The binding energy of nucleons in a nucleus can be affected by the pairwise Coulomb repulsion. Assum

Q: Why is the strong nuclear force considered charge-independent?

Experimental evidence from scattering experiments shows that the nuclear force between two protons, two neutrons, or a proton and a neutron is almost identical in strength, provided they are in the same quantum state. The strong force acts on 'color charge' (quarks), not electrical charge.

Q: Why does the formula use $Z(Z-1)$ instead of $Z^2$?

A proton cannot repel itself! If there are $Z$ protons, each proton interacts with the other $(Z-1)$ protons. The total number of unique interacting pairs is $Z(Z-1)/2$, which leads to the $Z(Z-1)$ dependence.

Q: If $E_b^n - E_b^p$ is positive, does that mean neutrons are always harder to remove?

Yes, in a given nucleus, it generally requires more energy to remove a neutron than a proton because the proton is already being pushed out by the Coulomb repulsion of the other protons.

Q: Why does the radius depend on $A^{1/3}$?

Nucleons are essentially incompressible and pack together like hard spheres. The total volume of the nucleus is proportional to the total number of nucleons $A$. Since Volume $\propto R^3$, it follows that $R \propto A^{1/3}$.

Q: In beta plus decay, if a proton turns into a neutron, doesn't the strong force change?

The total number of nucleons $A$ remains the same, so the overall strong nuclear binding is roughly constant. However, the loss of a positive charge significantly reduces the Coulomb repulsion, which is the primary reason the remaining protons become more tightly bound.

Question

The binding energy of nucleons in a nucleus can be affected by the pairwise Coulomb repulsion. Assume that all nucleons are uniformly distributed inside the nucleus. Let the binding energy of a proton be $E_b^p$ and the binding energy of a neutron be $E_b^n$ in the nucleus.
Which of the following statement(s) is(are) correct?

Options:

(A) $E_b^n - E_b^p$ is proportional to $Z(Z - 1)$ where $Z$ is the atomic number of the nucleus.

(B) $E_b^n - E_b^p$ is proportional to $A^{-\frac{1}{3}}$ where $A$ is the mass number of the nucleus.

(C) $E_b^n - E_b^p$ is positive.

(D) $E_b^p$ increases if the nucleus undergoes a beta decay emitting a positron.

Sigma Percentile · Accepted Answer

### The Heart of the Atom

Imagine you are shrinking down, diving deep into the center of an atom. You bypass the electron clouds and arrive at the incredibly dense, chaotic core: the nucleus.

Here, protons and neutrons are packed tightly together.

The glue holding this chaotic assembly together is the **strong nuclear force**. This force is incredibly powerful, but it has a beautiful simplicity: it is completely charge-independent. It treats a proton and a neutron exactly the same.

However, there is a catch. The protons carry a positive electric charge.

While the strong force is desperately trying to hold everything together, the **Coulomb force** is actively trying to tear the protons apart. This constant internal battle between the attractive strong force and the repulsive Coulomb force is the key to understanding nuclear stability.

### The Concept of Binding Energy

What happens if we try to pull a single nucleon out of this tightly bound system?

The energy we must supply to overcome the attractive forces and remove the nucleon is called its **binding energy**.

Let's think about extracting a neutron. To pull a neutron out, we only have to fight against the strong nuclear force. There is no electrical repulsion helping or hindering us.

Now, imagine extracting a proton. We still have to fight the strong nuclear force, but this time, the other protons in the nucleus are actually helping us! Their Coulomb repulsion pushes the proton outward.

Because the nucleus is already trying to expel the proton, the energy we need to supply—the proton's binding energy, $E_b^p$—is less than the neutron's binding energy, $E_b^n$.

Mathematically, the difference between their binding energies is exactly equal to the electrostatic potential energy of that single proton:

$$E_b^n - E_b^p = E_{	ext{Coulomb}}$$

### Calculating the Coulomb Penalty

To find out exactly how much this Coulomb penalty is, we need to look at the electrostatic potential energy of the nucleus.

If we model the nucleus as a uniformly charged sphere containing $Z$ protons, the total electrostatic potential energy $U$ is given by:

$$U = \frac{3}{5} \frac{1}{4\pi\epsilon_0} \frac{Z(Z-1)e^2}{R}$$

Notice the $Z(Z-1)$ term. This arises because each of the $Z$ protons interacts with the remaining $(Z-1)$ protons.

The Coulomb energy associated with just *one* proton is proportional to the total energy divided by $Z$. Therefore, the energy difference shares this exact proportionality:

$$E_b^n - E_b^p \propto \frac{Z(Z-1)}{R}$$

### The Role of the Nuclear Radius

We aren't quite finished yet. The radius of the nucleus, $R$, is not a constant. It depends on how many nucleons are packed inside.

The empirical formula for the nuclear radius is:

$$R = R_0 A^{1/3}$$

Here, $A$ is the mass number (total number of protons and neutrons), and $R_0$ is a constant (approximately $1.2 	ext{ fm}$).

Let's substitute this radius back into our energy difference proportionality:

$$E_b^n - E_b^p \propto \frac{Z(Z-1)}{R_0 A^{1/3}}$$

By bringing the $A^{1/3}$ term to the numerator, we get:

$$E_b^n - E_b^p \propto Z(Z-1) A^{-1/3}$$

This elegant result immediately proves that **Option (A)** and **Option (B)** are absolutely correct!

### The Positivity of the Difference

Let's take a step back and look at the physical meaning of our result.

Is the quantity $E_b^n - E_b^p$ always positive?

For any nucleus with more than one proton ($Z > 1$), the term $Z(Z-1)$ is strictly positive. The mass number $A$ is also positive.

Therefore, the Coulomb repulsion always provides a positive outward push. A neutron will always be bound more tightly than a proton in the same nucleus.

This confirms that **Option (C)** is also correct.

### The Beta Plus Decay Twist

Finally, let's analyze a dynamic scenario: beta plus ($\beta^+$) decay.

In a $\beta^+$ decay, a proton inside the nucleus converts into a neutron, emitting a positron ($e^+$) and an electron neutrino.

What happens to the environment inside the nucleus after this event?

The atomic number drops from $Z$ to $Z-1$. There is now one less proton contributing to the internal Coulomb repulsion.

With the repulsive forces weakened, the remaining protons are not being pushed apart as violently. The strong nuclear force can hold onto them more effectively.

Because the remaining protons are held more tightly, it becomes harder to extract them. Therefore, the binding energy of a proton, $E_b^p$, **increases**.

This brilliant piece of physical reasoning confirms that **Option (D)** is correct.

Select Answer:

Visualized Solution

The Nucleus and its Nucleons

Binding Energy Difference

Electrostatic Potential Energy

Radius and Mass Number

Positivity of the Difference

Beta Plus Decay

Conclusion

The Way Forward

The Sigma Insight: Nucleus and Nuclear Reaction

The Heart of the Atom

The Concept of Binding Energy

Calculating the Coulomb Penalty

The Role of the Nuclear Radius

The Positivity of the Difference

The Beta Plus Decay Twist

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Select Answer:

Visualized Solution

The Nucleus and its Nucleons

Binding Energy Difference

Electrostatic Potential Energy

Radius and Mass Number

Positivity of the Difference

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Conclusion

The Way Forward

The Sigma Insight: Nucleus and Nuclear Reaction

The Heart of the Atom

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The Role of the Nuclear Radius

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The Beta Plus Decay Twist

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