You're right on a lot of counts. The wavefunction of the system is indeed a function of the form
$$
\Psi=\Psi(\mathbf r_1,\mathbf r_2),
$$
and there's no separating the two, because of the cross term in the Schrödinger equation. This means that it is fundamentally impossible to ask for things like "the probability amplitude for electron 1", because that depends on the position of electron 2. So at least a priori you're in a huge pickle.
The way we solve this is, to a large extent, to try to pretend that this isn't an issue - and somewhat surprisingly, it tends to work! For example, it would be really nice if the electronic dynamics were just completely decoupled from each other:
$$
\Psi(\mathbf r_1,\mathbf r_2)=\psi_1(\mathbf r_1)\psi_2(\mathbf r_2),
$$
so you could have legitimate (independent) probability amplitudes for the position of each of the electrons, and so on. In practice this is not quite possible because the electron indistinguishability requires you to use an antisymmetric wavefunction:
$$
\Psi(\mathbf r_1,\mathbf r_2)=\frac{\psi_1(\mathbf r_1)\psi_2(\mathbf r_2)-\psi_2(\mathbf r_1)\psi_1(\mathbf r_2)}{\sqrt{2}}.
\tag1
$$
Suppose that the eigenfunction was actually of this form. What could you do to obtain this eigenstate? As a fist go, you can solve the independent hydrogenic problems and pretend that you're done, but you're missing the electron-electron repulsion. You could solve the hydrogenic problem for electron 1 and then put in its charge density for electron 2 and solve its single electron Schrödinger equation, but then you'd need to go back to electron 1 with your $\psi_2$. You can then try and repeat this procedure for a long time and see if you get something sensible.
Alternatively, you could try reasonable guesses for $\psi_1$ and $\psi_2$ with some variable parameters, and then try and find the minimum of $⟨\Psi|H|\Psi⟩$ over those parameters, in the hope that this minimum will get you relatively close to the ground state.
These, and similar, are the core of the Hartree-Fock methods. They make the fundamental assumption that the electronic wavefunction is as separable as it can be - a single Slater determinant, as in equation $(1)$ - and try to make that work as well as possible. Somewhat surprisingly, perhaps, this can be really quite close for many intents and purposes. (In other situations, of course, it can fail catastrophically!)
In reality, of course, there's a lot more to take into account. For one, Hartree-Fock approximations generally don't account for 'electron correlation' which is a fuzzy term but essentially refers to terms of the form $⟨\psi_1\otimes\psi_2| r_{12}^{-1} |\psi_2\otimes\psi_1⟩$. More importantly, there is no guarantee that the system will be in a single configuration (i.e. a single Slater determinant), and in general your eigenstate could be a nontrivial superposition of many different configurations. This is a particular worry in molecules, but it's also required for a quantitatively correct description of atoms.
If you want to go down that route, it's called quantum chemistry, and it is a huge field. In general, the name of the game is to find a basis of one-electron orbitals which will be nice to work with, and then get to work intensively by numerically diagonalizing the many-electron hamiltonian in that basis, with a multitude of methods to deal with multi-configuration effects. As the size of the basis increases (and potentially as you increase the 'amount of correlation' you include), the eigenstates / eigenenergies should converge to the true values.
Having said that, configurations like $(1)$ are still very useful ingredients of quantitative descriptions, and in general each eigenstate will be dominated by a single configuration. This is the sort of thing we mean when we say things like
the lithium ground state has two electrons in the 1s shell and one in the 2s shell
which more practically says that there exist wavefunctions $\psi_{1s}$ and $\psi_{2s}$ such that (once you account for spin) the corresponding Slater determinant is a good approximation to the true eigenstate. This is what makes the shells and the hydrogenic-style orbitals useful in a many-electron setting.
However, a word to the wise: orbitals are completely fictional concepts. That is, they are unphysical and they are completely inaccessible to any possible measurement. (Instead, it is only the full $N$-electron wavefunction that is available to experiment.)
To see this, consider the state $(1)$ and transform it by substituting the wavefunctions $\psi_j$ by $\psi_1\pm\psi_2$:
\begin{align}
\Psi'(\mathbf r_1,\mathbf r_2)
&=\frac{\psi_1'(\mathbf r_1)\psi_2'(\mathbf r_2)-\psi_2'(\mathbf r_1)\psi_1'(\mathbf r_2)}{\sqrt{2}}
\\&=\frac{
(\psi_1(\mathbf r_1)-\psi_2(\mathbf r_1))(\psi_1(\mathbf r_2)+\psi_2(\mathbf r_2))
-(\psi_1(\mathbf r_1)+\psi_2(\mathbf r_1))(\psi_1(\mathbf r_2)-\psi_2(\mathbf r_2))
}{2\sqrt{2}}
\\&=\frac{\psi_1(\mathbf r_1)\psi_2(\mathbf r_2)-\psi_2(\mathbf r_1)\psi_1(\mathbf r_2)}{\sqrt{2}}
\\&=\Psi(\mathbf r_1,\mathbf r_2).
\end{align}
That is, the Slater determinant that comes from linear combinations of the $\psi_j$ is indistinguishable from the one you get from the $\psi_j$ themselves. This extends to any basis change on that subspace with unit determinant; for more details see this thread. The implication is that labels like s, p, d, f, and so on are useful to describe the basis functions that we use to build the dominating configuration in a state, but they cannot be reliably inferred from the many-electron wavefunction itself. (This is as opposed to term symbols, which describe the global angular momentum characteristics of the eigenstate, and which can indeed be obtained from the many-electron eigenfunction.)
Your mass defect result 13.3 eV is indeed too low, it should be around 13.6 eV. Most of this discrepancy (0.3
eV) is most probably not real, otherwise basic textbooks would have to rewritten. CODATA gives proton and electron masses and those numbers should be quite reliable. They do not give hydrogen atom mass so it is from some other source which is most probably not guarranteed to be consistent with CODATA.
It is difficult to measure mass of neutral hydrogen atom to similarly high accuracy as the mass of the proton, because the atom is electrically neutral and quite unstable (forms $H_2$ molecules or ions). I do not think digits in 938.783073804 MeV/c2 after 6th decimal are certain to be correct.
To investigate this further I would go after how this number was determined, or how in general hydrogen atom H1 mass is best determined. Direct measurement of mass is probably problematic, resulting in low accuracy result. It may be that the more accurate way is actually the indirect way, assuming atom mass = mass of proton + mass of electron - ionization energy (known from spectroscopy).
Best Answer
How are you defining heat at the quantum level? The Schrodinger equation describes how objects behave at the quantum level and heat describes a form energy which is transferred between objects of different temperatures. Temperature is a macroscopic quantity and not microscopic. The equation above describes the energy of electrons in a specific shell and these electrons can change energy by the absorption or emission of photons, and photons do not possess temperature.
For exactly those reasons.
No they are not.
No. Heating hydrogen will not cause the absorption of photons which is needed to ionise hydrogen (other methods for ionisation of atoms exist, but I’m speaking in the context of this question). Moreover, the hydrogen will be ionised upon the absorption of photon with this energy and not any less. This is the crux of the term energy and other quantities are quantised at a microscopic level which gave rise to quantum mechanics.
Once again temperature is not relevant here. For ionisation to occur, a photon must be absorbed (there are other ways to ionise atoms too). So to answer your question, there is no such dependence.