The famous Einstein’s theory of special relativity established the fact that mass is another form of energy. Also, you can convert the mass-energy into other forms of energy. This opened the doors to a better understanding of nuclear masses and the interaction of nuclei with each other. In this article, we will look at the binding energy of a nucleus which is essential to understand the nuclear fission and fusion processes.

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According to Einstein’s mass-energy equivalence relation, we know that,

E = mc^{2} … (1)

where c is the velocity of light in vacuum and is ≅ 3 x 10^{8} m/s. Also, E is the energy equivalent of the mass, ‘m’. Experimental studies of nuclear reactions between nucleons, nuclei, electrons and other more recently discovered particles. According to the Law of Conservation of Energy, we know that in a reaction, the initial and final energy is the same provided we take the mass-energy into consideration.

## Nuclear Binding Energy

We know that the nucleus is made up of protons and neutrons. So, logically, the mass of the nucleus = the sum of masses of the protons and neutrons, right? Not really! The nuclear mass (M) is always less than this sum. To e=understand this better, let’s look at an example,

^{16}_{8}O has 8 protons and 8 neutrons. Now,

- Mass of 8 neutrons = 8 × 1.00866 u
- Mass of 8 protons = 8 × 1.00727 u
- Mass of 8 electrons = 8 × 0.00055 u

Therefore the expected mass of ^{16}_{8}O nucleus = = 8 × 2.01593 u = 16.12744 u.

We know from mass spectroscopy experiments that the atomic mass of 168O is 15.99493u. Subtracting the mass of 8 electrons from this, we get the experimental mass of ^{16}_{8}O nucleus = 15.99053u (15.99493u – [8 x 0.00055u]). Hence, we see that there is a difference between the two numbers of 0.13691u (16.12744 u – 15.99053u).

In simple words, the mass of the ^{16}_{8}O nucleus is less than the total mass of its constituents by 0.13691u. This difference in mass of a nucleus and its constituents is called the mass defect (ΔM) and is given by

ΔM = [Zm_{p} + (A – Z)m_{n}] – M … (2)

### So, what exactly does mass defect mean?

*The mass of an oxygen nucleus < the sum of masses of its protons and neutrons (in an unbounded state). **Therefore, t**he equivalent energy of the oxygen nucleus < the sum of the equivalent energies of its constituents.*

We can also say that if you want to break down an oxygen nucleus into 8 protons and 8 neutrons, then you must provide the extra energy (ΔMc^{2}). The relation between this energy (E_{b}) to the mass defect (ΔM) is derived from Einstein’s mass-energy equivalence relation {equation (1)}. Therefore,

E_{b} = ΔMc^{2} … (3)

In other words, if certain protons and neutrons are brought together to form a nucleus of a certain charge and mass, then an energy E_{b} is released in the process. This energy is called the **Binding Energy** of the nucleus. So, if we want to separate a nucleus into protons and neutrons, then we will need to provide an energy E_{b} to the particles.

## Binding Energy per Nucleon

A more useful measure of the binding between protons and neutrons is the binding energy per nucleon or E_{bn}. It is the ratio of the binding energy of a nucleus to the number of nucleons in the nucleus:

E_{bn} = E_{b}/A … (4)

where Eb is the binding energy of the nucleus and A is the number of nucleons in it. So, the binding energy per nucleon is the average energy per nucleon needed to separate a nucleus into its individual nucleons. Let’s look at a plot of the binding energy per nucleon versus the mass number for a large number of nuclei:

Here is what we can observe:

- The maximum binding energy per nucleon is around 8.75 MeV for mass number (A) = 56.
- The minimum binding energy per nucleon is around 7.6 MeV for mass number (A) = 238.
- For 30 < A < 170, E
_{bn}is nearly constant. - Ebn is low for both light nuclei (A < 30) and heavy nuclei (A > 170)

### Conclusion 01

The force is attractive in nature and very strong producing a binding energy of around a few MeV per nucleon.

### Conclusion 02

- Why is E
_{bn}nearly constant in the range 30 < A < 170? The answer is simple – the nuclear force is short-ranged. Now, imagine a really large nucleus. A nucleon (N_{A}) in this nucleus will be influenced only by some of its neighbours. These neighbours are those which lie in the short-range of the nuclear force. - This means that all nucleons beyond the range of the nuclear force form N
_{A}will have no influence on the binding energy of N_{A}. So, we can conclude that if a nucleon has ‘p’ neighbours within the range of the nuclear force, then its binding energy is proportional to ‘p’. - In the same large nucleus, if we increase the mass number (A) by adding nucleons, it will not change the binding energy of N
_{A}. Why? Because, in a large nucleus, most of the nucleons lie inside it and not on the surface. Hence, the change in binding energy, if any, would be negligibly small. - Remember, the binding energy per nucleon is a constant and is equal to pk, where k is a constant having the dimensions of energy. Also, the property that a given nucleon influences only nucleons close to it is also referred to as saturation property of the nuclear force.

### Conclusion 03

Next, imagine a very heavy nucleus having A = 240. This has a low binding energy. Therefore, if a nucleus A = 240 breaks down into two A = 120 nuclei, then the nucleons get bound more tightly. Right? Also, in the process energy is released. This concept is used in Nuclear Fission.

### Conclusion 04

On the other hand, imagine two very light nuclei with A < 10. If these two nuclei were to join to form a heavier nucleus, then the binding energy per nucleon of the fused and heavier nucleus is more than the E_{bn} of the lighter nuclei. Right? So, the nucleons are more tightly bound post fusion. And, energy is released in the process. This is how the Sun works!

## Solved Examples for You

Question: Obtain the binding energy (in MeV) of a nitrogen nucleus (^{14}_{7}N) , given m (^{14}_{7}N) =14.00307u.

Solution: A nucleus of Nitrogen contains 7 protons and 7 neutrons. Therefore, the mass defect of this nucleus is,

ΔM = 7m_{p} + 7m_{n} – m (^{14}_{7}N)

Now,

Mass of a proton = m_{p} = 1.007825u

Mass of a neutron = m_{n} = 1.008665u

Therefore,

ΔM = (7 x 1.007825u) + (7 x 1.008665u) – 14.00307u = 0.11236u

Now, we know that 1u = 931.5 MeV/c_{2}. Therefore,

ΔM = 0.11236 x 931.5 MeV/c^{2}

The binding energy of the nucleus is,

Eb = ΔMc^{2
}= 0.11236 x 931.5 (MeV/c^{2}) x c^{2
}= 0.11236 x 931.5 MeV

= 104.66334 MeV.

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