It is possible for a gamma ray photon of sufficient energy in the presence of a massive object such as an atomic nucleus to materialize into an electronpositron pair. This process of energy transforming into matter is known as pair production. Because the rest mass of an electron is about 0.51 MeV/c^{2}, pair production can only occur if the energy of the gamma ray photon is at least 1.02 MeV, which translates to a frequency of at least 2.5 x 10^{20} Hz, and a wavelength of about 1.2 picometers, well into the gamma ray range.
Pair production cannot occur in empty space because the massive object is needed to carry away part of the momentum of the gamma ray photon; the electronpositron pair must each have a momentum less than half of the original photon's momentum, or else they will be moving in the same direction and will annihilate each other, if conservation of energy and conservation of momentum are not to be violated. From conservation of energy, we have:
2
hf = 2mc
where h is Planck's constant, f is the frequency of the gamma ray photon, and mc^{2} is the total energy of either the electron or the positron that would materialize. The positron and the electron created must of course be moving away from each other for pair production to occur, otherwise they would just hit each other and produce two gamma ray photons instead in pair annihilation. A vector diagram of what would have to occur in empty space is as follows:
+
e , momentum p
^
/
/
/
/ θ
> p cos θ
>
> p cos θ
γ, p = hf/c \ θ
\
\
\
v

e , momentum p
The angles at which the electron and positron are moving away from each other must be equal so that momentum is conserved. From special relativity the momentum of the photon is hf/c, so in the direction of motion of the photon:
hf
__ = 2p cos θ
c
hf = 2pc cos θ
For the nonrelativistic case, for the electron and positron the momentum is p = mv, so we have:
2
hf = 2mc (v/c) cos θ
Because (v/c) must of course be less than 1 and cos θ is always less than or equal to 1, it must be that
2
hf < 2mc
But we had earlier hf = 2mc^{2}. So if no conservation laws are to be violated, for pair production to take place there needs to be some other massive object in the vicinity to take some of the extra photon momentum so that the electron and positron can go in different directions.
Pair production is more likely for highenergy photons that interact with heavy nuclei, such as lead. It is the most efficient way for ionizing radiation to lose energy, as the energy of the gamma rays produced is cut in half every time. Say we start with originally an 8 MeV gamma ray. This interacts with a lead nucleus and materializes into a positron and an electron with kinetic energies of at most 3.5 MeV each. The positron meets another electron somewhere, undergoing pair annihilation into two gamma rays of 1.25 MeV or so. In just one step the energy of the photons was cut more than half. The rest of the energy the photon could bleed off by Compton scattering. This is why lead lining is generally used for radiation shielding.