91Ó°ÊÓ

Formula used:

Fermi energy is a term in quantum mechanics usually referring to the energy difference between the highest and lowest occupied states of a single particle in a quantum system of non-interacting fermions at absolute zero temperature.

The Fermi energy E_Fis given by-

${\mathit{E}}_{\mathbf{F}}\mathbf{=}\frac{{\mathbf{Ï€}}^{\mathbf{2}}{\sout{\mathbf{h}}}^{\mathbf{2}}}{\mathbf{m}}{\left[\frac{3}{\left(2s+1\right)\mathrm{Ï€}\sqrt{2}}\frac{N}{V}\right]}^{\raisebox{1ex}{$\mathbf{2}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{3}$}\right.}$

Here, V is the volume, N is the number of particles, s is the spin of particles, D is the density, m is the mass of the particle

$\mathit{D}\mathbf{=}\frac{\mathbf{M}}{\mathbf{V}}$ ….. (2)

Convert the atomic mass of the lead-206 to standard units as below.

Nuclear density of lead- 206 , D = 17¹⁷kg/m³

$m=\left(206amu\right)\left(\frac{1.66×{10}^{-27}kg}{1amu}\right)=3.42×{10}^{-25}kg$

To get the Fermi energy of the neutrons, it would help to get a simpler expression.

Fermi energy first, Equation (1) is used, with s being (since as a fermion, it's a spin:

$\begin{array}{rcl}{E}_F& =& \frac{{\mathrm{Ï€}}^2\sout{{\mathrm{h}}^2}}{m}{\left[\frac{3}{\left(2s+1\right)\mathrm{Ï€}\sqrt{2}}\frac{N}{V}\right]}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\\ & =& \frac{{\mathrm{Ï€}}^2\sout{{\mathrm{h}}^2}}{m}{\left[\frac{3}{\left(2\left({\displaystyle \frac{1}{2}}\right)+1\right)\mathrm{Ï€}\sqrt{2}}\frac{N}{V}\right]}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\\ & =& \frac{\sout{{\mathrm{h}}^2}}{2m}{\left[\frac{3{\mathrm{Ï€}}^3\mathrm{N}}{2^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}\mathrm{Ï€³Õ}}\right]}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\\ E_F& =& \frac{\sout{{\mathrm{h}}^2}}{2m}{\left[\frac{3{\mathrm{Ï€}}^3\mathrm{N}}{2^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}\mathrm{Ï€³Õ}}\right]}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\end{array}$

Now equation (2) can be used to get an expression for the volume V, so that can be inserted for

V there:

$$\begin{gathered} D=\frac{M}{V} \\ V=\frac{M}{D} \end{gathered}$$

And plug that in the prior equation.

$$\begin{gathered} E_F=\frac{\sout{h^2}}{2m}{\left[\frac{3{\mathrm{Ï€}}^2}{V}\right]}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.} \\ =\frac{\sout{h^2}}{2m}{\left[\frac{3{\mathrm{Ï€}}^2\mathrm{N}}{\left({\displaystyle \frac{M}{D}}\right)}\right]}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.} \\ {E}_F=\frac{\sout{h^2}}{2m}{\left[\frac{3{\mathrm{Ï€}}^2\mathrm{ND}}{M}\right]}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}......\left(3\right) \end{gathered}$$

Let m is the mass of the neutron, and M is the mass of the entire nucleus. The N doesn't need be converted since the actual number of neutrons is known. So then insert the 124 for N then, along with 10¹⁷ kg/m³ for the density D of nuclear material 3.42x10^-25 kg for 1.67x10^-27 k, the mass of the neutron, m and 1.055x10^-34 J.s for $\sout{h^{}}$in equation (3).

$E_F=\frac{\sout{h^2}}{2m}{\left[\frac{3{\mathrm{Ï€}}^2\mathrm{ND}}{M}\right]}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}$

Substitute known numerical values in the above equation.

$\begin{array}{rcl}{E}_F& =& \frac{{\left(1.055×{10}^{-34}J.s\right)}^2}{2\left(1.67×{10}^{-27}kg\right)}{\left[\frac{3×{\left(3.14\right)}^2\left(124\right)\left({10}^{17}{\displaystyle \frac{kg}{m^3}}\right)}{3.42×{10}^{-25}kg}\right]}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\\ & =& 0.333×{10}^{-41}{\left[1072.447×{10}^{42}\right]}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\\ & =& 0.333×{10}^{41}\left[105.02×{10}^{28}\right]\\ & =& 3.49×{10}^{-12}J\end{array}$

They can be converted to MeV .

$\begin{array}{rcl}{E}_F& =& \left(3.49×{10}^{-12}J\right)\left(\frac{1eV}{1.6×{10}^{-19}J}\right)\left(\frac{1MeV}{{10}^{6}eV}\right)\\ & =& 21.8MeV\end{array}$

The Fermi energy of the neutrons in lead 206 is 21.8 MeV.

Nuclear density of lead is, D = 10¹⁷ kg/m³

The number of protons, N = 82

Fermi energy is given as,

$\begin{array}{rcl}{E}_F& =& \frac{\sout{h^2}}{2m}{\left[\frac{3{\mathrm{π}}^2\mathrm{ND}}{M}\right]}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\\ & =& \frac{{\left(1.055×{10}^{-34}J·s\right)}^2}{2\left(1.67×{10}^{-27}kg\right)}{\left[\frac{3×{\left(3.14\right)}^2\left(82\right)\left({10}^{17}{\displaystyle \frac{kg}{m^3}}\right)}{\left(3.42×{10}^{-25}kg\right)}\right]}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\\ & =& 0.333×{10}^{-41}\left[79.7×{10}^{28}\right]\\ & =& 2.65×{10}^{-12}J\end{array}$

That can be converted to MeV as follow.

$\begin{array}{rcl}{E}_F& =& \left(2.65×{10}^{-12}\right)\left(\frac{1eV}{1.6×{10}^{-19}J}\right)\left(\frac{1MeV}{{10}^{6}eV}\right)\\ & =& 16.5MeV\end{array}$

The Fermi energy of the protons in lead 206 is 16.5 MeV.

The two energies can possibly be made more similar if the repulsive energy for the protons is included in the calculations for the Femi energy. Given all the protons have the same charge and are very close together, there is a great deal of repulsive energy that exists between them. So that repulsive energy could add to the calculated Fermi energy for the protons to get the value closer to that of the neutrons, since it's observed that the two have them have about the same energy.