Wize University Physics Textbook (Master) > Modern Physics

The Compton Effect

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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=λf​−λi​=me​ch​ (1−cosθ) ​

Δλ\Delta\lambdaΔλ is the difference between the scattered wavelength (λf\lambda_fλf​) and the incident wavelength (λi\lambda_iλi​)
θ\thetaθ is the angle of the scattered photon measured relative to the original direction of motion
hhh is  Plank's constant 
mem_eme​is the mass of the electron
ccc is the speed of light

Wize Concept
The maximum change in wavelength happens when the angle is 180°180\degree180°.

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}p=λh​  for the momentum of the photon, and p=mvp=mvp=mv for the particle.
Use  E=hcλE=\dfrac{hc}{\lambda}E=λhc​ for the energy of the photon, and E=12mv2E=\dfrac{1}{2}mv^2E=21​mv2 for the particle.

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

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

a) If the angle of the scattered photon is 60°60\degree60°, 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)Δλ=λf​−λi​=me​ch​ (1−cosθ)

Isolate the λf\lambda_fλf​ to get:

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

      =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)=5×10−12+(9.11×10−31)(3×108)6.626×10−34​ (1−cos60°)

      =6.21×10−12=6.21\times10^{-12}=6.21×10−12  (m)

Part b)

The maximum Δλ\Delta \lambdaΔλ happens when 1−cos⁡θ1-\cos\theta1−cosθ is a maximum. Since −1≤cos⁡θ≤1-1\le\cos\theta\le1−1≤cosθ≤1, the maximum is 222 and it corresponds to cos⁡θ=−1\cos\theta=-1cosθ=−1 , i.e the angle is θ=180°\theta=180\degreeθ=180° (the photon turns straight back).

The equation becomes:

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

        =hmec (2)=\dfrac{h}{m_ec}\ (2)=me​ch​ (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)=(9.11×10−31)(3×108)6.626×10−34​ (2)

        =4.85×10−12=4.85\times10^{-12}=4.85×10−12  (m)

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

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)?

Scattered wavelength (in m, 2 decimal places, don't include units):

Momentum of scattered photon (in kgm/s, 2 decimal places, don't include units):

Energy of scattered photon (in J, 2 decimal places, don't include units):

Speed of scattered electron (in J, 2 decimal places, don't include units):

Momentum of scattered electron (in kgm/s, 1 decimal place, don't include units):

Direction of scattered electron (in rad, 1 decimal place, don't include units):

I don't know

Wize University Physics Textbook (Master) > Modern Physics

The Compton Effect

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The Compton Effect

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

Part a)

Part b)

Practice: Compton Effect