Can't really put 2-2 and together as to how having an inert degenerate He core translates to a He-flash. Also, at which points exactly do degeneracy and the He-flash start occurring?
[Physics] Why is the release of energy during the He-flash in stars almost explosive
astrophysicsstarsstellar-evolution
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Your first paragraph is not quite right. Gas pressure does not "stop" upon formation of an iron core, it is merely that the star cannot generate further heat from nuclear reactions and becomes unstable to collapse. i.e. The star does collapse! Perhaps what you mean is what halts the collapse (sometimes) before the star disappears inside its own event horizon and becomes a black hole? The answer is the degeneracy pressure of neutrons that are formed (endothermically) in electron capture events as the star collapses and also the repulsive (strong nuclear) force between neutrons in very dense nucleon gases with a small fraction of protons.
The analogy of filled "shells" is not too bad. In quantum mechanics we find that there are a finite number of possible quantum states per unit momentum per unit volume (often called "phase space"). In a "normal" gas, the occupation of these quantum states is governed by Maxwell-Boltzmann statistics - progressively fewer of these states are filled, according to $\exp(-E/kT)$.
In a Fermi gas at very high density or very low temperature, we reach a situation where the Pauli exclusion principle limits the occupation of these states to 2 particles per energy/momentum state (one for each spin); particles that might otherwise have occupied very low energy states are forced to occupy states of higher energy and momentum. In a "completely degenerate" gas, which is a good approximation for the electrons in a White Dwarf star or the neutrons in a Neutron star (the relevant case here), all the energy states are filled up to something termed the Fermi energy, with zero occupation at even higher energies.
Pressure is caused by particles having momentum (this is just basic kinetic theory). The large number of fermions with non-zero (even relativistic in some cases) momentum is the reason that a degenerate gas exerts a pressure, even if its temperature is reduced to close-to-zero. In fact, once a gas of fermions approaches complete degeneracy, a change in temperature has almost no effect on its pressure.
One point I will take issue with in your question, is the statement that "such particles cannot occupy the same small volume of space". In fact, the restriction is on the occupation of phase space. In a neutron star, the neutrons are almost touching each other, with separations of $\sim 10^{-15}$ m. You can cram lots of particles into a small volume, but only at the expense of giving them large momenta. If you like, this is a 3D version of the uncertainty principle: $$ (\Delta x \Delta p_x) (\Delta y \Delta p_y) (\Delta z \Delta p_z) = \Delta V (\Delta p)^3 \sim \hbar^3$$ This relationship tells you that particles can be packed tightly together but if they are, then they must have very different momenta. This large range of momenta is what leads to degeneracy pressure.
When it comes to halting the core-collapse of a massive star and supporting the resulting neutron star, degeneracy pressure is not the entire story. As I mentioned above, the separation between neutrons is of order $10^{-15}$ m, which is the approximate range of the strong nuclear force. This is not a coincidence. Degeneracy pressure alone is insufficient to halt the collapse or support a neutron star more massive than 0.7$M_{\odot}$ - the so called Tolman-Oppenheimer-Volkoff limit. Because the nuclear matter is highly asymmetric (many more neutrons than protons), there is an overall strong repulsive nuclear force above densities of about $3\times10^{17}$ kg/m$^3$ that is also very important in halting the collapse and supporting more massive neutron stars.
Best Answer
Short answer: it is a combination of (1) the ignition occurring in an electron-degenerate, isothermal core in which the equation of state is independent of temperature; and (2) the extreme temperature dependence of the triple alpha He fusion reaction.
Details:
The helium flash occurs at the tip of the first ascent red giant branch in stars with masses between 0.5 and about 2 solar masses. At this point the star consists of a helium core surrounded by a vigorously burning shell of hydrogen, surrounded by a very large convective envelope.
The core is the left over from core hydrogen burning, supplemented by helium produced by the hydrogen shell burning that takes over once the core hydrogen is exhausted. The inert core shrinks in radius from its main sequence size because it has more mass per particle, so must increase in density to maintain pressure. As it does so, the virial theorem demands that it also gets hotter. The shell burning drops more and more He into the core, the core shrinks further and gets hotter.
In stars > 2 solar masses, the core gets hot enough to ignite helium in the triple alpha process. This raises the core temperature, but not massively, because at the same time, the pressure increases, the core expands vigorously and the hydrogen burning shell is pushed outwards and extinguished.
In a lower mass star it is different. The He core density rises to the point that the core electrons become degenerate. Electron degeneracy pressure (EDP) dominates the total pressure of the gas and arises because at high enough densities, the electrons fill all low energy quantum states.EDP only depends on density, not temperature.
A core supported by EDP gets smaller and denser, the more massive it is, so becomes more degenerate as He ash is dropped onto it. But it is also sitting inside a hugely luminous H burning shell which heats it. In stars more massive than 0.5 solar masses, eventually the core gets hot enough ($\sim 10^{8}$ K) to ignite the He. Because degenerate electrons are extremely conductive, the core is almost isothermal, so ignition spreads rapidly through the core. This raises the core temperature, but crucially, not the core pressure (EDP is independent of temperature and, as Ken G explains in his answer, most of the heat gets absorbed by the non-degenerate ions that hardly contribute to the pressure). It just gets hotter and He fusion increases massively because it is very temperature sensitive (roughly proportional to $T^{40}$ !!). This runaway process is termed the "helium flash".
Eventually the temperature rises enough (to about $3\times10^{8}$ K) to break the electron degeneracy, the core expands rapidly, the H shell is extinguished and the core luminosity falls.