>>11035862I don't really understand all of this too much, but from what I can gather, the energies are all quantized, and the lowest energy for an electron in orbit is that far away from the nucleus. At this scale, you cannot think of the electron and proton as magnets; you only get those properties at a macroscopic level, so the assumption that the electron should fall in is unfounded. Now I believe the reason energy is quantized here comes from the Schrodinger equation, and the fact that the Hamiltonian operator is Hermitian (giving real eigenvalues), and that the wave functions live in Hilbert space (meaning that there are either finite or countably infinite eigenvectors) and hence, there are finite or countably infinite eigenvalues, i.e. possible energies. Therefore the energies are discrete. We shall ignore negative energies for now (let's just assume they are all filled up, which can happen due to the Pauli exclusion principle. Incidentally, the Pauli exclusion principle essentially comes from the fact that fermions have half-integer spin, so essentially a "rotation" of 360 degrees returns it not to its original state, but its negative. We describe them with antisymmetric operators, and when we apply an anti-symmetric operator on a state, we can think of it as a determinant. Since if any two rows or columns are equal the determinant vanishes, we can see that no two states can be the identical for fermions). Then there is a lowest energy level for the electron, and it turns out, this is at the ground state orbit. Let's suppose the electron did fall into the proton. Then I believe there would be a weak interaction resulting in the electron and proton to become a neutron. But the mass (energy) of a neutron is much greater than the sum of masses of a ground state electron and a proton, so you would need additional energy to make this happen (e.g. this happens at the core of the sun, although the electrons are not even attached to the nucleus in the sun).
