A photon having energy E0=0.880 MeV is scattered by a free electron initially at rest such that

A photon having energy E0=0.880 MeV is scattered by a free...

Key Concepts

Compton Scattering

This concept deals with the scattering of high-energy photons by electrons, where the photon loses energy and its wavelength increases. The analysis involves the Compton wavelength shift equation, which relates the change in wavelength of the photon to the scattering angle. It is a fundamental phenomenon that illustrates the particle nature of light and its interaction with matter.

Conservation of Energy and Momentum in Relativistic Collisions

In collisions involving photons and electrons at high energies, the laws of conservation of energy and momentum are applied under relativistic conditions. This means that both the relativistic energy (including the rest mass energy for electrons) and momentum must add up before and after the collision, enabling the determination of the final energy and momentum of the scattered particles.

Relativistic Kinematics

When dealing with particles moving at speeds comparable to the speed of light, classical kinematics is replaced by relativistic kinematics. This framework accounts for the effects of special relativity on energy, momentum, and mass, and is crucial for correctly calculating the outcomes, such as scattering angles, kinetic energies, and momenta for high-energy interactions like photon-electron scattering.

Transcript

00:01 In this exercise, we have a photon with an energy of 0 .88 megallectron volts, scattering of an electron.

00:11 And the electron, the scattering angle of the electron is theta, and so is the scattering angle of the photon.

00:20 Okay, so both are equal, and we want to find in question a, what is the value of theta? so what i'm going to do here first is to use conservation of momentum in order to find theta.

00:36 So the momentum in the x direction, the initial momentum is just the momentum of the incoming photon that i'm going to call p0.

00:45 That's in the x direction.

00:46 And i'm setting up this coordinate system here where the y -axis is upwards and the x -axis is horizontal.

01:00 So the momentum in the x direction, the final momentum in the x direction is the momentum of the electron that i'm going to call p .e.

01:12 Times the cosine of theta plus the momentum of the photon, the final momentum on the photon, then i'm going to call p gamma times the cosine of theta.

01:24 Okay, so this is p0 equals to p e plus p gamma times the cosine of theta.

01:38 Okay, and in the y direction, so first i'm going to just highlight this because we're going to need this later.

01:49 And in the y direction, we have zero momentum in the wide direction for the incoming photon, because the photon is traveling horizontally.

02:07 And for the electron, we have a momentum of minus p .e sine theta.

02:13 And for the photon, we have p gamma sine theta.

02:18 And from here we have that p e equals p gamma.

02:22 This means that the magnitude of the momentum of the electron is equal to the magnitude of the momentum of the photon.

02:29 Okay, and i'm going to substitute this in the equation that i found before for p0.

02:36 So we have p0 equals 2, 2 p gamma cosine of theta.

02:46 Okay.

02:48 And i have to find what p0 and p gamma are.

02:52 So let's do it.

02:56 P0 is the momentum of the photon, of the incoming photon.

03:02 That's just h, the plunks constant, over the wavelength of the incoming photon, lambda.

03:09 Is going to be equal to 2h over lambda prime, which is the momentum, which is the wavelength of the scattered photon times the cosine of theta.

03:23 And according to, and i'm also going to highlight this because you're going to need this.

03:31 And in order to solve this equation, all i have to do is to find gamma prime as a function of gamma, and for that we have compton's formula that tells us that the wavelength of the scattered fulton lambda prime is equal to lambda plus h over mc 1 minus cosine of theta.

03:54 Okay, so i'm going to just substitute this into the previous equation.

04:02 So what we have is h over lambda is equal to 2h over lambda prime, and lambda prime is lambda plus h over mc 1 minus cosine of theta.

04:18 All this times cosine of theta.

04:23 So the h is cancel out.

04:26 And we have gamma plus mc1 minus the cosine of theta equals lambda cosine of theta.

04:38 And i'm going to isolate cosine of theta.

04:42 So we have lambda.

04:44 I'm sorry, there's a two here right.

04:46 So lambda plus h over mc 1 minus cosine of theta.

04:54 I'm sorry, i'm trying to isolate cosine of theta.

04:58 So this is going to be just lambda plus h over mc.

05:02 This is going to be equal to 2 lambda plus h over mc times cosine of theta.