Nuclear Reaction

  

A nuclear reaction may result in more than two products, the following is called a knock-out reaction: Neutron Absorption: Sometimes two nuclear particles stick together. The following reactions are called neutron absorption: 27 Al is the common isotope of aluminum; 235 U is a rare isotope of uranium. The products, aluminum-28 and uranium-236. Define nuclear reaction. Nuclear reaction synonyms, nuclear reaction pronunciation, nuclear reaction translation, English dictionary definition of nuclear reaction. A reaction, as in fission, fusion, or radioactive decay, that alters the energy, composition, or structure of an atomic nucleus.

Changes of nuclei that result in changes in their atomic numbers, mass numbers, or energy states are nuclear reactions. To describe a nuclear reaction, we use an equation that identifies the nuclides involved in the reaction, their mass numbers and atomic numbers, and the other particles involved in the reaction.

Learning Objectives

  • Identify common particles and energies involved in nuclear reactions
  • Write and balance nuclear equations

Changes of nuclei that result in changes in their atomic numbers, mass numbers, or energy states are nuclear reactions. To describe a nuclear reaction, we use an equation that identifies the nuclides involved in the reaction, their mass numbers and atomic numbers, and the other particles involved in the reaction.

Types of Particles in Nuclear Reactions

Many entities can be involved in nuclear reactions. The most common are protons, neutrons, alpha particles, beta particles, positrons, and gamma rays, as shown in Figure (PageIndex{1}). Protons ( (ce{^{1}_{1}p}), also represented by the symbol (ce{^1_1H})) and neutrons ( (ce{^1_0n})) are the constituents of atomic nuclei, and have been described previously. Alpha particles ( (ce{^4_2He}), also represented by the symbol (ce{^{4}_{2}alpha})) are high-energy helium nuclei. Beta particles ( (ce{^{0}_{−1}beta}), also represented by the symbol (ce{^0_{-1}e})) are high-energy electrons, and gamma rays are photons of very high-energy electromagnetic radiation. Positrons ( (ce{^0_{+1}e}), also represented by the symbol (ce{^0_{+1}β})) are positively charged electrons (“anti-electrons”). The subscripts and superscripts are necessary for balancing nuclear equations, but are usually optional in other circumstances. For example, an alpha particle is a helium nucleus (He) with a charge of +2 and a mass number of 4, so it is symbolized (ce{^4_2He}). This works because, in general, the ion charge is not important in the balancing of nuclear equations.

Note that positrons are exactly like electrons, except they have the opposite charge. They are the most common example of antimatter, particles with the same mass but the opposite state of another property (for example, charge) than ordinary matter. When antimatter encounters ordinary matter, both are annihilated and their mass is converted into energy in the form of gamma rays (γ)—and other much smaller subnuclear particles, which are beyond the scope of this chapter—according to the mass-energy equivalence equation (E = mc^2), seen in the preceding section. For example, when a positron and an electron collide, both are annihilated and two gamma ray photons are created:

[ce{^0_{−1}e + ^0_{+1}e } rightarrow gamma + gamma label{21.3.1}]

Gamma rays compose short wavelength, high-energy electromagnetic radiation and are (much) more energetic than better-known X-rays. Gamma rays are produced when a nucleus undergoes a transition from a higher to a lower energy state, similar to how a photon is produced by an electronic transition from a higher to a lower energy level. Due to the much larger energy differences between nuclear energy shells, gamma rays emanating from a nucleus have energies that are typically millions of times larger than electromagnetic radiation emanating from electronic transitions.

Balancing Nuclear Reactions

A balanced chemical reaction equation reflects the fact that during a chemical reaction, bonds break and form, and atoms are rearranged, but the total numbers of atoms of each element are conserved and do not change. A balanced nuclear reaction equation indicates that there is a rearrangement during a nuclear reaction, but of subatomic particles rather than atoms. Nuclear reactions also follow conservation laws, and they are balanced in two ways:

  1. The sum of the mass numbers of the reactants equals the sum of the mass numbers of the products.
  2. The sum of the charges of the reactants equals the sum of the charges of the products.

If the atomic number and the mass number of all but one of the particles in a nuclear reaction are known, we can identify the particle by balancing the reaction. For instance, we could determine that (ce{^{17}_8O}) is a product of the nuclear reaction of (ce{^{14}_7N}) and (ce{^4_2He}) if we knew that a proton, (ce{^1_1H}), was one of the two products. Example (PageIndex{1}) shows how we can identify a nuclide by balancing the nuclear reaction.

Example (PageIndex{1}): Balancing Equations for Nuclear Reactions

The reaction of an (α) particle with magnesium-25 ( (ce{^{25}_{12}Mg})) produces a proton and a nuclide of another element. Identify the new nuclide produced.

Solution

The nuclear reaction can be written as:

[ce{^{25}_{12}Mg + ^4_2He rightarrow ^1_1H + ^{A}_{Z}X} nonumber]

where

  • (ce A) is the mass number and
  • (ce Z) is the atomic number of the new nuclide, (ce X).

Because the sum of the mass numbers of the reactants must equal the sum of the mass numbers of the products:

[mathrm{25+4=A+1} nonumber]

so

[ mathrm{A=28} nonumber]

Similarly, the charges must balance, so:

[mathrm{12+2=Z+1} nonumber]

so

[mathrm{Z=13} nonumber]

Check the periodic table: The element with nuclear charge = +13 is aluminum. Thus, the product is (ce{^{28}_{13}Al}).

Exercise (PageIndex{1})

The nuclide (ce{^{125}_{53}I}) combines with an electron and produces a new nucleus and no other massive particles. What is the equation for this reaction?

Answer

[ce{^{125}_{53}I + ^0_{−1}e rightarrow ^{125}_{52}Te} nonumber]

Following are the equations of several nuclear reactions that have important roles in the history of nuclear chemistry:

  • The first naturally occurring unstable element that was isolated, polonium, was discovered by the Polish scientist Marie Curie and her husband Pierre in 1898. It decays, emitting α particles: [ce{^{212}_{84}Po⟶ ^{208}_{82}Pb + ^4_2He}nonumber]
  • The first nuclide to be prepared by artificial means was an isotope of oxygen, 17O. It was made by Ernest Rutherford in 1919 by bombarding nitrogen atoms with α particles: [ce{^{14}_7N + ^4_2α⟶ ^{17}_8O + ^1_1H} nonumber]
  • James Chadwick discovered the neutron in 1932, as a previously unknown neutral particle produced along with 12C by the nuclear reaction between 9Be and 4He: [ce{^9_4Be + ^4_2He⟶ ^{12}_6C + ^1_0n} nonumber]
  • The first element to be prepared that does not occur naturally on the earth, technetium, was created by bombardment of molybdenum by deuterons (heavy hydrogen, (ce{^2_1H})), by Emilio Segre and Carlo Perrier in 1937: [ ce{^2_1H + ^{97}_{42}Mo⟶2^1_0n + ^{97}_{43}Tc}nonumber]
  • The first controlled nuclear chain reaction was carried out in a reactor at the University of Chicago in 1942. One of the many reactions involved was: [ ce{^{235}_{92}U + ^1_0n⟶ ^{87}_{35}Br + ^{146}_{57}La + 3^1_0n} nonumber]

Summary

Nuclei can undergo reactions that change their number of protons, number of neutrons, or energy state. Many different particles can be involved in nuclear reactions. The most common are protons, neutrons, positrons (which are positively charged electrons), alpha (α) particles (which are high-energy helium nuclei), beta (β) particles (which are high-energy electrons), and gamma (γ) rays (which compose high-energy electromagnetic radiation). As with chemical reactions, nuclear reactions are always balanced. When a nuclear reaction occurs, the total mass (number) and the total charge remain unchanged.

Glossary

alpha particle
(α or (ce{^4_2He}) or (ce{^4_2α})) high-energy helium nucleus; a helium atom that has lost two electrons and contains two protons and two neutrons
antimatter
particles with the same mass but opposite properties (such as charge) of ordinary particles
beta particle
((β) or (ce{^0_{-1}e}) or (ce{^0_{-1}β})) high-energy electron
gamma ray
(γ or (ce{^0_0γ})) short wavelength, high-energy electromagnetic radiation that exhibits wave-particle duality
nuclear reaction
change to a nucleus resulting in changes in the atomic number, mass number, or energy state
positron ((ce{^0_{+1}β}) or (ce{^0_{+1}e}))
antiparticle to the electron; it has identical properties to an electron, except for having the opposite (positive) charge

Contributors and Attributions

  • Paul Flowers (University of North Carolina - Pembroke), Klaus Theopold (University of Delaware) and Richard Langley (Stephen F. Austin State University) with contributing authors. Textbook content produced by OpenStax College is licensed under a Creative Commons Attribution License 4.0 license. Download for free at http://cnx.org/contents/[email protected]).

Balancing Nuclear Equations

To balance a nuclear equation, the mass number and atomic numbers of all particles on either side of the arrow must be equal.

Learning Objectives

Produce a balanced nuclear equation

Key Takeaways

Key Points

  • A balanced nuclear equation is one where the sum of the mass numbers (the top number in notation) and the sum of the atomic numbers balance on either side of an equation.
  • Nuclear equation problems will often be given such that one particle is missing.
  • Instead of using the full equations, in many situations a compact notation is used to describe nuclear reactions.

Key Terms

  • baryon: A heavy subatomic particle created by the binding of quarks by gluons; a hadron containing three quarks. They have half-odd integral spin and are thus fermions.

Nuclear reactions may be shown in a form similar to chemical equations, for which invariant mass, which is the mass not considering the mass defect, must balance for each side of the equation. The transformations of particles must follow certain conservation laws, such as conservation of charge and baryon number, which is the total atomic mass number. An example of this notation follows:

[latex]^6_3text{Li} + ^2_1text{H}rightarrow ^4_2text{He} + ?[/latex]

To balance the equation above for mass, charge, and mass number, the second nucleus on the right side must have atomic number 2 and mass number 4; it is therefore also helium-4. The complete equation therefore reads:

[latex]^6_3text{Li} + ^2_1text{H}rightarrow ^4_2text{He} + ^4_2text{He}[/latex]

Or, more simply:

[latex]^6_3text{Li} + ^2_1text{H}rightarrow 2 ^4_2text{He}[/latex]

Lithium-6 plus deuterium gives two helium-4s.: The visual representation of the equation we used as an example.

Compact Notation of Radioactive Decay

Instead of using the full equations in the style above, in many situations a compact notation is used to describe nuclear reactions. This style is of the form A(b,c)D, which is equivalent to A + b gives c + D. Common light particles are often abbreviated in this shorthand, typically p for proton, n for neutron, d for deuteron, α representing an alpha particle or helium-4, β for beta particle or electron, γ for gamma photon, etc. The reaction in our example above would be written as Li-6(d,α)α.

How do nuclear reactors work

Balancing a Radioactive Decay Equation

In balancing a nuclear equation, it is important to remember that the sum of all the mass numbers and atomic numbers, given on the upper left and lower left side of the element symbol, respectively, must be equal for both sides of the equation. In addition, problems will also often be given as word problems, so it is useful to know the various names of radioactively emitted particles.

Nuclear reaction

Example

[latex]^{ 235 }_{ 92 }text{U} rightarrow ^{ 231 }_{ 90 }text{Th} + ?[/latex]

This could be written out as uranium-235 gives thorium-231 plus what? In order to solve, we find the difference between the atomic masses and atomic numbers in the reactant and product. The result is an atomic mass difference of 4 and an atomic number difference of 2. This fits the description of an alpha particle. Thus, we arrive at our answer:

[latex]^{ 235 }_{ 92 }text{U} rightarrow ^{ 231 }_{ 90}text{Th} + ^{ 4 }_{ 2 }text{He}[/latex]

Example

[latex]^{ 214 }_{ 84 }text{Po} + 2 ^{ 4 }_{ 2 }text{He} + 2^0_{-1}text{e}rightarrow ?[/latex]

This could also be written out as polonium-214, plus two alpha particles, plus two electrons, give what? In order to solve this equation, we simply add the mass numbers, 214 for polonium, plus 8 (two times four) for helium (two alpha particles), plus zero for the electrons, to give a mass number of 222. For the atomic number, we take 84 for polonium, add 4 (two times two) for helium, then subtract two (two times -1) for two electrons lost through beta emission, to give 86; this is the atomic number for radon (Rn). Therefore, the equation should read:

[latex]^{ 214 }_{ 84 }text{Po}+2^{ 4 }_{ 2 }text{He}+2^0_{-1}text{e}rightarrow ^{222}_{86}text{Rn}[/latex]

Writing nuclear equations: Describes how to write the nuclear equations for alpha and beta decay.

Nuclear Binding Energy and Mass Defect

A nucleus weighs less than its sum of nucleons, a quantity known as the mass defect, caused by release of energy when the nucleus formed. Citrix workspace osx.

Learning Objectives

Calculate the mass defect and nuclear binding energy of an atom

Key Takeaways

Nuclear

Key Points

  • Nuclear binding energy is the energy required to split a nucleus of an atom into its components.
  • Nuclear binding energy is used to determine whether fission or fusion will be a favorable process.
  • The mass defect of a nucleus represents the mass of the energy binding the nucleus, and is the difference between the mass of a nucleus and the sum of the masses of the nucleons of which it is composed.

Key Terms

  • nucleon: One of the subatomic particles of the atomic nucleus, i.e. a proton or a neutron.
  • strong force: The nuclear force, a residual force responsible for the interactions between nucleons, deriving from the color force.
  • mass defect: The difference between the calculated mass of the unbound system and the experimentally measured mass of the nucleus.

Binding Energy

Nuclear binding energy is the energy required to split a nucleus of an atom into its component parts: protons and neutrons, or, collectively, the nucleons. The binding energy of nuclei is always a positive number, since all nuclei require net energy to separate them into individual protons and neutrons.

Nuclear Reaction Examples

Mass Defect

Nuclear binding energy accounts for a noticeable difference between the actual mass of an atom’s nucleus and its expected mass based on the sum of the masses of its non-bound components.

Recall that energy (E) and mass (m) are related by the equation:

[latex]text{E}=text{mc}^2[/latex]

Here, c is the speed of light. In the case of nuclei, the binding energy is so great that it accounts for a significant amount of mass.

The actual mass is always less than the sum of the individual masses of the constituent protons and neutrons because energy is removed when when the nucleus is formed. This energy has mass, which is removed from the total mass of the original particles. This mass, known as the mass defect, is missing in the resulting nucleus and represents the energy released when the nucleus is formed.

Mass defect (Md) can be calculated as the difference between observed atomic mass (mo) and that expected from the combined masses of its protons (mp, each proton having a mass of 1.00728 amu) and neutrons (mn, 1.00867 amu):

[latex]text{M}_text{d}=(text{m}_text{n}+text{m}_text{p})-text{m}_text{o}[/latex]

Nuclear Binding Energy

Once mass defect is known, nuclear binding energy can be calculated by converting that mass to energy by using E=mc2. Mass must be in units of kg.

Once this energy, which is a quantity of joules for one nucleus, is known, it can be scaled into per-nucleon and per- mole quantities. To convert to joules/mole, simply multiply by Avogadro’s number. To convert to joules per nucleon, simply divide by the number of nucleons.

Nuclear binding energy can also apply to situations when the nucleus splits into fragments composed of more than one nucleon; in these cases, the binding energies for the fragments, as compared to the whole, may be either positive or negative, depending on where the parent nucleus and the daughter fragments fall on the nuclear binding energy curve. If new binding energy is available when light nuclei fuse, or when heavy nuclei split, either of these processes result in the release of the binding energy. This energy—available as nuclear energy—can be used to produce nuclear power or build nuclear weapons. When a large nucleus splits into pieces, excess energy is emitted as photons, or gamma rays, and as kinetic energy, as a number of different particles are ejected.

Nuclear binding energy is also used to determine whether fission or fusion will be a favorable process. For elements lighter than iron-56, fusion will release energy because the nuclear binding energy increases with increasing mass. Elements heavier than iron-56 will generally release energy upon fission, as the lighter elements produced contain greater nuclear binding energy. As such, there is a peak at iron-56 on the nuclear binding energy curve.

Nuclear binding energy curve: This graph shows the nuclear binding energy (in MeV) per nucleon as a function of the number of nucleons in the nucleus. Notice that iron-56 has the most binding energy per nucleon, making it the most stable nucleus.

The rationale for this peak in binding energy is the interplay between the coulombic repulsion of the protons in the nucleus, because like charges repel each other, and the strong nuclear force, or strong force. The strong force is what holds protons and neutrons together at short distances. As the size of the nucleus increases, the strong nuclear force is only felt between nucleons that are close together, while the coulombic repulsion continues to be felt throughout the nucleus; this leads to instability and hence the radioactivity and fissile nature of the heavier elements.

Example

Reactions

Calculate the average binding energy per mole of a U-235 isotope. Show your answer in kJ/mole.

First, you must calculate the mass defect. U-235 has 92 protons, 143 neutrons, and has an observed mass of 235.04393 amu.

[latex]text{M}_text{d}=(text{m}_text{n}+text{m}_text{p})-text{m}_text{o}[/latex]

Md = (92(1.00728 amu)+143(1.00867 amu)) – 235.04393 amu

Md = 1.86564 amu

Nuclear Reaction

Calculate the mass in kg:

1.86564 amu x [latex]frac{1 text{kg}}{6.02214times10^{26} text{amu}}[/latex] = 3.09797 x 10-27 kg

Now calculate the energy:

E = mc2

E = 3.09797 x 10-27 kg x (2.99792458 x 108[latex]frac{text{m}}{text{s}}[/latex])2

E =2.7843 x 10-10 J

Now convert to kJ per mole:

Nuclear Reaction Worksheet

[latex]2.7843times10^{-10}frac{text{Joules}}{text{atom}} times frac {6.02times10^{23} text{atoms}}{text{mole}}times frac{1 text{kJ}}{1000 text{joules}} =[/latex] 1.6762 x 1011[latex]frac{text{kJ}}{text{mole}}[/latex]