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Pair annilation. Consider the case where an electron $\mathrm{e}^{-}$ and a positron $\mathrm{e}^{+}$ annihilate each other and produce photons. Assume that these two particles collide head-on with equal, but small, speeds. (a) Show that it is not possible for only one photon to be produced. (Hint: Consider the conservation law that must be true in any collision.) (b) Show that if only two photons are produced, they must travel in opposite directions and have equal energy. (c) Calculate the wavelength of each of the photons in part (b). In what part of the electromagnetic spectrum do they lie?

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a. conservation of momentum is violated. We require another particle to make the net momentum zero, so there must be two photons.b. As each photon has the same momentum, to get a net momentum of zero, the two photons must travel in opposite directions.c. gramma rays

Physics 103

Chapter 30

Nuclear and High-Energy Physics

Atomic Physics

Nuclear Physics

Particle Physics

University of Washington

University of Sheffield

University of Winnipeg

McMaster University

Lectures

02:42

Atomic physics is the field of physics that studies atoms as an isolated system of electrons and an atomic nucleus. It is primarily concerned with the arrangement of electrons around the nucleus and the processes by which these arrangements change. The theory of quantum mechanics, a set of mathematical rules that describe the behaviour of matter and its interactions, provides a good model for the description of atomic structure and properties.

02:26

In physics, nuclear physics is the field of physics that studies the constituents and interactions of atomic nuclei. The most commonly known applications of nuclear physics are nuclear power generation and nuclear weapons technology, but the research has provided application in many fields, including those in nuclear medicine and magnetic resonance imaging, ion implantation in materials engineering, and radiocarbon dating in geology and archaeology.

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Pair Annihilation. Conside…

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Consider the case where a…

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Two equal-energy photons c…

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Under the right circumstan…

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An electron and a positron…

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09:30

A gamma-ray photon changes…

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00:45

A positron and an electron…

for this question, Part A. It says to show that it's not possible for only one photo on to be produced. So to show that it's not possible for only one proton to be produced, we're gonna consider a conservation law. Which conservation law, then, is the question that we we should be considering. So initially there are only two particles the electron of the positron, and they're moving in opposite directions to annihilate with one another. But the masses the same since a positron is just a positive electron. So maybe since the masses are the same, we know they're gonna have the same momentum Is just gonna be in opposite directions. So this is a momentum conservation. So how queen? How we can answer this is we can say M C V minus. So we'll call that the mass of the electron is equal to M 70 plus the massive deposit trunk. Therefore, this leads to the mo mentum of the electron. We'll call this P E minus and it's a vector because it has direction is equal and opposite two, the momentum of the positron. So since the moment Tums are conserved, if one of these, um it's a photo on the other must also admit of photons because if one emits a photon and the other does not, then there the momentum would no longer be conserved. So we can answer this by saying, Since Mo mentum is conserved, uh, two photons must be omitted if the moment, um is to continue to be conserved in the system. Okay, so all of that would be the answer to part a part B is another conceptual question. It says, Um show that if only two photons air produced, they must have. They must travel in opposite directions and have equal energy. Okay, So again, duke do conservation of momentum. The momentum of the photon will call gamma that's coming from the negative. Electrons will call This game of minus has to be equal and opposite to the foe time that's coming from the positron. We'll call that gamma Plus. Since the masses are the same, they must have opposite. They must be travelling in opposite directions. So we can say since masses are the same, they must travel in opposite directions. Okay, It also says to show that they have to have the same energy well, how can we show that they have the same energy? So to continue part B, the energy here is equal to what is it? 1/2 and b squared. Okay, but in here is the same for both. Right, But what is what is Mo mentum? This is also equal to P squared over to end. Since momentum is mass times velocity. So therefore it both have the same mo mentum. The energies are the same. You can say, Since Mo mentum is equal, the energies are the same. And lastly part see says to calculate the wavelength of each of the photons and part B. And what part of the electromagnetic spectrum do they lie? Okay, So to calculate the wavelength, we need to find the frequency of the of the photo. This is because the wavelength is equal to see the speed of light divided by the frequency when we know the speed of light. So we just need to find the frequency. So to find the frequency we need to find the energy. Uh, the energy here is just gonna be the mass of the system times the speed of light squared. The mass of the system is a mass of the electron, plus the massive of the positron times the speed of light squared. But we said that the masses of the two are the same. So we can say this is two times the mass of the electron times the speed of light squared energy here. This is how we're gonna find the frequency. Energy here is also related to frequency because it's two times h, which is planks constant times the frequency F. Therefore, the frequency is equal to the energy which is to times m sabi time C squared, right, divided by two times place constant h. So now we have everything we need to solve for Lambda. So we complaining in this value for f will simplify it real quick because these twos can cancel. So we complain in this value for the frequency and to our expression, and we confined the wavelength. So Lamda here is equal to see uh, divided by the frequency. But the frequency here would then give us we go back to this page, it's gonna give us any time C squared on the bottom. Times clicks, constant on the top again, we can simplify the square here is gonna cancel with the sea. What? We're left with this place Constant divided by Emi time. See, everything here is just a constant. So we just have to plug in the values and calculate the answer. You can look at these values or I could just go ahead and give them to you. The value for uh value for the speed of light see, is 2.99 times 10 to the eight meters per second. The value for the mass of the positron or the mass of the electron is 9.1 times 10 to the minus 31 kilograms and the value for place constant eight is equal to 6.62 times 10 to the minus 34 meter squared times kilogram her second. If you plug all of these values and to this expression, we find that this is equal to two point for two times 10 to the minus 12 meters so we can box that in. But we're also asked what, uh what wavelength business in So you can look this up. But this is in the gamma ray wavelength of the spectrum. So this is a gamma ray, very high energy and weaken box. This inn is the solution to part see of our question

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