What is the minimum gamma ray energy needed to produce a positron?

Answer 1

The minimum photonic energy required to create an electron-positron pair is 1.022 MeV. This energy is the equivalent of the rest masses of the pair of particles created. There is a bit more to this, as conservation of momentum must be observed. Pair production will occur, when it occurs, in the vicinity of an atomic nucleus when a high energy gamma ray zips in. The nucleus provides some "help" to "balance the equations" that describe the event, and that atomic nucleus will allow symmetry to be preserved. Specifically, that atomic nucleus acts in the conservation of momentum. That's why pair production won't occur as gamma rays are flying through the vacuum of space; there are (virtually) no atoms out there to facilitate the event. It may sound complex, but it's not all that difficult to get a handle on the phenomenon. A link is provided to our friends at Wikipedia. Knowledge there is free, and pair production isn't all that tough to understand. The article is brief and readable, and the complex equations that might have been posted are absent. Check it out.

Answer 2

The mass-energy of a positron is 0.511 MeV, so this is the minimum energy of a photon that would be required to produce it. But we don't see a lone positron created by a gamma ray. The answer is a little more complex than that, so let's look.

We see both a positron and an electron produced in an event we call pair production. The minimum energy required to produce the electron-positron pair would double that required to create the positron alone, or 1.022MeV (2 x 0.511 MeV). Lev Landau covered this in his 1934 paper, in which he describes the formation of particles from two-photon interactions from the narrow cone of gamma rays that may be formed when high energy particles collide.

To answer the question directly, the minimum gamma ray energy required to produce a positron is 1.022MeV. But you'll get a positron and an electron. Again, you'll recall that only half of that minimum energy is converted into a positron, the other half being used to produce an electron.

Answer 3

You must add the mass of the electron and the photon, and then multiply by the square of the speed of light to convert it to energy units.

Conveniently, the Wikipedia lists the mass of particles in units of MeV/c2 (among others); in the case of the electron, 0.510998928(11) MeV/c2. Note: the number within parentheses indicates an uncertainty in the last two digits. Twice this mass, and omitting the c-squared, gives you the energy in MeV, a unit commonly used in nuclear physics.