# Compton Effect, frequency question

• Kennedy111
In summary, the frequency of the deflected X-ray is 6.274871536 x 10^-19 Hz when an X-ray with a frequency of 3.74 x 10^20 Hz is incident on a thin piece of metal and observed deflected at 48°. The equations used to solve this problem include Δλ= (h/mc)(1-cosθ), E = hf, and λ= hc/E.
Kennedy111

## Homework Statement

An X-ray with a frequency of 3.74 x 10^20 Hz is incident on a thin piece of metal. The lower frequency X ray on the other side is observed deflected at 48°. What is the frequency fo the deflected X ray?

fi = 3.74 x 10^20 Hz
Deflected at 48°

## Homework Equations

I'm not sure... I would guess..

Δλ= (h/mc)(1-cosθ)
E = hf
E = hc/λ
(hc/λ)=hf

## The Attempt at a Solution

I really have no idea where to begin... If someone could just start me off that would be awesome :) This is my attempt anyway..

Ei = hf
= (6.63 x 10^-34 Js)(3.74 x 10^20 Hz)
= 2.47962 x 10^-13 J

E = hc/λ
λ= hc/E
= (6.63 x 10^-34 js)(3.00 x 10^8 m/s) / (2.47962 x 10^-13 J)
= 8.02139037 x 10^-13 m

Δλ = λf - λi
λf = Δλ + λi
= ((6.63 x 10^-34 Js) / (9.11 x 10^-31 kg)(3.00 x 10^8 m/s))(1- cos48) + 8.02139037 x 10^-13 m
=4.78097437 x 10^-12 m

Ef = hc/λ
=(6.63 x 10^-34 Js)(3.00 x 10^8 m/s) / (4.78097437 x 10^-12 m)
=4.16023983 x 10^-14 J

E = hf
f = E/h
= (4.16023983 x 10^-14 J) / (6.63 x 10^-34 Js)
= 6.274871536 x 10^-19 Hz

Yeah... I don't think this is right..

Kennedy111 said:
Ei = hf
= (6.63 x 10^-34 Js)(3.74 x 10^20 Hz)
= 2.47962 x 10^-13 J

E = hc/λ
λ= hc/E
= (6.63 x 10^-34 js)(3.00 x 10^8 m/s) / (2.47962 x 10^-13 J)
= 8.02139037 x 10^-13 m
x-rays are electromagnetic radiation. So, the easiest way to get back and forth between λ and f is to use the basic wave relation c = λf. Thus, λ = c/f. You can check that this will give the same result for λi.
Δλ = λf - λi
λf = Δλ + λi
= ((6.63 x 10^-34 Js) / (9.11 x 10^-31 kg)(3.00 x 10^8 m/s))(1- cos48) + 8.02139037 x 10^-13 m
=4.78097437 x 10^-12 m
Looks set up correctly, but I don't get the same value for λf.
Ef = hc/λ
=(6.63 x 10^-34 Js)(3.00 x 10^8 m/s) / (4.78097437 x 10^-12 m)
=4.16023983 x 10^-14 J

E = hf
f = E/h
= (4.16023983 x 10^-14 J) / (6.63 x 10^-34 Js)
= 6.274871536 x 10^-19 Hz

Again, you can just use f = c/λ.

Wait... was your λf value 1.06771397 x 10^-12 ??

My calculator was in radian mode...

Kennedy111 said:
Wait... was your λf value 1.06771397 x 10^-12 ??

My calculator was in radian mode...

I get about 1.605 x 10^-12 m

Hmm... I'm not sure what I'm doing wrong then...

Nevermind! I got it now! :D Thanks :)

## 1. What is the Compton Effect?

The Compton Effect, also known as Compton scattering, is the phenomenon where a photon (particle of light) collides with an electron and transfers some of its energy to the electron, causing it to recoil and change direction.

## 2. What causes the Compton Effect?

The Compton Effect is caused by the interaction between a photon and an electron. When a photon collides with an electron, it transfers some of its energy to the electron, causing it to recoil and change direction.

## 3. How does the frequency of a photon change during the Compton Effect?

The frequency of a photon decreases after it undergoes the Compton Effect. This decrease in frequency is known as the Compton shift and is directly proportional to the energy transferred to the electron.

## 4. What is the significance of the Compton Effect?

The Compton Effect played a crucial role in the development of quantum mechanics as it provided evidence for the particle-like behavior of light. It also helped in understanding the dual nature of light as both a wave and a particle.

## 5. How is the Compton Effect used in practical applications?

The Compton Effect is used in various practical applications, such as in X-ray machines and PET scanners, to produce high-energy photons. It is also used in materials testing and in studying the structure of matter through scattering experiments.

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