As you rightly note, all of the doubly excited states of helium are above the single-ionization threshold, and this means that they are auto-ionizing: they have enough energy to slingshot one of the electrons into the continuum and, given half the chance, will proceed to do so.
The only reason that this doesn't happen in the naive theory is that the electrons simply don't 'know' about the existence of the other guy to an extent that's sufficient for the transfer of actual quantum mechanical energy between them. That is, you can add self-consistent mean-value-of-the-density contributions to the (now nonlinear and nonlocal) hamiltonian for each electron, but you are still, by the simple fact that you're solving single-particle Schrödinger equations, explicitly forbidding the kinds of configuration interactions that make autoionizing states spring to life.
As such, any sort of entangling interaction between the two electrons will act as a link between your (formerly) bound states and the continuum, and it will force the eigenstates to be a linear combination of both: that is, the bound state becomes a non-square-integrable resonance embedded in the continuum, with an imaginary part of the eigenvalue which requires the state to decay over time as population goes away towards infinity.
The example you mentioned, $\mathrm{sech}(|\mathbf r_1-\mathbf r_2|)$, is a good candidate, as would be a softened Coulomb potential, but you could also use e.g. a contact interaction, $\delta(\mathbf r_1-\mathbf r_2)$, and get much the same results. (In fact, those would be pretty interesting to see, but I'm not aware of any references and it sounds like a hard thing to search for.)
If you want the abstract theory, there's probably no place better than the original Fano paper on Fano resonances,
U. Fano. Effects of configuration interaction on intensities and phase shifts. Phys. Rev. 124, 1866 (1961), eprints.
It has aged a bit, but it is still pretty readable, and the depth of its arguments makes up for its slightly outdated notation. In particular, it is very general regarding what sorts of particle-particle coupling would give rise to the autoionizing character of those states: all you need is a nonzero entangling matrix element, and it will happen. (Even better, so long as that matrix element is reasonably flat along the relevant bit of continuum, you also get the Fano line profile as a universal characteristic, with only the $q$ parameter tuning the shape.)
Best Answer
$$He \longrightarrow He^+ +e^- \ \ \ \ \ .....\Delta H_1$$ $$He^+\longrightarrow He^{+2} +e^-\ \ \ \ .......\Delta H_2$$
We know $\Delta H_2$ by the Bohr's model of H-like atom .
Ionization energy of a H-like atom with atomic number $Z$ is $\Delta H_2=13.6Z^2/n^2$ ($e^-$ initially in $n^{th}$ orbit)
Also, $\Delta H_1 +\Delta H_2=79\text{ eV}$
So, we can get $\Delta H_1$.