Question

What is the binding energy per nucleon? Explain the observations and draw the conclusions by plotting a graph between the binding energy per nucleon versus the mass number for many nuclei.

Accepted Answer

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

$E_{b n} = \frac{E_{b}}{A}$

where $E_{b}$ 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 nucleons. Let’s look at a plot of the binding energy per nucleon versus the mass number for many 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.
$E_{bn}$ 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 binding energy of around a few $Mev$ per nucleon.
Conclusion 02
Why is $E_{b n}$ 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 (NA) 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 from NA will have no influence on the binding energy of NA. 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 NA. 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 referred to as the saturation property of the nuclear force.
Conclusion 03
Next, imagine a very heavy nucleus having $A = 240$ . This has 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 $E_{b n}$ 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!

What is the binding energy per nucleon? Explain the observations and draw the conclusions by plotting a graph between the binding energy per nucleon versus the mass number for many nuclei.

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