Compton Effect, frequency question

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.
  • #1
Kennedy111
27
0

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..

Please help!
 
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  • #2
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.
 
  • #3
Wait... was your λf value 1.06771397 x 10^-12 ??

My calculator was in radian mode...
 
  • #4
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
 
  • #5
Hmm... I'm not sure what I'm doing wrong then...
 
  • #6
Nevermind! I got it now! :D Thanks :)
 

Related to Compton Effect, frequency question

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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