2.3: Nuclear Reactions and Nuclear Equations

Learning Objectives

• To write nuclear equations to represent nuclear reactions.
• To identify a missing particle when provided with the rest of a nuclear equation.

Reaction Formats

As radioisotopes decay, they tend to follow predictable formats where the original atom gives off, or emits, radiation and at the same time changes into a different element. Some of the common formats are described here. 

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:

\[\ce{^{235}_{92}U \rightarrow \,_2^4\alpha + \, _{90}^{231}Th} \label{Eq2}\]

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 \(\ce{^{4}_{2}\alpha}\) and use subtraction (222 − 4 = 218 and 86 − 2 = 84) to identify the daughter isotope as an isotope of polonium, \(\mathrm{^{218}_{84}Po}\):

\(\ce{_{86}^{222}Rn\rightarrow \, _2^4\alpha + \, _{84}^{218}Po}\)

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

\(\ce{_{84}^{209}Po\rightarrow \, _2^4\alpha + \, _{82}^{205}Pb}\)

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:

\[\ce{_6^{14}C \rightarrow \, _7^{14}N + \, _{-1}^0\beta} \label{Eq3}\]

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 beta particle is assigned an “atomic number” of 1−, equal to its charge. This helps us keep track of the atomic numbers as a proton changes into a neutron.)

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:

\[\ce{_{86}^{222}Rn\rightarrow \, _{84}^{218}Po + \, _2^4\alpha + \gamma} \label{Eq4}\]

This may not seem like much energy, but if 1 mole of radon atoms were to decay, the gamma ray energy would be 49 million kJ!

Electron Capture

The format is different than the ones described above. Here an electron is absorbed into the nucleus of an atom causing a proton to change into a neutron. Because the number of protons changes, the atom changes into a different element. For example, aluminum-26 undergoes electron capture to become magnesium-26.

\[\ce{_{13}^{26}Al +_{-1}^{0}e\rightarrow \, _{12}^{26}Mg } \label{Eq5}\]

Using the symbol \(_{-1}^{0}e\) helps us keep track of the changing atomic numbers as a proton changes into a neutron. 

Table \(\PageIndex{1}\): Summary of nuclear reaction formats

Description	Format	Examples
A parent atom decays and emits alpha, beta, or positron radiation	parent \(\rightarrow\) daughter + radiation	alpha decay, beta emission, etc.
An atom absorbs an electron into the nucleus (and a proton changes into a neutron)	parent + e^- \(\rightarrow\) daughter	electron capture

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 a beta particle, \(\ce{^{0}_{-1}\beta}\). 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:

\(\ce{_5^{12}B\rightarrow \, _6^{12}C + \, _{-1}^0\beta + \gamma}\)

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

\(\ce{_53^{131}I\rightarrow \, _54^{131}Xe + \, _{-1}^0\beta + \gamma}\)

Bombardment

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

\[\ce{_{92}^{235}U \rightarrow \, _{56}^{139}Ba + \, _{36}^{94}Kr + 2^1_0n} \label{Eq6}\]

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.

To write the nuclear equation for a reaction: 

1. Write the symbols for each particle on the correct side of the arrow. Starting materials that are present before the reaction go on the left. Products that are made during the reaction go on the right. If there is an unknown particle, i.e. one that is mentioned but not identified specifically, leave a space for it. 
2. Find the sum of the mass numbers on both sides of the equation. If they are not equal, that helps you identify the value of the mass number for the unknown particle. Repeat this step for the atomic numbers.  
3. Identify the symbol for the unknown particle. If it is an element, get the 1 - 2 letter symbol for the element from the periodic table by looking up the atomic number. Important: Do not use the mass number to identify an element! Many elements can have isotopes with the same mass number, but an atomic number uniquely identifies an element. The unknown particle may also be a subatomic particle (\(_1^1p\), \(_0^1n\), or \(_{-1}^0e\)) or radiation (\(_2^4\alpha\), \(_{-1}^0\beta\), or \(_{+1}^0\beta\)), so consider whether the numbers you solved for match one of those particles. 

Example \(\PageIndex{3}\)

The bombardment of curium-245 with calcium-48 produces livermorium-290 and some number of neutrons. Write the nuclear equation, including the proper number of neutrons. 

Solution

1. First determine whether each particle should be before or after the arrow and fill in the known numbers based on the mass numbers provided in the problem and the atomic numbers found on the periodic table. The phrase "bombardment of ____ with ____" is indicating how the reaction starts. These particles go on the left. The word "produces" indicates what will be after the arrow. \(\ce{_{96}^{245}Cm + _{20}^{48}Ca \rightarrow \, _{116}^{290}Lv + ^1_0n}\)
2. Next, solve for the number of neutrons that are produced. The sum of the mass numbers on each side of the arrow should be equal. On the left, 245 + 48 = 293. On the right, livermorium has a mass number of 290 and the neutrons make up the difference. There must be three neutrons in order to reach a total of 293 on the right. 
3. The symbol for a neutron is \(^1_0n\). Reactions like this produce multiple separate neutrons. They are not grouped together into one particle. Therefore, we use a coefficient in front of the neutron symbol to indicate the number of copies produced. 

\(\ce{_{96}^{245}Cm + _{20}^{48}Ca \rightarrow \, _{116}^{290}Lv + 3^1_0n}\)

Example \(\PageIndex{4}\)

Uranium-238 can be hit by a small atom to produce neptunium-238 and two neutrons. Write the nuclear equation for this reaction, including the symbol for the unidentified atom. 

Solution

1. First write the symbols for the particles that you know on the correct side of the arrow. Leave a space for the unknown particle, if relevant. In this case a collision initiates the reaction so the uranium and "small atom" are on the left. The neptunium and neutrons are produced, so they are written after the arrow.  \(\ce{_{92}^{238}U + ? \rightarrow \, _{93}^{238}Np + 2^1_0n}\)
2. Next, determine the missing mass number and atomic number. Remember that the sums need to be the same on both sides of the arrow. For the mass numbers 238 + ? = 238 + (2)1 where the mass number for the neutron is multiplied by two because there are two neutrons. The missing mass number is a 2. For the atomic numbers, 92 + ? = 93 + (2)0 so the unknown atom has an atomic number of 1. 
3. Finally, write the missing symbol. The atomic number of 1 tells us that the unknown element is hydrogen. Remember not to use the mass number to identify an element.  

\(\ce{_{92}^{238}U + _{1}^{2}H \rightarrow \, _{93}^{238}Np + 2^1_0n}\)

Exercise \(\PageIndex{3}\)

A lead-208 atom collides with another atom to form darmstadtium-269 and a neutron. Write the nuclear equation for this reaction, including the symbol for the unidentified atom. 

Answer

\(\ce{_{82}^{208}Pb + _{28}^{62}Ni \rightarrow \, _{110}^{269}Ds + ^1_0n}\) 

Summary

• Particles present before the reaction start are written before/to the left of the arrow. Particles that result from a reaction are written after/to the right of the arrow. Note: The arrow always points from left to right. 
• 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
  • The sums of the mass numbers should be the same on both sides of the equation
• If an unknown particle needs to be identified, figure out its mass number and atomic number first, then identify the 1-2 letter symbol that represents it.
  • Use the atomic number, not the mass number to identify an element. 
  • Use both numbers to identify a subatomic particle or radiation. 

Concept Review Exercise

1. What are the major types of radioactivity? Write chemical equations demonstrating each type.

Answer

1. 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)