We wanted to find the frequency of a gamma radiation particle. We know that if it had undergone beta decay, it would have released the same amount of energy, but by releasing 2 beta radiation particles, both traveling at half of c. The equation to find the energy of these particles is shown below, where m0 is the particle’s rest mass energy, ≈0.511 meV. It gives ≈8.5 microjoules of energy per particle, or ≈17 microjoules in all. We then use the second equation to find the frequency of a photon with ≈17 microjoules of energy. Our answer is thus that the frequency of the photon is ≈2.566×1024 rotations per second.
E\approx\sqrt{\frac{(m_0\times0.5c)^2}{1-\frac{v^2}{c^2}}c^2+(m_0c^2)^2}\approx 8.5\times10^{-6}J\\[32pt] r=\frac Eh\approx\frac{17\times10^{-6}J}{6.626\times10^{-34}Js}\approx\frac{2.566\times10^{28}}s