Yes, the particles are just moving back and forth, not moving in any particular direction on average. Here's an animation to make this more clear: http://www.acoustics.org/press/151st/Lindwall.html (first one on the page)
However, there are at least two weird, unphysical things about that animation. For one, the particles are only moving in response to the sound wave, in perfect synchronization, and they aren't also moving constantly in random directions. If you had some material that would remain a gas at absolute zero, then a sound wave in that absolute-zero gas would look more or less like the animation. But for an everyday-volume sound wave in room-temperature air, the random thermal motion (which is always happening whether or not there is a sound wave) is much stronger than the motion caused by the sound wave. The only reason we notice a sound wave at all is because it is an ordered motion that carries energy in a particular direction. If you followed the motion of a single air molecule, it would look entirely random and there would be no trace of the sound wave. The sound wave only becomes apparent when you look at the large-scale pattern of density variations.
The other weird thing about the animation is that the molecules stop moving and turn around without colliding with anything! It should be obvious that in a real gas made of electrically neutral particles, there is no long-distance force that would cause this to happen, and a moving particle would not change direction unless it actually collided with another particle.
To answer your particular questions, yes, the whole medium is "vibrating" in that the density and pressure are increasing and decreasing periodically, as the gas flows back and forth. However, I would not refer to it as a "tunnel", because, as you can see from the animation, there doesn't have to be any well-defined cross-sectional shape for the sound wave. In fact, the simplest geometry to consider is a plane wave, which extends infinitely in all directions perpendicular to the direction of propagation. It's actually impossible to make a "beam" of sound that will propagate forever without spreading out.
In beta decay, the mass difference between the parent and daughter particles is converted to the kinetic energy of the daughter particles. For instance, in the decay of the free neutron,
$$
\rm n \to p + e^- + \bar\nu_e, \tag{$\beta^-$ decay}
$$
the difference between the mass on the left and the mass on the right is about $0.78\,\mathrm{MeV}/c^2$, and this is the energy liberated in the decay. (If you're a chemistry person, an eV is a useful energy unit; the $E=mc^2$ conversion is roughly $1000\,\mathrm{MeV}\approx 1\,\mathrm{amu}\times c^2$.) Equivalent processes like
$$
\rm p + \bar\nu_e \to n + e^+ \tag{neutrino capture}
$$
don't occur unless the kinetic energy on the left side is already large enough to account for the extra mass on the right side. Since the electron/positron mass is about $0.51\,\mathrm{MeV}/c^2$, neutrino capture on protons at rest is impossible for neutrinos with less than $1.80\rm\,MeV$ kinetic energy. This means, among other things, that neutrinos emitted from neutron decay at rest will never have enough energy to cause positron emission on protons at rest elsewhere.
You get $\beta^-$ decay from free neutrons because free neutrons are heavier than free protons. However it's not the case for all nuclei that the more positive isobars are less massive. For instance, the mass difference between postassium-40 and argon-40 is about $1.50\,\mathrm{ MeV}/c^2$, with potassium (19 protons) heavier than argon (18 protons), so the decay
$$
\rm ^{40}_{19}K \to {}^{40}_{18}Ar^- + \beta^+ + \nu_e + 0.48\, MeV
$$
is allowed (though rarer than some other branches) and merrily proceeding inside the bananas on your kitchen counter.
Best Answer
The oxygen molecule is unusual in having a ground state that has unpaired electrons, and this means it has a net spin of one. Technically the ground state is a triplet. This is unusual because ground states normally have spin zero i.e. they are singlets. For example the $\mathrm{N}_2$ molecule has a singlet ground state.
Anyhow, the non-zero spin of the oxygen molecule causes the ground state to split into a couple of closely separated levels. This type of splitting is known as fine structure. The spacing between these levels is 0.246 meV, which corresponds to a frequency of 59.4 GHz i.e. microwaves with a frequency of 59.4 Ghz have just the right amount of energy to cause transitions between the fine structure levels.
And this is what happens when oxygen absorbs microwaves at 60 GHz. The absorption spectrum is actually very complex because the oxygen molecule changes its rotation as well as jumping between the fine structure levels, so we get a complicated cluster of lines centred around 60 GHz. At atmospheric pressure the lines are broadened by collisions between oxygen molecules and we see only the characteristic broad absorption hump at 60 GHz. To resolve the many lines within this hump requires the absorption to be measured at very low pressures where collisions between the molecules are infrequent.