# Compton Scattering of Electron

• satchmo05
In summary: Use the fact that the energies of the photon and electron before and after the collision are equal.In summary, the change in energy of a photon scattering at 110 degrees from a free electron can be calculated using the formula ΔE = E2 - E1 = hc/λ2 - hc/λ1, where λ2 is the wavelength of the scattered photon and λ1 is the initial wavelength. The direction of the recoiling electron's velocity vector can be determined relative to the initial direction of the photon. The energies of the scattered photon and recoiling electron can be calculated using the equation E=hc/λ, where λ is the wavelength. The change in energy can also be calculated using the formula ΔE = E2 - E1
satchmo05

## Homework Statement

Suppose a 0.511[MeV] photon from a positron-electron annihilation scatters at 110 degrees from a free electron. What are the energies of the scattered photon and the recoiling electron? Relative to the initial direction of the 0.511[MeV] photon, what is the direction of the recoiling electron's velocity vector?

## Homework Equations

λ2 - λ1 = (h/mc)(1-cos(theta))
change in energy = (hc/(delta(λ)) = h*delta(f)

## The Attempt at a Solution

~ λ2 - λ1 = (h*c)/(m*c2)*(1 - cos(110)) = 0.00326[nm] = delta(λ)
~ variating change in energy formula --> delta(f) = c/(delta(λ)) = 9.2025e19 [Hz]
~ Multiplying frequency by Planck's constant, I get the change in energy = 6.1e-19 [J] = 0.380732094 [MeV]

I am confused at what I do from here to determine the energies of either the scattered photon and the recoiling electron. Since the photon's initial energy is 0.511[MeV], and using change in energy answer from above, would this mean that the energy of recoiling electron be equal to (0.380732094 + 0.511) [MeV]?

Thank you for all help in advance. Hopefully this work I have shown in clear!

Your equation for the change in energy is wrong. Use the fact that

$$E_\gamma = h\nu = \frac{hc}{\lambda}$$

Vela, thanks for the reply. There is nothing wrong with that formula that I used earlier. Planck's constant is in units of J*s, and frequency is in s-1. Ergo, energy is in Joules, so my units are fluid. I am not understanding what you're saying. From what I see in your formula, your units for energy would be a Joule*meter, which does not work. Please explain. Thanks.

The formulas I cited are correct. I'm not sure how you're getting joule-meter.

The units in your formula work okay, but algebraically, it's just wrong. The energy of the photons are given by E1=hc/λ1 and E2=hc/λ2, so the difference in energy is

$$\Delta E = E_2 - E_1 = \frac{hc}{\lambda_2} - \frac{hc}{\lambda_1} \ne \frac{hc}{\Delta \lambda}$$

Ah, I see what you're saying now. So what would I do with the change in energy value? That was my original question in the statement above. Thanks.

Try looking in your text where it derives the formula for the change in wavelength. Compton scattering is an elastic collision between a photon and an electron.

Vela, I am not finding anything of use to me right now.

Yes, I just read over the text on Compton scattering, and no mention of a formula deriving the change in wavelength. Any help you would be able to give me? Thanks again.

It's an elastic collision, so kinetic energy is conserved.

## What is Compton Scattering of Electron?

Compton scattering of electron is a phenomenon in which an electron collides with a photon, resulting in a change in the energy and direction of both the electron and the photon.

## How does Compton Scattering of Electron occur?

Compton scattering occurs when an electron interacts with a high energy photon, such as an X-ray or gamma ray. The electron absorbs the energy of the photon, causing it to recoil and change direction. The photon is then emitted in a new direction with lower energy.

## What is the significance of Compton Scattering of Electron?

Compton scattering is a crucial phenomenon in understanding the behavior of matter and radiation. It provides evidence for the particle nature of light and also helps in studying the structure of atoms and molecules.

## What are the applications of Compton Scattering of Electron?

Compton scattering is used in various fields such as medical imaging, material science, and astrophysics. It is used in X-ray and gamma ray imaging techniques to study the internal structure of objects. It is also used in nuclear physics to study the properties of atomic nuclei.

## How is Compton Scattering of Electron different from other scattering processes?

Compton scattering is different from other scattering processes because it involves a change in the energy of both the electron and the photon. In other scattering processes, only the direction of the particles may change, but not their energy.

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