Zero magnetic quantum number in Classical physics I was reading a book on Atomic Physics where it was stated that

"In space quantisation of an electron the state with magnetic quantum
  number $(m)=0$ was always excluded on the ground that application of
  electric field would case the electron to collide with the nucleus.
  But later we will see that in quantum mechanics, the states $m=0$ are
  allowed."

So can somebody explain a bit more extensively why classical mechanics forbids the state with $m=0$ but quantum mechanics doesn't?
 A: 
Zero magnetic quantum number in Classical physics

There are no quantum numbers in classical physics, so there is a contradiction in the title.
$m$ is a quantum number dependent on $l$, it is a projection of the angular momentum  on the z axis.
$L_z$, the magnitude of the angular momentum in the z direction, is given by 
$L_z=m.ℏ$
Exploring quantized states with the Bohr  and semi-classical models,  before the theory of quantum mechanics developed, orbits were a planetary model, one would  exclude the $l=0$ quantum states , because they would pass through the nucleus. The orbit would be a straight line through it because of zero angular momentum. In the planetary model this meant the electron would annihilate on the proton , and there would be  no atoms.
Edit after given link by J.Murray:
For higher $l$ values , example $l=1$the three projections m=1, m=0, and m=-1 would add orbits, differentiating them and allowing a better description  of the observed atomic spectra.
The argument in the text for excluding $m=0$ for all $l$, is that such a state in an atom would change with the application of an electric field , so even though the angular momentum would be $1$ (in $ℏ$ units), the electron orbit would pass through the nucleus and annihilate. ( I have not checked the claim, that an electric field would force the orbit through the nucleus for $m=0$ for all $l$, I expect it is correct as an electric  or magnetic field would induce a rotation to a classical orbt). As atoms are stable this situation had to be excluded.
The development of the wave equations of quantum mechanics and the Born rule (postulate)  which directly connected measurements with probabilities, give orbitals and not orbits for  the electrons in an atom, i.e. probability distributions. So there exists a probability for an electron to exist in the nucleus, but it is very small. The dimensions of the nucleus are orders of magnitude smaller than the orbital space:  fermi (~$10^{-15}$ meters) for the  size of nucleus, microns (~$10^{-6}$ meters) for orbitals.
Electron capture in the nucleus shows that in certain nuclei the probability is measurable.
