JEE Advance - Physics (2022 - Paper 1 Online - No. 14)

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?

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

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

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

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

Explanation

Total binding energy (without considering repulsions),

$${E_b} = [Z{m_p} + (A - Z){m_n} - {m_x}]{c^2}$$

Where, $$_Z^AX$$ is the nuclei under consideration.

Now, considering repulsion :

Number of proton pairs $$ = {}^z{C_2}$$

$$\Rightarrow$$ This repulsion energy $$ \propto \,{{Z(Z - 1)} \over 2} \times {1 \over {4\pi { \in _0}}}{{{e^2}} \over R}$$

Where R is the radius of the nucleus

$$ \Rightarrow E_b^p - E_b^n \propto Z(Z - 1)$$ $$\therefore$$ there will be no repulsion term for neutrons.

Also, since $$R = {R_0}{A^{1/3}}$$

$$ \Rightarrow E_b^p - E_b^n \propto {A^{ - 1/3}}$$

Because of repulsion among protons,

$$E_b^p < E_b^n$$

Since in $$\beta^+$$ decay, number of protons decrease $$\Rightarrow$$ repulsion would decrease

$$ \Rightarrow E_b^p$$ increases

JEE Advance - Physics (2022 - Paper 1 Online - No. 14)

Explanation

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