Is it possible that matter/antimatter collisions emit resonance frequencies of H/He? Is it possible that matter/antimatter collisions emit resonance frequencies of H/He? 
I am researching a high energy phenomenon that occurs between the frequencies of 1200-1580 MHz. This come from a very very long way away though, so the signal does change a little bit. I'm trying to find out if matter/antimatter collisions could be the cause, and my mass estimates are asteroid sized masses of matter and anti-matter colliding, and it all seems very promising. Other than the fact that such collisions probably emit high energy Gamma Ray Bursts, not Radio Bursts. However, I found this to be interestingly near the resonance frequency of hydrogen (1420MHz) and I saw a figure that said Helium was 1400, but I'm not sure if that is correct. Is it possible for Matter-Antimatter collisions to emit the resonance frequency of there respected elements? Such as Hydrogen-antiHydrogen collision, emitting a radio signal at 1420 MHz?
 A: No.
When matter and antimatter annihilate, they do so particle by particle. Each electron annihilates with an anti-electron, protons with anti-protons, etc. And in each annihilation the total rest mass of the two particles is converted to energy in photons. Even the lightest annihilation, that of an electron with an anti-electron, must put over $1\ \mathrm{MeV}$ into photon energy.
Case of 1 photon: The energy is simply the sum of the rest masses of the two particles. But momentum conservation prohibits this from ever taking place.
Case of 2 photons: Momentum conservation says they will have equal and opposite momenta. Thus electron/anti-electron annihilation will produce a $511\ \mathrm{keV}$ photon going in one direction, and another $511\ \mathrm{keV}$ photon going in the opposite direction. These photons have a frequency of over $10^{20}\ \mathrm{Hz}$.
Case of 3 or more photons: Energy/momentum conservation no longer uniquely specifies the allowed end state frequencies. In fact, there is a continuum rather than a well-defined line. In theory radio emission can come from this, but in practice it won't be much, and certainly there won't be any identifiable feature in a narrow radio range. The more photons in the final state, the more likely some of them will be low frequency. But for each additional photon the probability of getting that many photons is reduced by a lot.
Furthermore, nothing is altered by the annihilating particles initially being in bound states of hydrogen or helium. The binding energies are much smaller than the annihilation energies. This is the same reason gamma rays Compton scattering in a material "see" free electrons -- the binding to nuclei is relatively insignificant. Moreover, the $1420\ \mathrm{MHz}$ line is from a hyperfine transition whose energy is even more miniscule than atomic transitions: about $6\ \mathrm{μeV}$.

Footnote on the actual data: It's not clear why you are looking at annihilation at all. You have a radio signal that you yourself associated with perhaps the most famous line in astronomy. If all you see is emission characteristic of ground-state hydrogen, why assume antimatter is involved in any way?
