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 [2022]

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? [2022]

Smart Achievers · Accepted Answer

(1, 2, 4)

Binding energy of proton and neutron due to nuclear force is the same. So, the difference is due to electrostatic potential repulsion energy, and it is positive.

So $E_{b}^{p} - E_{b}^{n} =$ Electrostatic potential energy

Number of proton pairs $= {}^{Z}C_{2} = \frac{Z (Z - 1)}{2}$

So, Repulsion energy $\propto \frac{Z (Z - 1)}{2} \times \frac{1}{4 π ε_{0}} \cdot \frac{e^{2}}{R},$

$R =$ radius of the nuclei

$E_{b}^{p} - E_{b}^{n} \propto Z (Z - 1)$ $\Rightarrow (1) is correct.$

As, $R = R_{0} A^{1 / 3}$

So, $E_{b}^{p} - E_{b}^{n} \propto A^{- 1 / 3}$ $\Rightarrow (2) is correct.$

Because of electrostatic repulsion,

$E_{b}^{p} < E_{b}^{n}$ $\Rightarrow (3) is incorrect.$

Since in $β^{+}$ decay the number of protons decreases, electrostatic repulsion decreases.

$\Rightarrow E_{b}^{p} increases$ $\Rightarrow (4) is correct.$

Solution

1 $E_{b}^{p} - E_{b}^{n}$ is proportional to $Z (Z - 1)$ where $Z$ is the atomic number of the nucleus.

2 $E_{b}^{p} - E_{b}^{n}$ is proportional to $A^{- \frac{1}{3}}$ where $A$ is the mass number of the nucleus.

3 $E_{b}^{p} - E_{b}^{n}$ is positive.

4 $E_{b}^{p}$ increases if the nucleus undergoes a beta decay emitting a positron.

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Solution

1 Ebp-Ebn is proportional to Z(Z-1) where Z is the atomic number of the nucleus.

2 Ebp-Ebn is proportional to A-13 where A is the mass number of the nucleus.

3 Ebp-Ebn is positive.

4 Ebp increases if the nucleus undergoes a beta decay emitting a positron.

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1 $E_{b}^{p} - E_{b}^{n}$ is proportional to $Z (Z - 1)$ where $Z$ is the atomic number of the nucleus.

2 $E_{b}^{p} - E_{b}^{n}$ is proportional to $A^{- \frac{1}{3}}$ where $A$ is the mass number of the nucleus.

3 $E_{b}^{p} - E_{b}^{n}$ is positive.

4 $E_{b}^{p}$ increases if the nucleus undergoes a beta decay emitting a positron.