A 0.880 -MeV photon is scattered by a free electron...

A 0.880 -MeV photon is scattered by a free electron initially at rest such that the scattering angle of the scattered electron is equal to that of the scattered photon $(\theta=\phi \text { in }$ Fig. $40.13 ) .$ (a) Determine the angles $\theta$ and $\phi$ . (b) Determine the energy and momentum of the scattered photon. (c) Determine the kinetic energy and momentum of the scattered electron.

Key Concepts

Compton Scattering

This concept describes the phenomenon in which a photon scatters off a free electron, leading to a change in the photon's wavelength (energy) and a recoil of the electron. It is fundamental to understanding photon-electron interactions and is derived using quantum mechanics combined with relativistic kinematics.

Conservation of Energy and Momentum

These conservation laws are crucial in analyzing scattering events. They require that the total energy and the total momentum before and after the collision remain constant. In Compton scattering, the initial energy and momentum of the photon and the electron must equal the sum of the energy and momentum of the scattered photon and recoiling electron, forming the basis for deriving the relationships between scattering angles and particle energies.

Photon Properties

Photons are massless particles characterized by their energy and momentum, related by the equation E = pc, where p is momentum and c is the speed of light. These properties are essential for analyzing interactions in scattering processes, as changes in photon energy imply changes in momentum which are transferred to the electron during the collision.

Relativistic Kinematics

Since the energies involved in Compton scattering can be comparable to the rest mass energy of the electron, relativistic effects become significant. Relativistic kinematics provide the framework to relate energies, momenta, and scattering angles accurately in such high-energy interactions.

Scattering Geometry

The relation between the scattering angles of the photon and the electron plays a critical role in determining the kinematics of the interaction. In many scattering problems, symmetry or specific angular relationships are used to simplify the analysis and calculate quantities like energy transfer and momentum components.

Transcript

00:01 Hi everyone this is the problem based on scattering of photon by electron.

00:07 Here it is given a photon of energy .88 mega electron bolt get scattered by electron at rest.

00:24 If angle of scattering of photon and electron are to be same we have to find the value of theta and 5 in b.

00:40 We have to find energy and momentum of photon after scattering that we have to find we have to find energy kinetic energy and momentum of electron after scattering we have to apply conservation of momentum if we apply the conservation of momentum for x -axis momentum of momentum of photon is called to momentum photon after scattering along x -axis that is cost of theta plus momentum of electron into cause of 5 as given theta is equal to 5 so you can write this momentum will be h upon lambda not this is h upon lambda dash cause of theta p p e cause of theta so from here we can write h upon lambda 0 is equal to h upon lambda d s plus momentum of electron into cost data say equation number one now applying conservation of momentum for y -axis in the y direction initial momentum is zero momentum of photo after scattering along y direction plus so it should be negative because net momentum must be zero so from here you will get p e is equal to p photon dash that is h upon lambda dash so from equation one and two we can write so cos of b can write lambda s is cut to twice of lambda not cause of theta equation lambda dash minus lambda not is called to h upon mass of electron into c one minus cost theta substituting the value of lambda not lambda dash sorry to two lambda not cost of theta minus lambda not h upon mass of electron into c one minus cost theta now now we have to solve for.

04:19 So from here you can write 2 cos of theta minus 1 is equal 2 as c by lambda 0 into 1 upon m c squared 1 minus cos theta.

04:47 So we can say 2 cost of theta minus 1 is equal 2 energy of photon upon mc 2 x squared.

04:58 1 minus cost theta so for solving theta we can write m e into c square plus energy of photo upon twice of m e into c square plus energy of photo say equation number four now substituting the value we will get that theta so cost of theta value of m e into c square in electron voltage 5 .11 5 mega electron volt 0 .88 mega electron volt 2 into 0 .551 plus 0 .88 so it is to be 0 .732 so theta or 5 to be equal to 43 .0 degree this is the answer of part a now part b energy...

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