How far will an electron-positron pair travel in a lead shield if the incident photon is 5 MeV and the pair both have the same energy?

Answer 1

== If the question can be imagined and framed properly, it can be asked, even if it has no "real world" applications. But let's get real with this one. The answer is "not very far" into the lead shield. A sheet of aluminum foil would stop the pair of particles. We've got problems with this one. There isn't a handy table for looking up the slowing down length of lead for positrons across a range of positronic energies, but let's take a short detour. We need some review. A positron and an electron are created in pair production. You're already fairly familiar with the electron. It has a negative charge, doesn't weigh very much (has little mass), and when it's moving, it will have to contend with the orbiting electrons of all the atoms in its path of travel. (Even in air, there are a tons and tons of "speed bumps" in the way of our little electron.) The electron can't cope well with these "flocks" of critters of its own kind, even if it is of high energy. It will scatter and lose energy at just about every atom it encounters along its trajectory (if we can use that term). It takes little time, that is, it can't travel very darn far, before the "encounters" it has with anything it "bumps into", i.e., those scattering events, "suck" all the energy out of our little electron and it's left hanging. It has a very short mean free path in air. In any kind of liquid or solid, it's even shorter. A lot shorter. And it's the same with a positron - except that it will join an electron at some point along its journey and the pair will be mutually annihilated. A couple of hot gamma rays will leave the scene of the event. Even with a few MeV of energy, the mean free path of an electron and a positron is extremely short in air. Oh, and these particles will be traveling in oppositedirections when they are created. It isn't like they'd be moving as a pair in the same direction like cars in adjacent lanes on a freeway. What about plumbum? We're talking mean free path here - the mean free path of an electron and a positron in lead. What is the slowing down length for a positron or an electron at a couple of MeV in lead? Short. Very short. Very, very short. Laying out the problem mathematically would just be an exercise in probability and statistics. And forget about setting up an experiment to "prove" the calculated answer. Got a clever little application for running Monte Carlo calculations on your computer? Start identifying and defining your variables. And don't forget to include factors that deal with both inelastic andelastic scattering. You'll have both happening here 'cause it's "real world" stuff. So no fudging and leaving out elastic scattering possibilities. It's a physics grad student's nightmare. Good luck with all that. What about conducting an experiment? The only thing close to being effective at "looking" at the penetration power of the positron is probably a spectroscope or PET imager, but how are you going to use them? The spectroscope would be difficult to apply for the purpose here (impossible, probably), and the PET units can't "resolve" the tiny distances we're talking about. By the way, it is true that you'd only be looking at the annihilation events resulting from the positron's recombination with an electron in PET imaging. And the ability of those machines is a long way from having the kind of resolution you'd need to "see" results that you could measure. You'd doubtless have better luck with just calculating an answer.

Answer 2

The range of the electron-positron pair in lead can be estimated using the Bethe formula, which depends on the initial energy of the particle. For a 5 MeV photon pair-produced in lead, the range for electron-positron pairs will be on the order of millimeters to centimeters. The exact range will depend on the specific energy of the electron-positron pair and the exact composition and thickness of the lead shield.