Compton Effect, angle of deflection

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SUMMARY

The discussion centers on calculating the angle of deflection in the Compton Effect using X-ray frequencies. The initial frequency (Fi) is 4.50 x 1019 Hz, and the final frequency (Ff) is 4.32 x 1019 Hz. The calculated change in wavelength (Δλ) is 1.26262626 x 10-13 m, leading to an angle of deflection (θ) of approximately 18.6°. A participant identified a potential calculation error in the wavelength determination, suggesting a different initial wavelength value.

PREREQUISITES
  • Understanding of the Compton Effect and its equations
  • Familiarity with wavelength and frequency calculations
  • Knowledge of Planck's constant (h) and the mass of an electron (m)
  • Basic trigonometry for angle calculations
NEXT STEPS
  • Review the derivation of the Compton wavelength shift formula
  • Practice calculations involving frequency and wavelength conversions
  • Explore the implications of the Compton Effect in quantum mechanics
  • Investigate common errors in wavelength and frequency calculations
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Students studying physics, particularly those focusing on quantum mechanics and the Compton Effect, as well as educators looking for problem-solving strategies in wave-particle interactions.

Kennedy111
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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...
 
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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.
 
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