Nuclear Forces, Defect of Mass and Binding Energy

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Nucleur Physics Learning Material
Tutorial IDTitleTutorialVideo
20.2Nuclear Forces, Defect of Mass and Binding Energy

In this Physics tutorial, you will learn:

  • What is nuclear force? Where does it act? What kind of force is it (attractive or repulsive)?
  • What force does balance the nuclear force?
  • What is mass-energy equivalence? When does it occur?
  • What is defect in mass? How to calculate it?
  • What is binding energy? How to calculate it?
  • How does the number of neutrons N vary with the atomic number Z in various elements?
  • What is the excess of neutrons? When does it occur?


Despite in micro-world there is another approach on natural laws compared to classical physics, we still rely on classical laws to explain the new phenomena observed in modern physics. We will explain in this article new concepts such as nuclear force, mass-energy equivalence, binding energy, mass defect, excess of neutrons, nuclear field etc., which are all specific but have their logics based on previous information explained in classical physics. Therefore, it is better to take a look at concepts such as mass, energy, momentum and impulse we have discussed in other sections before reading this article.

Nuclear Force

Since in any atomic nucleus there are Z positively charged protons, it is evident that an attracting force must exist to keep the nucleons held together (i.e. to keep the nucleus stable), as opposed to repelling electric force caused by like charged protons. This force known as nuclear force, must balance the electric force acting between protons. Obviously, in order to perform its function (i.e. to keep nucleons inside the nuclei), nuclear force must act in very short distances, not more than 10-15 m, as it must not exceed the dimensions of atomic nuclei.

Despite nucleons may move in respect to each other, there is a constant equilibrium between the above two forces, otherwise nucleons would collide with each other. Experiments show that the net force between nucleons is more repulsive at short distances (electric force is greater than nuclear force) and more attractive at medium distances (nuclear force is greater than electric force). In other words, there are two forces acting between nucleons: one is the electric (repulsive) force and the other is the nuclear (attractive) one. The distance between nucleons determines which of these forces overcomes the other. Logically, there must be an equilibrium position in which the nucleus acquires stability and where the magnitudes of the above two forces are equal (the net force is zero).

As noted above, the two forces (attractive and repulsive) act at distances smaller than the dimensions of nuclei. For bigger distances, both forces - especially the attracting (nuclear) force we are concerned with in this article - point towards zero. The nuclear force vs distance correspondence is given in the graph below.

Physics Tutorials: This image provides visual information for the physics tutorial Nuclear Forces, Defect of Mass and Binding Energy

The effect of (repelling) electric force increases with the increase in the atomic number Z. For this reason, elements heavier than Uranium (Z > 92) manifest lack of stability in their nuclei and tend to break down in various ways. In those nuclei with such a high number of protons, the electric repulsion decreases less than nuclear force with the increase in distance compared to lighter nuclei. As a result, a given proton - while experiencing the electric repulsion caused by all the other protons in the nucleus - can experience the nuclear attraction caused only by the adjacent nuclei. Hence, the nucleus becomes unstable, as the proton tends to leave it.

Obviously, this approach is too simplified; the stability of nuclei depends on many other factors as well. The quantum analysis of all phenomena occurring inside atomic nuclei proves that the above rule of stability is not always valid; there are some heavier nuclei (Z = 114) in specific conditions that can be stable too. The main drawback of the above explanation on nuclear force consists on the fact that it cannot provide answer to a question that can naturally arise in everybody's mind: "Giving that one proton and one neutron are able to create a stable system (for example in hydrogen atoms), why two neutrons cannot create a system with such a stability?" This looks quite strange, given that two neutrons do not exert any electric force between them, as both of them are neutral. This question tells us that the properties of nuclear forces are more complex than we may think and they do not depend only from distance. Other factors such as the type of nucleon, orientation of spin (net nuclear angular momentum) etc., affect the magnitude of nuclear forces and the behavior of nucleons in general.

Since nucleons always have a certain non-zero distance between them (see the graph above), their interaction occurs in distance, that is, it takes place by means of distant forces (we already know that electric force is a force that acts in distance). As such, the nuclear interaction occurs not through direct contact between nucleons but by means of fields and their corresponding carriers. We will see in Section 21 that these carriers are some elementary particles known as quarks, which are smaller than protons themselves (quarks are particles that carry fractional electric charges). In fact, a single proton contains a certain number of quarks. Moreover, other elementary particles such as mesons are involved in holding the nucleons together. Therefore, it is evident that the more we delve into the study of the nucleus, the more we realize that our initial concept consisting in the consideration of atom as the smallest particle of matter was wrong; it was too far from the truth. The number of such elementary particles that are smaller than protons, neutrons and electrons is too large that nowadays there are special tables designated with the purpose to classify them, similarly to classification of elements in the periodic table.

Mass - Energy Equivalence

Consider an object of mass m moving linearly at a very high uniform speed, very close to the speed of light. Obviously, we denote this speed by c. Since the speed is uniform, there is a constant force F acting on the object during its motion. As a result, an energy E and momentum p are induced in the object, where

E = F ∙ d


p = m ∙ c

Here, d represents the distance travelled by the object during the process. We have used the scalar versions of formulae as we assumed the motion as linear.

Giving that the momentum gained during the process is equal to the impulse of object, we have

p = F ∙ t

where t is the duration of moving process. Thus, rearranging the last formula for F and substituting it in the first formula, we obtain

F = p/t ⇒ E = p ∙ d/t
= p ∙ c
= m ∙ c ∙ c

Thus, we obtain

E = m ∙ c2

This is the famous equation of mass-energy equivalence, introduced by Einstein. It implies that, despite the total mass of a system may change, the total energy and momentum remain constant. For example, if an electron and a proton collide with each other, this process destroys the mass of both particles but generates a large amount of energy in the form of photons. Now, you understand where does the m0 · c2 term discussed in article 18.6 when dealing with relativistic energy comes from. To avoid confusion we denoted this energy by ε instead of E and explained that it represents the stored energy in the object when it is at rest (we called it rest energy).

As you know, there are other forms of energy than just the rest energy and the kinetic energy. There is heat energy, chemical energy, binding energies of atoms and nuclei, etc. It turns out that all forms of energy are reflected in the total mass of the body. So although we have justified E = mc2 in terms of the kinetic energy, the mass-energy equivalence is quite a bit more general.

Example 1

How much energy a 200 g apple at rest would release if it was totally destroyed (not existed anymore)?

Solution 1


m = 200 g = 0.2 kg
c = 300 000 km/s = 3 × 108 m/s
E = ?

The problem implies that the entire mass of the apple converts to energy. From the formula of mass-energy equivalence

E = m ∙ c2

we obtain after substitutions:

E = (0.2 kg) ∙ (3 × 108 m/s)2
= 1.8 × 1016 J

This value is enormous; it is 30 times greater than the energy released during the atomic bombing of Hiroshima (6 × 1014 J). However, it is not easy to convert mass into energy; this process requires very special appliances and the cost of this operation is extremely high.

Binding Energy

We know that the conversion between atomic mass unit (u) and kilogram (kg) is

1 u = 1.66054 × 10-27 kg

From the mass-energy equivalence, we have for its energy equivalent (giving that mass of proton is mp = 1.6726 × 10-27 kg and that of neutron is mn = 1.6749 × 10-27 kg):

εp = mp ∙ c2
= (1.6726 × 10-27 kg) ∙ (3 × 108 m/s)2
= 1.50534 × 10-10 J


εn = mn ∙ c2
= (1.6749 × 10-27 kg) ∙ (3 × 108 m/s)2
= 1.50741 × 10-10 J

Sometimes, the mass of micro particles is expressed in MeV/c2 (Mega electronvolt per square of the light speed). Thus, the conversion factor between atomic mass unit u and MeV/c2 is

1 u = 931.494 MeV/c2

(try to prove it), while the mass of a free proton or neutron is

mp ≈ mn = 938.272 MeV/c2

In the microscopic world, the way things work (the laws governing microscopic phenomena) are very different from those we have discussed in classical physics. Let's consider a stable nucleus. This means its nucleons cannot be considered as free particles as they are interacting inside the nucleus by means of nuclear forces caused by the nuclear field, which makes them possess nuclear energy. Therefore, a nucleus with A nucleons possesses energy in two forms: in the mass of nucleons and energy of nuclear field.

From the law of energy conservation and the mass-energy equivalence, we expect a difference in mass between the sum of free individual nucleons and the mass of nucleus made by the same nucleons. The process of free nucleons merging to form a stable nucleus results requires that part of energy conserved in the form of mass converts into pure nuclear energy. In simpler words, some of the nucleons mass is converted into energy during the merging process of individual nucleons to form an atomic nucleus. Such a difference in mass is also confirmed experimentally. In scientific terms, it is known as "mass defect" or "binding energy", depending on the approach. From the law of energy conservation, it is clear that binding energy, Eb is calculated through the equation:

Eb = ∆m·c2 = Z ∙ mp ∙ c2 + (A-Z) ∙ mn ∙ c2 - M ∙ c2

where mp is the mass of proton, mn the mass of neutron and M is the mass of the whole nucleus. Simplifying both sides by c2, we obtain the formula for the mass defect:

∆m = Z ∙ mp + (A - Z) ∙ mn - M

Based on the above analysis, it is clear that the system is bound when the mass defect is positive (Δm > 0), i.e. when the mass of nucleus is smaller than the sum of masses of the individual nucleons. Dividing the mass defect by the atomic mass A, we obtain the average binding energy per nucleon Eb(A), which is function of atomic mass. The following figure shows the dependence of binding energy from atomic mass for several chemical elements.

Physics Tutorials: This image provides visual information for the physics tutorial Nuclear Forces, Defect of Mass and Binding Energy

Binding energy shows how stable the nucleus is. The higher the binding energy the more stable a nucleus is. From the graph you can easily find that metal elements such as iron (Fe), nickel (Ni), etc., are the most stable elements, as they have a higher binding energy per nucleon and to break this stability, an energy at least equal to binding energy is required from an external source. These elements are in the middle part of periodic table.

Considering the nuclei stability as an equilibrium between attractive nuclear forces and electric repulsive forces, it is clear that for two isobars (Z is different and A is equal), the nucleus with the lowest atomic number Z (the one with the highest N therefore) is more stable. This is because such nuclei contain less protons and as a result, the electric force between them is smaller. Hence, the nuclear force needed to balance the system is smaller as well. Remember that the most stable system is the one which possesses the least energy possible. The graph below shows how the number of neutrons N varies with the atomic number Z in various elements.

Physics Tutorials: This image provides visual information for the physics tutorial Nuclear Forces, Defect of Mass and Binding Energy

From this graph, it is obvious that in general N > Z. This fact is known as excess of neutrons. As you see, the range of radioactive isotopes is much greater than non-radioactive ones. This is obvious, since radioactive materials are less stable. We will elaborate further this property in the next article.

Example 2

Calculate the binding energy (in joules) for the nucleus of helium (42He) atoms. Use the periodic table in article 20.1 as a reference.

Solution 2

As we know, the binding energy Eb of an atomic nucleus is:

Eb = ∆m ∙ c2

where Δm is the mass defect and c is the light speed in vacuum.

Likewise, we know that mass defect of nucleus is:

∆m = Z ∙ mp + N ∙ mn - M

where N is the number of neutrons and M is the mass of nucleus.

However, in the tables of isotopes only the values of neutral atomic nuclei are given and not those of nuclear masses. Therefore, the above formula must be written in terms of neutral atomic mass ma. Giving that

M = ma - Z ∙ me

where me is the mass of electron, we write

∆m = Z ∙ mp + N ∙ mn - (ma-Z ∙ me )
= Z ∙ mp + Z ∙ me + N ∙ mn - ma
= Z ∙ (mp + me ) + N ∙ mn - ma

Giving that

mp + me = m(11H)

we obtain:

∆m = Z ∙ m(11H) + N ∙ mn - m(42He)

We have:

Z = 2


N = A - Z
= 4 - 2
= 2

Likewise, we have the following constants taken from periodic table and other data given in the last two articles:

m(11H ) = 1.008 u
mp = 1u
mn = 1.008 u
m(42He) = 4.003 u

Hence, after substitutions we obtain for the mass defect of helium nucleus:

∆m = 2 ∙ 1.008 u + 2 ∙ 1.008 u - 4.003 u
= 4.032 u - 4.003 u
= 0.029 u

When converted into kilograms, this value becomes

∆m = 0.029 u ∙ 1.66054 × 10-27 kg/u
= 4.8156 × 10-29 kg

Therefore, the binding energy of helium nuclei is

Eb = ∆m ∙ c2
= (4.8156×10-29 kg) ∙ (3 × 108 m/s)2
= 4.334 × 10-12 J

Remark! The mass values in atomic mass units (u) for proton, neutron and hydrogen atom are all rounded. In this example, the mass of neutron is given as equal to the mass of hydrogen atom. However, the mass of neutron is slightly greater than the mass of hydrogen atom (which contains only one proton in the nucleus and one electron revolving in the orbit).

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