# Compton Effect, angle of deflection

• Kennedy111
In summary, the incident X-ray frequency was changed to 4.50 x 10^19 Hz, and the deflected X-ray frequency was measured to be 4.32 x 10^19 Hz. By using the equations Δλ = λf - λi and Δλ = (h/mc)(1-cosθ), the angle of deflection was found to be 18.6°. However, there may have been a calculation error in finding the wavelength of the X-ray before deflection.
Kennedy111

## Homework Statement

The scientist changes the frequency of the incident X-ray to 4.50 x 10^19 Hz and measures the deflected X ray frequency of 4.32 x 10^19 Hz. What was the angle of deflection?

Fi = 4.50 x 10^19 Hz
Ff = 4.32 x 10^19 Hz

## Homework Equations

Δλ = λf - λi
Δλ = (h/mc)(1-cosθ)
λ = c/f

## The Attempt at a Solution

First, I found the wavelength of the X ray before and after it is deflected.

λi = c/f
= (3.00 x 10^8 m/s) / (4.50 x 10^19 Hz)
= 6.818181812 x 10^-12 m

λf = c/f
= (3.00 x 10^8 m/s) / (4.32 x 10^19 Hz)
= 6.9444444444 x 10^-12 m

Then found the change in wavelength

Δλ = λf - λi
= (6.944444444 x 10^-12 m) - (6.818181812 x 10^-12 m)
= 1.26262626 x 10^-13 m

Then I used the change in wavelength to find the angle

Δλ = (h/mc)(1-cosθ)
1.26262626 x 10^-13 m = ((6.63 x 10^-34 Js) / (9.11 x 10^-31 kg)(3.00 x 10^8 m/s))(1-cosθ)
1.26262626 x 10^-13 m = (2.4259056 x 10^-12)(1-cosθ)
0.052047623 = 1-cosθ
cosθ = 0.947952377
cos^-1(0.947952377) = 18.56692499° = 18.6°

I'm just really unsure of the process that I took...

Kennedy111 said:

## Homework Statement

The scientist changes the frequency of the incident X-ray to 4.50 x 10^19 Hz and measures the deflected X ray frequency of 4.32 x 10^19 Hz. What was the angle of deflection?

Fi = 4.50 x 10^19 Hz
Ff = 4.32 x 10^19 Hz

## Homework Equations

Δλ = λf - λi
Δλ = (h/mc)(1-cosθ)
λ = c/f

## The Attempt at a Solution

First, I found the wavelength of the X ray before and after it is deflected.

λi = c/f
= (3.00 x 10^8 m/s) / (4.50 x 10^19 Hz)
= 6.818181812 x 10^-12 m

λf = c/f
= (3.00 x 10^8 m/s) / (4.32 x 10^19 Hz)
= 6.9444444444 x 10^-12 m

Then found the change in wavelength

Δλ = λf - λi
= (6.944444444 x 10^-12 m) - (6.818181812 x 10^-12 m)
= 1.26262626 x 10^-13 m

Then I used the change in wavelength to find the angle

Δλ = (h/mc)(1-cosθ)
1.26262626 x 10^-13 m = ((6.63 x 10^-34 Js) / (9.11 x 10^-31 kg)(3.00 x 10^8 m/s))(1-cosθ)
1.26262626 x 10^-13 m = (2.4259056 x 10^-12)(1-cosθ)
0.052047623 = 1-cosθ
cosθ = 0.947952377
cos^-1(0.947952377) = 18.56692499° = 18.6°

I'm just really unsure of the process that I took...
Hey I'm not positive but i did all the same calculations as you and yet i got a different answer. I believe this is because you made a calculation error when you multiplied (3.00 x 10^8 m/s) / (4.50 x 10^19 Hz). You see when you did this you got 6.818181812 x 10^-12 m but i got 6.666666666*10^-12m. correct me if I'm wrong but i believe this is the only mistake.

TSny

## 1. What is the Compton Effect and how does it relate to the angle of deflection?

The Compton Effect, also known as Compton scattering, is a phenomenon in which a photon (particle of light) collides with an electron, causing the photon to lose energy and change direction. The angle of deflection refers to the angle at which the scattered photon is deflected from its original path. This angle is directly related to the energy and momentum of the photon and electron involved in the collision.

## 2. How is the angle of deflection calculated in the Compton Effect?

The angle of deflection in the Compton Effect can be calculated using the following formula: θ = arccos[(1-λ'/λ)cos(θi)], where λ is the wavelength of the incident photon, λ' is the wavelength of the scattered photon, and θi is the angle of incidence (the angle between the incident photon and the electron's direction of motion).

## 3. What is the significance of the angle of deflection in the Compton Effect?

The angle of deflection in the Compton Effect is significant because it provides valuable information about the energy and momentum of the particles involved in the collision. By measuring the angle of deflection, scientists can determine the wavelength of the scattered photon and use this information to study the properties of matter and radiation.

## 4. Can the angle of deflection be affected by factors other than the energy and momentum of the particles involved?

Yes, the angle of deflection in the Compton Effect can also be influenced by factors such as the angle of incidence, the density and composition of the material the particles are interacting with, and the strength of the electromagnetic field in which the collision occurs. These factors can cause slight variations in the calculated angle of deflection.

## 5. How is the Compton Effect and angle of deflection used in practical applications?

The Compton Effect and angle of deflection have a wide range of practical applications in various fields such as medical imaging, materials science, and astronomy. In medical imaging, they are used in X-ray technology to produce detailed images of the body's internal structures. In materials science, they are used to study the atomic structure of materials and identify their composition. In astronomy, they are used to analyze the composition of distant objects and study the properties of cosmic rays.

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