Problem 2

 

2.) (10 points) Compton scattered photons off electrons that were initally at rest. He used photons of wavelngth 0.0711 nm in his experiments.
(a) What is the energy of these photons?
(b) What is the wavelength of the photons scattered at $\theta = 180{}^{\circ}$?
(c) What is the energy of the photons scattered at this angle?
(d) Use momentum conservation to find the momentum of the recoil electron.

Answer:
a) To find the energy of the photons, we have
E = hf
using the fact that $c = f \lambda$, we have
$E = \frac{hc}{\lambda}$
I gave you that the initial wavelength was
$\lambda_{0} = .0711 nm$
So, the energy is given by
$E = 2.79 \times 10^{-15} J$
E = 17.4 keV
 
b) The change in wavelength due to compton scattering through an angle $\theta$ is given by
$\Delta \lambda = \frac{h}{m_{e}c} (1 - \cos \theta)$
Here, $\theta = 180{}^{\circ}$, so we have
$\Delta \lambda = \frac{h}{m_{e}c} (2)$
$\Delta \lambda = .00485 nm$
The wavelength after the scattering takes place is given by
$\lambda' = \lambda_{0} + \Delta \lambda$
$\lambda' = .0711 nm + .00485 nm$
$\lambda' = .0760 nm$
 
c) the energy of the scattered photon then is given by
$E' = \frac{hc}{\lambda'}$
$E' = 2.615 \times 10^{-15} J$
E' = 16.3 keV
 
To find the momentum of the recoiling electron, we look at conservation of momentum.
$\vec{p_{i}} = \vec{p_{f}}$
$\vec{p_{\lambda}} = \vec{p_{electron}} + \vec{p_{\lambda'}}$
Note that the inital momentum is not zero, but given by the momentum carried in my the incident photon. The momentum of the electron is then given by
$\vec{p_{electron}} = \vec{p_{\lambda}} - \vec{p_{\lambda'}}$
Now, keep in mind the momentum is a vector quantity. The incident momentum is in the positive $\hat{x}$ direction, but the photon that leaves is scattered $180{}^{\circ}$ the other way, or in the negative $\hat{x}$ direction. Therefore, if we just look at magnitudes, we have
$p_{electron} = p_{\lambda} + p_{\lambda'}$
The momentum of a photon is given by
$p_{\lambda} = \frac{E}{c}$
So, we have
$p_{electron} = \frac{1}{c} (E + E')$
$p_{electron} = \frac{1}{c} (17.4 keV + 16.3 keV)$
$p_{electron} = \frac{33.7 keV}{c}$
$p_{electron} = 1.806 \times 10^{-23} kg m/s$