What are the important inferences from the average binding energy curve ?

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Q. What are the important inferences from the average binding energy curve ?

Puducherry TnboardTamil Nadu HSC (DGE) Board 2020Subjective· 3mImportance★★★★★

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Concept understanding — Nuclear Binding Energy per Nucleon

The Intuition: Why Are Nuclei Stuck Together?

Imagine a nucleus as a tight cluster of protons and neutrons. Protons all carry positive charge, so they should be violently repelling each other. Yet the nucleus holds together. That means there must be an even stronger attractive force — the strong nuclear force — acting between nucleons (protons and neutrons). But here's the catch: this force is extremely short-range. A nucleon only feels the pull from its immediate neighbours, not from nucleons far across the nucleus.

So the nucleus is a tug-of-war. The strong force pulls nucleons together, but the electrostatic repulsion between protons tries to blow the nucleus apart. For a nucleus to be stable, the strong force must win.

Now, if you want to break a nucleus apart into its individual protons and neutrons, you have to do work against the strong force — you have to supply energy. That energy, once supplied, is stored in the separated nucleons as extra mass. This is the core idea: the mass of a stable nucleus is always less than the sum of the masses of its individual protons and neutrons. The missing mass is called the mass defect, and the energy equivalent of that missing mass (via $E = mc^2$ ) is the binding energy.

Note

Binding energy is the energy you must put in to completely separate a nucleus into its constituent nucleons. It is not energy stored inside the nucleus like fuel; it is the energy that holds the nucleus together.

The Precise Statement

For a nucleus with $Z$ protons, $N$ neutrons, and mass $m_{\text{nucleus}}$ :

\text{Mass defect } \Delta m = Z m_p + N m_n - m_{\text{nucleus}}

where $m_p$ and $m_n$ are the masses of a free proton and neutron. Then the total binding energy is:

\text{BE} = \Delta m \, c^2

But a big nucleus has more nucleons, so its total binding energy will naturally be larger. To compare how tightly bound different nuclei are, we use binding energy per nucleon:

\text{BE/A} = \frac{\text{BE}}{A}

where $A = Z + N$ is the mass number. This is the average energy you'd need to remove one nucleon from the nucleus. A higher BE/A means a more stable nucleus.

The Famous Curve: Why Iron is Special

If you plot BE/A against mass number $A$ , you get a curve that rises steeply for light nuclei, peaks at iron-56 (about 8.8 MeV per nucleon), and then slowly falls for heavier nuclei.

Important

Iron-56 has the highest binding energy per nucleon of any nuclide. It is the most stable nucleus in nature.

This shape tells you two things:

1. Fusion of light nuclei releases energy. If you take two very light nuclei (like hydrogen isotopes) and smash them together to form a medium-mass nucleus, the product has a higher BE/A than the reactants. The difference in binding energy is released as kinetic energy of the products. This is how stars burn.

2. Fission of heavy nuclei releases energy. If you split a very heavy nucleus (like uranium-235) into two medium-mass fragments, those fragments also have higher BE/A than the original. Again, the difference comes out as energy. This is how nuclear reactors work. …

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Figure — The entire question is 'inferences from the average binding-energy curve', and fig-13-1 is exactly that curve

The average binding-energy-per-nucleon curve peaks around iron ( $A\approx56$ ), explaining why nuclear fusion of light elements and fission of heavy elements both release energy, and why medium-mass nuclei are the most stable.

Sharp initial rise, then a broad, nearly flat maximum: for very light nuclei ( $A<20$ ), the binding energy per nucleon $(BE/A)$ rises rapidly and irregularly with $A$ ; it then rises more gradually and reaches a broad maximum of about $8.7-8.8$ MeV/nucleon for nuclei with mass number roughly between $30$ and $120$ , peaking near $A\approx56$ (iron group).
Gradual decrease for heavy nuclei: beyond about $A=120$ , $BE/A$ decreases slowly, falling to about $7.6$ MeV/nucleon for the heaviest natural nuclei (like uranium) — this decrease is due to the increasing Coulomb repulsion among the growing number of protons, which is not compensated by the short-ranged attractive nuclear force.
Nuclei of intermediate mass number are the most stable, since they have the highest binding energy per nucleon.
Nearly constant plateau ⇒ saturation of nuclear force: the fact that $BE/A$ is roughly constant over a wide range of $A$ shows that each nucleon interacts (via the strong nuclear force) with only a limited number of its immediate neighbours, not with every other nucleon in the nucleus — i.e. the nuclear force is short-ranged and saturates. …

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Q.What are the important inferences from the average binding energy curve ?

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