What is the Wavelength of Scattered Radiation in a Compton Effect Experiment?

In summary, the wavelength of the scattered radiation in a Compton effect experiment using x-rays with a wavelength of 4.5*10^-12m and scattered at an angle of 53.0 degrees is 5.47*10^-12m. The formula used for this calculation was Δλ = h/mc*(1-cosθ) and the answer was obtained by adding the change in wavelength (Δλ) to the initial wavelength (λi).
  • #1
skg94
79
0

Homework Statement



If x ray with a weavelength of 4.5*10^-12m are used in a Compton effect experiment and are scattered at an anlge of 53.0 degree, the wavelength of the scattered radiation is?

ANS: 5.47*10^-12

Homework Equations





The Attempt at a Solution



The only thing i could think of was Δλ=h/mc*(1-cosθ)
which doesn't work seeing as how the xray didnt impact anything, i even tried it as if it impaced an electron using the mass of the electron. What formula or concept am i missing?
 
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  • #2
skg94 said:

Homework Statement



If x ray with a wavelength of 4.5*10^-12m are used in a Compton effect experiment and are scattered at an anlge of 53.0 degree, the wavelength of the scattered radiation is?

ANS: 5.47*10^-12

Homework Equations



The Attempt at a Solution



The only thing i could think of was Δλ=h/mc*(1-cosθ)
which doesn't work seeing as how the x-ray didn't impact anything, i even tried it as if it impacted an electron using the mass of the electron. What formula or concept am i missing?
In Compton scattering, the x-rays are usually scattered by electrons.

With that, I get the given answer.
 
Last edited:
  • #3
skg94 said:

Homework Statement



If x ray with a weavelength of 4.5*10^-12m are used in a Compton effect experiment and are scattered at an anlge of 53.0 degree, the wavelength of the scattered radiation is?

ANS: 5.47*10^-12

Homework Equations





The Attempt at a Solution



The only thing i could think of was Δλ=h/mc*(1-cosθ)
which doesn't work seeing as how the xray didnt impact anything, i even tried it as if it impaced an electron using the mass of the electron. What formula or concept am i missing?

For m did you use the rest mass of the electron? 'Cause the formula is correct.
 
  • #4
I did

6.63*10^-34/ 9.11*10^31 * 3*10^8 * (1-cos53)= 9.65959164*10^-13

Then ANS - 4.5*10^-12 which yields 3.53*10^-12.

what did i do wrong?
 
  • #5
skg94 said:
I did

6.63*10^-34/ 9.11*10^31 * 3*10^8 * (1-cos53)= 9.65959164*10^-13

Then ANS - 4.5*10^-12 which yields 3.53*10^-12.

what did i do wrong?
How about ANS + 4.5*10^-12 ?
 
  • #6
SammyS said:
How about ANS + 4.5*10^-12 ?

oh, how come its plus isn't Δλ= λi-λf, or is it plus? isn't the change in something usually subtract?
 
  • #7
skg94 said:
oh, how come its plus isn't Δλ= λi-λf, or is it plus? isn't the change in something usually subtract?
You start with λi and finish with λf , so λi + Δλ = λf . In other words, Δλ = λf - λi , like with anything else.
 
  • #8
ah.. silly mistakes are the ones that always get you thank you
 

Related to What is the Wavelength of Scattered Radiation in a Compton Effect Experiment?

What is the Compton effect?

The Compton effect, also known as the Compton scattering, is a phenomenon in which a photon (electromagnetic radiation) collides with an electron, transferring some of its energy to the electron and resulting in a change in wavelength and direction of the photon.

Who discovered the Compton effect?

The Compton effect was discovered by American physicist Arthur Compton in 1922. He conducted experiments to observe the change in wavelength of X-rays after they interacted with matter, which led to the discovery of the Compton effect.

What is the significance of the Compton effect?

The Compton effect provided strong evidence for the particle-like nature of light, as well as for the concept of photons. It also helped explain the absorption and scattering of X-rays and other forms of electromagnetic radiation by matter.

What is the difference between the Compton effect and the photoelectric effect?

Both the Compton effect and the photoelectric effect involve the interaction of photons and electrons. However, the Compton effect involves a change in the wavelength and direction of the photon, while the photoelectric effect involves the ejection of an electron from a metal surface due to the absorption of a photon.

How is the Compton effect used in practical applications?

The Compton effect is used in fields such as X-ray imaging and spectroscopy. It is also used in medical imaging techniques such as computed tomography (CT) scans, which use X-rays and the Compton effect to create detailed images of internal body structures.

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