Lesson Explainer: The Atomic Nucleus Chemistry

In this explainer, we will learn how to describe the stability of the atomic nucleus through the strong force and through the ratio of protons to neutrons and calculate nuclear binding energy.

The atomic nucleus is the central section of an atom that is made up of two different types of particles. It contains positively charged protons and neutrally charged neutrons. Atomic nuclei should be unstable due to intense electrostatic repulsion between positively charged protons, but they sometimes end up being stable for more than a billion years. The stability of nuclei suggests the presence of an attractive force between protons and neutrons.

The stability of any one atomic nucleus can be ascribed to the strong nuclear force, also known as the residual strong force. This is a fundamental force of nature that binds neutrons and protons together. It is used to explain the otherwise inexplicable stability of the atom and its nucleus. It can be used to explain the existence of all atoms and, by extension, every single type of simple and complex molecule that has ever existed.

Definition: Strong Nuclear Force

The strong nuclear force is the attractive force that holds protons and neutrons together in an atomic nucleus.

The following figure shows that there is electrostatic repulsion between protons in the atomic nucleus. It also shows that there are attraction forces between nucleons (protons and neutrons). The attraction forces can be explained in terms of the strong nuclear force.

The strong nuclear force is the strongest fundamental force of nature. It is significantly stronger than the electrostatic force. It makes neutrons and protons bind together in the atomic nucleus, and it can be used to understand why so many atomic nuclei are stable for unimaginably long timescales.

The strong nuclear force is quite an unusual force of nature because it is incredibly strong at a distance of roughly one proton diameter and practically nonexistent at longer distances. It is powerfully attractive between nucleons at a distance of roughly 1×10 metres, known as 1 femtometre (fm), but essentially insignificant at distances greater than this.

Example 1: Identifying the Force That Binds Neutrons and Protons Together in the Atomic Nucleus

What is the name of the force that binds neutrons and protons together in the atomic nucleus?

  1. Electrostatic force
  2. Strong nuclear force
  3. Gravitational force
  4. Weak nuclear force
  5. Magnetic force

Answer

There are different fundamental forces that mediate interactions between all known forms of physical matter. The electrostatic force acts between charged particles and it can drive them away from each other; it can also make them attract each other. The gravitational force is the weakest fundamental force of nature, and it is responsible for governing the gravitational interactions between objects that have mass. The weak nuclear force is weaker than the strong nuclear force, and it is responsible for different types of radioactive decay. The magnetic force is responsible for the attractive and repulsive interactions between magnets. The strong nuclear force is the strongest fundamental force of nature, and it binds protons and neutrons together in the atomic nucleus. We can use these statements to determine that option B is the correct answer to this question.

The strong nuclear force can be compared with the electrostatic force because it can only affect certain types of particles. It affects groups of protons, and it affects groups of neutrons. It also affects adjacent neutron and proton particles. Quarks are similarly affected by the strong nuclear force. The strong nuclear force makes quarks bind together and form neutrons and protons.

Example 2: Identifying the Subatomic Particle Pairs That Are Affected by the Strong Nuclear Force

Which of the following pairs of subatomic particles have an attractive interaction that does not involve the strong nuclear force?

  1. Electrons and protons
  2. Quarks and quarks
  3. Protons and neutrons
  4. Protons and protons
  5. Neutrons and neutrons

Answer

The strong nuclear force does not affect all particles. It affects quarks and pairs of nucleons in the atomic nucleus. This means that it affects pairs of protons in the atomic nucleus. It also means that it affects pairs of neutrons in the atomic nucleus. It means that it affects pairs of protons and neutrons in the atomic nucleus, and it also means that it affects pairs of quarks as well. The strong nuclear force does not affect electrons. These statements can be used to determine that option A is the correct answer for this question.

Any one nucleon essentially experiences the maximum strong nuclear force when it is surrounded by a couple layers of nucleon particles. This is because the strong nuclear force has an incredibly short range. It is essentially insignificant at distances more than 1 fm, and individual protons have a diameter of approximately 0.8 fm. It has little-to-no effect over a distance that it is greater than a couple of proton particle diameters.

The electrostatic force has an infinite range. Positively charged protons cannot experience an effective maximum force of electrostatic repulsion. They experience a large force of repulsion if they are surrounded by a couple layers of nucleons. They experience an even larger force of repulsion if they are surrounded by more than a couple layers of nucleons. Large atomic nuclei can become incredibly unstable if they contain too many positively charged protons. This is illustrated in the following image.

This line of reasoning can be used to understand why small atomic nuclei can be stable if they have essentially the same number of protons and neutrons. It can also be used to understand why the largest atomic nuclei are only stable if they have about one-and-a-half times as many neutrons as protons. The potentially overwhelming electrostatic force interactions of very large nuclei can only be minimized if the atomic nuclei are mostly made up of neutrally charged neutrons, which do not generate any electrostatic repulsion forces whatsoever.

This rationale is confirmed in graphs that show the relationship between nuclear stability and the neutron-to-proton 𝑁𝑍 number ratio. The following graph uses black dots to represent stable atomic nuclei that do not undergo radioactive decay. This band of stable nuclei is sometimes called the band of stability, but it can also be called the belt, zone, or valley of stability.

The graph shows that small atomic nuclei are stable if they contain either exactly or approximately the same number of protons and neutrons. All of the smallest atomic nuclei are stable if their neutron-to-proton ratio is exactly equal to one 𝑁𝑍=1.0 or approximately equal to one 𝑁𝑍1.0. All of the larger atomic nuclei have higher neutron-to-proton number ratios. The stable medium-sized iron (Iron-56) and silver isotope nuclei (silver-107) are shown as having 𝑁𝑍 values of 1.15 and 1.28. The largest stable nuclei are shown as having 𝑁𝑍 values that are approximately equal to 1.50.

The graph also shows how unstable atomic nuclei can undergo different forms of radioactive decay. The type of radioactive decay depends on neutron-to-proton ratios. Small and medium-sized unstable atomic nuclei tend to undergo 𝛽 decay processes if they have too many protons to be stable. Small and medium-sized unstable atomic nuclei tend to undergo 𝛽 decay if they could gain a more stable 𝑁𝑍 ratio by having one of their neutrons transform into a proton. The green box shows the area of 𝛽 decay. It shows how small-to-medium sized nuclei tend to undergo 𝛽 decay if they have too many protons to be stable. The blue box shows the area of 𝛽 decay. It shows how small-to-medium sized nuclei tend to undergo 𝛽 decay if they could gain a more stable 𝑁𝑍 ratio by having one of their neutrons transform into a proton.

The heaviest unstable atomic nuclei tend to emit an 𝛼 particle (He)422+ that is made up of two protons and two neutrons. The area of alpha decay processes is represented with a red-colored box. It shows that large atomic nuclei tend to emit alpha particles. The figure has shown that unstable atomic nuclei can attain a more stable 𝑁𝑍 number ratio through different forms of radioactive decay. Unstable nuclei tend to undergo the type of radioactive decay that moves them toward the band of stability.

Example 3: Determining the Neutron-to-Proton Ratio for the Zirconium-94 Isotope

9440Zr is a stable isotope of zirconium. What is the neutron-to-proton ratio of this isotope?

  1. 0.431
  2. 1.351
  3. 1.741
  4. 2.351
  5. 0.741

Answer

The stability of a nucleus is primarily determined by its neutron-to-proton 𝑁𝑍 number ratio. The smallest atomic nuclei have a neutron-to-proton number ratio that is exactly or approximately equal to the number one 𝑁𝑍=1.0𝑁𝑍1.0or. Medium-sized atomic nuclei tend to be stable when 𝑁𝑍1.25.

The neutron-to-proton 𝑁𝑍 number ratio is obtained by dividing one number by another number. The first number is the number of neutrons in an isotope. The second number is its number of protons. Both of these values are easily obtained. The number of protons is always equal to the atomic number of an element. The number of neutrons is the difference of the atomic number and the mass number of an isotope. Atomic number and mass number are expressed with the X notation. The subscript 𝑍 term represents the atomic number. The 𝐴 term represents the superscript mass number.

The zirconium-94 isotope is represented with the 9440Zr notation. It has an atomic number of 40 and a mass number of 94. The 9440Zr nucleus contains 40 protons and 54 neutrons. The number of neutrons was obtained as the difference between the atomic number and the mass number. The 9440Zr nucleus must contain 54 neutrons because 9440=54.

The neutron-to-proton 𝑁𝑍 number ratio is the quotient of two numbers. It is the quotient of mass and atomic numbers for the 9440Zr isotope. The 𝑁𝑍 ratio for 9440Zr is determined as follows: 𝑁𝑍=5440=1.35.

The quotient matches the value in option B. We can use these statements to determine that option B is the correct answer to this question.

Example 4: Identifying Areas in Nuclear Stability Graphs and Determining How an Isotope Might Decay Based on Its Position on the Graph

The given plot shows the number of protons and neutrons for all the stable nuclei known to exist.

  1. What name is given to the area of the graph within which all stable nuclei are found?
    1. Belt of elements
    2. Magic number
    3. Valley of decay
    4. Strong nuclear zone
    5. Band of stability
  2. The orange circle on the plot represents the unstable isotope 13855Cs. How might this isotope decay to form a more stable nucleus?
    1. 𝛽 decay
    2. 𝛼 decay
    3. Electron capture
    4. 𝛽 decay
    5. Gamma emission

Answer

Part 1

The stability of any one nucleus is primarily determined by its neutron-to-proton 𝑁𝑍 number ratio. The smallest atomic nuclei are stable when they have a neutron-to-proton that is exactly or approximately equal to the number one 𝑁𝑍=1.0𝑁𝑍1.0or. The medium-sized atomic nuclei tend to be stable when 𝑁𝑍1.25, and the largest atomic nuclei are stable when 𝑁𝑍1.5. Graphs can be shown to represent the 𝑁𝑍 values of atomic nuclei that are stable. The area that includes data for the 𝑁𝑍 values of the stable atomic nuclei is sometimes called the band of stability. We can use these statements to determine that option E is the correct answer for this question.

Part 2

Unstable atomic nuclei will tend to undergo the type of radioactive decay that gives them the 𝑁𝑍 ratio of a stable atomic nucleus. Small and medium-sized unstable atomic nuclei tend to undergo 𝛽 decay if they have too many protons to be stable. Small and medium-sized unstable atomic nuclei tend to undergo 𝛽 decay if it could gain a more stable 𝑁𝑍 ratio by having one of their neutrons transform into a proton. The heaviest atomic nuclei tend to emit 𝛼 particles, which are made up of two protons and two neutrons. We can use these statements to determine that the 13855Cs isotope nucleus would undergo 𝛽 decay because it is a medium-sized atomic nucleus that has too many neutrons to be stable on long timescales. This line of reasoning suggests that option A is the correct answer to this question.

When dealing with individual particles or atomic nuclei, it is useful to have a system of units for mass and energy. The unified atomic mass unit is frequently used for the mass of protons, neutrons, electrons, and atomic nuclei. Unified atomic mass units are given the symbol (u) and can be converted into units of kilograms using the following relationships: 1()=1.66×10.uniedatomicmassunitukg

Using this system, the masses of a proton and a neutron can be stated as follows: 𝑚=1.007276,𝑚=1.008664.pnuu

There also exists a relationship between mass and energy, according to Einstein’s equation: Δ𝐸=Δ𝑚𝑐.

In this equation, Δ𝑚 is the change in mass in kg, Δ𝐸 is the change in energy in J, and 𝑐 is the speed of light in a vacuum 3×10/ms.

Often, the energies associated with atomic nuclei and particles are incredibly small. To make the numbers easier to work with, scientists developed a unit called the electron volt (eV): 1=1.60×10.eVJ

For great energies, the mega-electron volt can also be used: 1=1×10=1.60×10.MeVeVJ

By using Einstein’s equation, the unified atomic mass unit, the speed of light in a vacuum, and the electron volt, the following relationship can be determined as follows: Δ𝐸=Δ𝑚𝑐.

If Δ𝑚=1=1.66×10,ukg then Δ𝐸=1.66×10×3×10/Δ𝐸=1.49×10.kgmsJ

Converting from joules to mega-electron volts (MeV) gives us the following: 1.49×10×11.60×10=931.25.JMeVJMeV

Therefore, 1931.uMeV

Every single atomic nucleus is made up of some combination of protons and neutrons. It would be reasonable to assume that the mass of an atomic nucleus is equal to the cumulative mass of the individual protons and neutrons that combined together to make the nucleus. Scientists have found, however, that the mass of any one nucleus is lower than the cumulative mass of the individual protons and neutrons that combined together to make that nucleus.

The difference between the mass of the nucleus and the mass of individual nucleons is the mass defect. The mass defect (Δ𝑚) can be calculated with the following equation that uses the 𝑚p and 𝑚n symbols to represent the mass of individual protons and neutrons: Δ𝑚=𝑍𝑚+(𝐴𝑍)𝑚𝑚.pnnuc

The 𝑍 and (𝐴𝑍) terms represent the number of protons and neutrons in the atomic nucleus, and the 𝑚nuc term represents the mass of the atomic nucleus. The nuclear binding energy is the energy that holds the atomic nucleus together. By definition, the nuclear binding energy refers to the minimum energy required to break apart a nucleus into its individual proton and neutron constituents.

Definition: Nuclear Binding Energy

Nuclear binding energy is the minimum energy required to disassemble an atomic nucleus into unbound protons and neutrons.

The image below illustrates the definition of nuclear binding energy.

Using the equations described so far, we can determine the mass defect for the nucleus of a helium-4 atom and then determine the binding energy that holds the nucleus together.

The experimental mass for the nucleus of a helium-4 atom is 4.0015 u. In order to calculate the mass defect, the mass based on its constituent nucleons must first be calculated. The nucleus of an atom of helium-4 is composed of two neutrons and two protons; therefore, we get the following: Massofnucleonsuuu=2×𝑚+(2×𝑚)=[(2×1.00728)+(2×1.00866)]=4.03188.pn

The mass defect (Δ𝑚) is equal to the difference between the calculated mass of the helium-4 nucleus (4.0015 u) and the experimental value (4.03188 u): Massdefectcalculatedmassofnucleusexperimentalmassofnucleusmassdefectuuu(Δ𝑚)=(Δ𝑚)=4.031884.00150=0.03038.

Having calculated the mass defect, the binding energy for the nucleus can now be determined. There are two ways to do this. Having established the relationship 1931uMeV, we can simply multiply the mass defect by 931 MeV to obtain a binding energy of 28.3 MeV. However, let’s work through the full calculation.

To convert the mass defect in energy, we can use Einstein’s equation: Δ𝐸=Δ𝑚𝑐.

In this equation, Δ𝑚 is the change in mass and 𝑐 is the speed of light in a vacuum. However, Δ𝑚 needs to be in units of kilograms. As 1=1.66×10ukg, then the mass defect can be converted from unified atomic mass units to kilograms through the following: (0.03038)×1.66×101=5.0381×10.ukgukg

Substituting the values into Einstein’s equation gives the following: Δ𝐸=5.0381×103×10/Δ𝐸=4.53429×10.kgmsJ

This value can then be converted from joules into electron volts (eV). The conversion can be achieved by dividing the energy term by the number of joules in a single electron volt (1.60×10 J): 𝐸=4.53429×10×1(1.60×10)𝐸=28303933.JeVJeV

This value can also be expressed in units of MeV if it is divided by a value of 1×10 because 1=1×10MeVeV: 𝐸=28303933×11×10𝐸=28.3(1).eVMeVeVMeVtodecimalplace

However, if we want the binding energy per nucleon, then we need to divide this value by the total number of neutrons and protons in the nucleus. For a helium-4 nucleus, there are four nucleons—2 protons and 2 neutrons. Therefore, the binding energy per nucleon can be calculated as follows: bindingenergypernucleonMeVMeVdecimalplace=28.34=7.1(1).

The value seems to be absolutely enormous when compared with the energy that is released during chemical reactions. The formation of one water molecule from hydrogen (H)2 and oxygen atoms (O)122 releases about 3.0 eV of energy, and the formation of one carbon dioxide (CO)2 molecule releases about 4.1 eV of energy. The formation of a helium nucleus from individual proton and neutron particles releases over a million times more energy than the energy released during the formation of a single water molecule or a single carbon dioxide molecule.

Example 5: Determining the Binding Energy per Nucleon for an Atom of Lithium-7

What is the average binding energy per nucleon in units of mega-electron volts for an atom of lithium-7 with an observed mass of 7.01435 u? Give your answer to 2 decimal places. Take the masses of a proton and a neutron to be 1.00728 u and 1.00866 u respectively.

Answer

In order to determine the binding energy per nucleon, the mass defect for an atom of lithium-7 must first be determined.

Each atomic nucleus of lithium-7 has to be made up of a total of seven nucleons. As the atomic number of lithium is 3, then three of these nucleons are protons. Therefore, each nucleus must contain four neutrons because 73=4.

The mass of a neutron in unified atomic mass units is given in the question. Using this, the mass of four neutrons is as follows: 4×(1.00866)=4.03464.uu

The mass of a proton in unified atomic mass units is also given in the question. The mass of three protons is therefore as follows: 3×(1.00728)=3.02184.uu

The total mass of four neutrons and three protons is found by adding these two mass values together: 4.03464+3.02184=7.05648.uuu

The calculated atomic mass for an atom of lithium-7 is therefore 7.05648 u. The mass defect for the atomic nucleus of a lithium-7 atom can be calculated if the cumulative mass of three protons and four neutrons is taken away from the observed lithium-7 atomic mass value of 7.01435 u: Massdefectuuu=7.056487.01435=0.04213.

The mass defect can be converted into kg units to make it easier to calculate energy values, using the conversion that 1=1.66×10ukg: 0.04213×1.66×101=6.99358×10.ukgukg

The Einstein relation Δ𝐸=Δ𝑚𝑐 can be used to determine the associated energy value: 𝐸=6.99358×10×3×10/.kgms

This energy value is shown below: 𝐸=6.294222×10.J

This value can be converted into electron volts if it is divided by 1.60×10 J: 𝐸=6.294222×10×1(1.60×10)=39289775.28.JeVJeV

This value can be converted into MeV if it is divided by a value of 1×10: 𝐸=39289775.28×11×10=39.28977528.eVMeVeVMeV

The energy per nucleon can be determined if this value is divided by the total number of nucleons: BindingenergypernucleonMeVMeV=39.289775287=5.61282504.

The binding energy per nucleon is 5.61 MeV to two decimal places.

Key Points

  • The strong nuclear force makes neutrons and protons bind together in the atomic nucleus.
  • The strong nuclear force is powerfully attractive at very short ranges.
  • The strong nuclear force affects groups of protons, groups of neutrons, and adjacent neutron and proton particles.
  • The stability of an atomic nucleus is primarily determined by its neutron-to-proton 𝑁𝑍 ratio.
  • The band of stability is the domain of neutron vs proton number graphs that shows the 𝑁𝑍 ratio of stable atomic nuclei.
  • Unified atomic mass units (u) can be used for the mass of nucleons, while energy can be given in units of electron volts (eV) and mega-electron volts (MeV).
  • 1=1.66054×10ukg, and 1=1.60×10eVJ.
  • The difference between the mass of the nucleus and the mass of individual protons and neutrons is the mass defect.
  • The Einstein relation Δ𝐸=𝑚𝑐 can be used to determine the binding energy of an atomic nucleus from its mass defect value.

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