Question

Equation (9-42) gives the Fermi energy for a collection of identical fermions packed into the lowest energies allowed by the exclusion principle. Argue that if applied to neutrons or protons (ignoring their repulsion) in a nucleus. the equation suggests that the Fermi energy is roughly the same for all nuclei. Making the rough approximation that the spacing between quantum levels is a constant in a given nucleus, argue that this spacing should vary from one nucleus to another in proportion to${\mathit{A}}^{\mathbf{-}\mathbf{1}}$ .

Short answer

The spacing must be around $\frac{E_F}{A}$, which is proportional to $A^{-1}$.

Step-by-step solution

Given data

The equation bellowed (9-42) gives the Fermi energy for identical collection fermions packed into the lowest energies that are allowed by the exclusion principle.

$E_F=\frac{{\pi }^2{\hslash }^2}{m}{\left(\frac{3}{(2s+1)\pi \sqrt{2}}\frac{N}{V}\right)}^{2/3}$

Formula of Semi empirical binding Energy

Semi empirical binding Energy is given by the expression,

$BE=c_{1}A-c_2{A}^{2/3}-c_3\frac{Z(Z-1)}{A^{1/3}}-c_4\frac{{(N-Z)}^2}{A}$

Calculate the number of nucleons

We know that the spacing between quantum levels is constant, the formula for Fermi energy is.

$E_F=\frac{{\pi }^2{\hslash }^2}{m}{\left(\frac{3}{(2s+1)\pi \sqrt{2}}\frac{N}{V}\right)}^{2/3}$

Assume each nucleon as sphere. So, the volume of each nucleon is $\frac{4}{3}\pi R_0^3$.

So, the volume of a nucleus with A number of nucleons Is $V=A\times \frac{4}{3}\pi R_0^3$.

Since there are roughly states with equal energy spacing between spacing between 0 energy and Fermi energy.

So the spacing must be around $\frac{E_F}{A}$, which is proportional to $A^{-1}$.