High School
SAT
ACT
University
MCAT
LSAT
DAT
Wize High School Grade 12 Physics Textbook > Light as a Particle

The Compton Effect (Light = Particle)

Photons Previous Section
Photoelectric Effect (Light = Particle)Next Section
Popular Courses
Find My Course
0:00 / 0:00

The Compton Effect


To observe the Compton effect, photons are shined onto a gas (e.g. gas of electrons) and they are observed to scatter at different angles with different wavelengths.


Wize Concept
This happens because when a photon hits a particle, it transfers some of its momentum to it, so it ends up having less momentum and therefore a longer wavelength. This is similar to an elastic collision between hard objects: momentum and energy are both conserved.





The difference in wavelengths for the incident and scattered photons is given by:

 Δλ=λf−λi=hmec (1−cos⁡θ) \boxed{ \ \Delta \lambda=\lambda_f-\lambda_i=\frac{h}{m_ec}\ (1-\cos \theta) \ }

  • Δλ\Delta\lambda is the difference between the scattered wavelength (λf\lambda_f) and the incident wavelength (λi\lambda_i)
  • θ\theta is the angle of the scattered photon measured relative to the original direction of motion
  • hh is Plank's constant
  • mem_eis the mass of the electron
  • cc is the speed of light

Wize Concept
The maximum change in wavelength happens when the angle is 180°180\degree.



Watch Out!
The particle doesn't have to be an electron! So use the mass of the particle you're given (e.g. a proton) in the formula above.



Exam Tip
To find information about the electron, use conservation of momentum and/or conservation of energy.
  • Use p=hλp=\dfrac{h}{\lambda} for the momentum of the photon, and p=mvp=mv for the particle.
  • Use E=hcλE=\dfrac{hc}{\lambda} for the energy of the photon, and E=12mv2E=\dfrac{1}{2}mv^2 for the particle.

0:00 / 0:00

Example: Maximum Δλ\bco{\Delta \lambda}


A 5×10−125\times10^{-12} m photon strikes an electron at rest.

a) If the angle of the scattered photon is 60°60\degree, find the wavelength of the scattered photon.
b) Find the maximum change in wavelength.


Part a)


Δλ=λf−λi=hmec (1−cos⁡θ)\Delta \lambda=\lambda_f-\lambda_i=\dfrac{h}{m_ec}\ (1-\cos \theta)

Isolate the λf\lambda_f to get:

λf=λi+hmec (1−cos⁡θ)\lambda_f=\lambda_i+\dfrac{h}{m_ec}\ (1-\cos \theta)

=5×10−12+6.626×10−34(9.11×10−31)(3×108) (1−cos⁡60°)=5\times10^{-12}+\dfrac{6.626\times10^{-34}}{(9.11\times10^{-31})(3\times10^8)}\ (1-\cos 60\degree)

=6.21×10−12=6.21\times10^{-12} (m)


Part b)


The maximum Δλ\Delta \lambda happens when 1−cos⁡θ1-\cos\theta is a maximum. Since −1≤cos⁡θ≤1-1\le\cos\theta\le1, the maximum is 22 and it corresponds to cos⁡θ=−1\cos\theta=-1 , i.e the angle is θ=180°\theta=180\degree (the photon turns straight back).

The equation becomes:

Δλ=hmec (1−cos⁡θ)\Delta \lambda=\dfrac{h}{m_ec} \ (1-\cos \theta)

=hmec (2)=\dfrac{h}{m_ec}\ (2)

=6.626×10−34(9.11×10−31)(3×108) (2)=\dfrac{6.626\times10^{-34}}{(9.11\times10^{-31})(3\times10^8)}\ (2)

=4.85×10−12=4.85\times10^{-12} (m)

Practice: Compton Effect

Light with a wavelength of 1.8 ×10−12 1.8\ \times10^{-12\ } m is shined towards a gas of electrons, and the scattered light is observed at angle of π4\dfrac{\pi}{4} with respect to the direction of the incident light. (Use h=6.626×10−34h=6.626\times 10^{-34} and me=9.11×10−31m_e=9.11\times10^{-31})

a) What is the wavelength of the scattered light at this angle?

b) What is the momentum of the scattered photon?

c) What is the energy of the scattered photon?

d) What is speed of the scattered electron?

e) What is momentum of the scattered electron (magnitude only)?

f) What is the direction of the scattered electron (the angle below the original photon's direction of motion)?