotope. Answer \(\ce{_{84}^{209}Po\rightarrow \, _2^4He + \, _{82}^{205}Pb}\) Beta Emission The net effect of beta particle emission on a nucleus is that a neutron is converted to a proton. The overall mass number stays the same, but because the number of protons increases by one, the atomic number goes up by one. Carbon-14 decays by emitting a beta particle: \[\ce{_6^{14}C \rightarrow \, _7^{14}N + \, _{-1}^0e} \label{Eq3}\] Again, the sum of the atomic numbers is the same on both sides of the equation, as is the sum of the mass numbers. (Note that the electron is assigned an “atomic number” of 1−, equal to its charge.) Gamma Emission For example, in the radioactive decay of radon-222, both alpha and gamma radiation are emitted, with the latter having an energy of 8.2 × 10 −14 J per nucleus decayed: \[\ce{_{86}^{222}Rn\rightarrow \, _{84}^{218}Po + \, _2^4He + \gamma} \label{Eq4}\] This may not seem like much energy, but if 1 mol of radon atoms were to decay, the gamma ray energy would be 49 million kJ! Example \(\PageIndex{2}\): Boron-12 Write the nuclear equation that represents the radioactive decay of boron-12 by beta particle emission and identify the daughter isotope. A gamma ray is emitted simultaneously with the beta particle. Solution The parent isotope is \(\ce{^{12}_{5}B}\) while one of the products is an electron, \(\ce{^{0}_{-1}e}\). So that the mass and atomic numbers have the same value on both sides, the mass number of the daughter isotope must be 12, and its atomic number must be 6. The element having an atomic number of 6 is carbon. Thus, the complete nuclear equation is as follows: \[\ce{_5^{12}B\rightarrow \, _6^{12}C + \, _{-1}^0e + \gamma}\] The daughter isotope is \(\ce{^{12}_6 C}\). Exercise \(\PageIndex{2}\): Iodine-131 Write the nuclear equation that represents the radioactive decay of iodine-131 by beta particle emission and identify the daughter isotope. A gamma ray is emitted simultaneously with the beta particle. Answer \[\ce{_53^{131}I\rightarrow \, _54^{131}Xe + \, _{-1}^0e + \gamma}\] Occasionally, an atomic nucleus breaks apart into smaller pieces in a radioactive process called spontaneous fission (or fission). Typically, the daughter isotopes produced by fission are a varied mix of products, rather than a specific isotope as with alpha and beta particle emission. Often, fission produces excess neutrons that will sometimes be captured by other nuclei, possibly inducing additional radioactive events. Uranium-235 undergoes spontaneous fission to a small extent. One typical reaction is \[\ce{_{92}^{235}U \rightarrow \, _{56}^{139}Ba + \, _{36}^{94}Kr + 2^1_0n} \label{Eq5}\] where \(\ce{_0^1n}\) is a neutron. As with any nuclear process, the sums of the atomic numbers and the mass numbers must be the same on both sides of the equation. Spontaneous fission is found only in large nuclei. The smallest nucleus that exhibits spontaneous fission is lead-208. Fission is the radioactive process used in nuclear power plants and one type of nuclear bomb. Key Takeaway Nuclear equations should account for all of the protons and neutrons involved in a nuclear reaction. The sums of the atomic numbers should be the same on both sides of the equation and the sums of the mass numbers should be the same on both sides of the equation. Concept Review Exercise What are the major types of radioactivity? Write chemical equations demonstrating each type. Answer The major types of radioactivity are alpha decay, beta decay, and gamma ray emission; alpha decay with gamma emission: \(\mathrm{_{86}^{222}Rn \rightarrow \, _{84}^{218}Po + \, ^4_2He + \gamma}\); beta decay: \(\ce{_6^{14}C \rightarrow _7^{14}N + ^0_{-1}e}\) (answers will vary)