How does the Hamiltonian operator, in the Schrödinger equation for potential and kinetic energy, apply to a general element from the periodic table? I have been studying the separable differential equation form of the Hamiltonian equation. It is something like: $$ \hat{H} = -\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2} $$
What I am looking for is a more in depth explanation of how this equation can, if possible, be applied to a given mass of an element from the periodic table (using units of atomic mass unit).
I am confused because I am missing some piece of the puzzle about something. To me, it seems as if you could just substitute the atomic mass for $m$ and that would give the result. I realize, however, that that likely isn’t the full story: the equation involves the sophisticated Hamiltonian and partial differential operators (and the higher formalized forms involve bras and kets).
EDIT: The question has been answered sufficiently.
 A: The basic ingredients for a (non-relativistic) Hamiltonian describing an atom are: $K_e,K_n$, the kinetic energy of the electrons and of the nucleus. Then you have the electrostatic (repulsive) interaction between electrons $V_{e-e}$ and the electrostatic (attractive) interaction between electrons and the nucleus $V_{e-n}$.
Since the mass of the nucleus is much larger than that of the electrons, its kinetic energy can at first be discarded. The electron-electron interaction is smaller than the electro-nucleus interaction due to the large charge of the nucleus. If one neglects this term too one obtains an Hydrogen-like Hamiltonian that can be explicitly solved.
The prediction based on this model constitute and excellent starting point for the explanation of the periodic table.
A: For a system of $N$ particles with masses $m_i$ and charges $q_i$, interacting via Coulombd interaction, the Hamiltonian can be written as
$$
H=-\sum_{i=1}^N\frac{\hbar^2\nabla_i^2}{2m_i} + \frac{1}{2}\sum_{i=1}^N\sum_{j=1,\\j\neq i}^N\frac{q_i q_j}{|\mathbf{r}_i-\mathbf{r}_j|}
$$
Note that I mentioned only the Coulomb interaction above - to explain the properties of atoms one usually treats the nucleus as a single particle, of mass $M$ equal to the sum of the masses of protons and neutrons for this chemical isotope and charge $Ze$, whereas the remaining particles are electrons with mass $m_e$ and charge $-e$. Since protons and neutrons are each about a thousand times heavier than the electrons, the mass of the atom is approximately that of the nucleus.
The exact solution is possible only for the hydrogen and its isotopes, whereas for the atoms with more electrons and molecules one usually uses approximate methods.
