Thomas and Fermi independently developed a theory of complex many electrons atom or ion by choosing
fermi gas of electron in ground state.The electrons confined in space by a central potential V(r) where V(r) is zero for large distance(r) from the nucleus.
Apart from this assumption they also assumed the variation of potential is very slow in distance very much larger than de'Broglie wavelengths of electron and the thermal energy kT is much smaller than the potential energy in the boundary of the atom.In the near of nucleas potential remain constant .And for large distance its value is zero.Then Schrödinger equation becomes
$$\frac{-\hbar^2\nabla^2\Psi}{2m}=E\Psi$$
Which solution gives
$$\Psi=C\sin\left({k_x x}\right)\sin\left({k_y y}\right)\sin\left({k_z z}\right)$$
$$k^2=\frac{2mE}{\hbar^2}={k_x}^2+{k_y}^2+{k_z}^2$$
Where we are using the boundary conditions
$$\Psi(x,y,0)=\Psi(x,y,L)=\Psi(x,0,z)=\Psi(x,L,z)=\Psi(0,y,z)=\Psi(L,y,z)=0$$ where the solution confined in a cube of side length L. Where C is a normalization constant.
Now $$k_x=\frac{n_x\pi }{L} \space,k_y=\frac{n_y\pi }{L} \space,k_z=\frac{n_z\pi }{L} \space,n_x,n_y,n_z \space are \space integers.$$
Now finding the value of C following:
$$\int{\vert\Psi(x,y,z)\vert}^2d^3x=1$$
$$\vert{C}\vert^2\int_{-L/2}^{L/2}\sin(k_x x)dx \int_{-L/2}^{L/2} \sin (k_y y) dy \int_{-L/2}^{L/2} \sin(k_z z)dz =1$$
$$\space which \space gives\space C=(2/L)^{\frac{3}{2}} \space therefore \space\Psi= (2/L)^{\frac{3}{2}}\sin\left({k_x x}\right)\sin\left({k_y y}\right)\sin\left({k_z z}\right)$$
$$k^2=\frac{2mE}{\hbar^2}={k_x}^2+{k_y}^2+{k_z}^2 $$
The maximum kinetic energy of electron $$\frac{p_0^2}{2m}=-eV(r) \Rightarrow p_0=\sqrt{-2meV(r)}$$
Let $$V(r)=-\frac{Ze}{4\pi\epsilon_0}\chi$$
Now the poisson's equation be$$\nabla^2V(r)=\frac{-\rho}{\epsilon_0}$$
The number of electronic state in a cell
$$2(\frac{L}{2\pi})^3\int_0^{p_0/ \hbar}\int_0^{\pi}\int_0^{2\pi}k^2\sin \theta dk d\theta d\phi=2(\frac{L}{2\pi})^3\frac{p_0^3}{3}4\pi=\frac{p_0^3L^3}{3\pi ^2\hbar^3}$$
$$Therefore \space charge \space density \space \rho=e\frac{p_0^3L^3}{3\pi ^2\hbar^3}=\frac{L^3[\sqrt{-2meV(r)}]^3}{3\pi^2\hbar^3}$$
$$Let \space V(r)=-\frac{Ze\chi}{4\pi\epsilon_0 r}$$
$$\frac{dV(r)}{dr}=\frac{Ze\chi}{4\pi\epsilon_0 r^2}-\frac{Ze}{4\pi\epsilon_0 r}\frac{d\chi}{dr}$$
$$\Rightarrow \frac{d}{dr}(r^2\frac{dV(r)}{dr})=-\frac{Zer}{4\pi\epsilon_0 }\frac{d^2\chi}{dr^2}$$
Now substituting above values in poisson's equation we get
$$\frac{1}{r^2}\frac{d}{dr}(r^2\frac{dV(r)}{dr})=-\left(\sqrt{2m\frac{Ze^2\chi}{4\pi\epsilon_0 r}}\right)^3e\frac{L^3}{3\pi ^2\hbar^3\epsilon_0}$$
$$\Rightarrow -\frac{Ze }{4\pi\epsilon_0 r}\frac{d^2\chi}{dr^2} = \left(2m\frac{Ze^2\chi}{4\pi\epsilon_0 r}\right)^{3/2}e\frac{L^3}{3\pi ^2\hbar^3\epsilon_0}$$
$$\Rightarrow \frac{d^2\chi}{dr^2}=\left(\frac{ze^2}{4\pi\epsilon_0}\right)^{1/2}\frac{\chi^{3/2}}{r^{1/2}}(2m)^{3/2}e^2\frac{L^3}{3\pi ^2\hbar^3\epsilon_0}$$
Now let r=bx where x is dimionsless and b has dimensions of length and substituting this into in above equation we get
👆$$\Rightarrow \frac{d^2\chi}{b^2dx^2}= \left(\frac{ze^2}{4\pi\epsilon_0}\right) \frac{\chi^{3/2}} {b^{1/2} x^{1/2}} (2m)^{3/2} e^2 \frac{L^3}{3\pi ^2\hbar^3\epsilon_0}$$
👆$$\Rightarrow \frac{d^2\chi}{dx^2}=\sqrt{\left ( \frac{ze^2}{4\pi\epsilon_0}\right)}\frac{\chi^{3/2}b\sqrt{b}}{ x^{1/2}}(2m)^{3/2}e^2\frac{L^3}{3\pi ^2\hbar^3\epsilon_0}$$
Now let $$\frac{1}{b\sqrt{b}}= \sqrt{\left ( \frac{ze^2}{4\pi\epsilon_0}\right)}(2m)^{3/2}e^2\frac{L^3}{3\pi ^2\hbar^3\epsilon_0}$$
Then
👆$$ \frac{d^2\chi}{dx^2}=\frac{\chi^{3/2}}{x^{1/2}}$$
Which is known as Fermi-Thomas equation.
where b=
