Differences

This shows you the differences between two versions of the page.

--- the_compton_effect [2021/02/01 22:21] – created admin
+++ the_compton_effect [2022/10/13 17:36] (current) – admin
@@ Line 1: / Line 1: @@
-In 1923, Compton found that the wavelength of X-rays scattered off free electrons was larger than the wavelength of the incident radiation.  Classically, X-rays are too energetic to be absorbed by free electrons.  They would provide an oscillating electromagnetic field that would cause the electrons to oscillate at the same frequency.  Therefore, they should emit X-rays with the same frequency as the incident radiation.
+In 1923, Arthur Compton found that the wavelength of X-rays scattered off free electrons was larger than the wavelength of the incident radiation.  Classically, X-rays are too energetic to be absorbed by free electrons.  They would provide an oscillating electromagnetic field that would cause the electrons to oscillate at the same frequency.  Therefore, they should emit X-rays with the same frequency as the incident radiation.
-Compton was able to explain this effect by assuming that X-rays consist of a stream of particles (photons) of energy $E=h\nu$ that scatter elastically off the electrons.
+Compton was able to explain this effect by assuming that X-rays consist of a stream of particles (photons) of energy $E=h\nu$ that scatter elastically off the electrons.  Using conservation of momentum and energy, together with the quantum postulate, we can derive the wavelength and angle of the scattered photons.  The setup is illustrated in the following figure.
-Insert diagram here.
+{{ :compton.png?direct&600 |}}
 By conservation of momentum
@@ Line 11: / Line 11: @@
 Taking the scalar product of this with itself gives
-\begin{align*}
+\begin{align}
-  P_e^2 & = p^2 + {p'}^2 - 2  \vec{p} \cdot \vec{p}' \\
+  P_e^2  & = p^2 + {p'}^2 - 2  \vec{p} \cdot \vec{p}' \\
-  & = p^2 + {p'}^2 - 2pp' \cos(\theta).
+  & = p^2 + {p'}^2 - 2pp' \cos\theta. \qquad\qquad\qquad (1)
-\end{align*}
+\end{align}
+By conservation of energy
+\[E + E_0 = E' + E_e.\]
+We will use the relativistic energy-momentum relation $E^2 = p^2c^c + m_0c^4$.  Photons are massless, so for the incident photon we have $E = pc$ and for the scattered photon we have $E' = p'c$.  The electron is initially at rest, so $E_0 = m_e c^2$, where $m_e$ is the rest mass of the electron.  For the recoiling electron, we have $E_e = \sqrt{m_e^2 c^4 + P_e^2 c^2}$ or, equivalently $E_e = \sqrt{E_0^2 + P_e^2 c^2}$.  Substituting all of this into the conservation of energy formula gives
+\[pc + E_0 = p'c + \sqrt{E_0^2 + P_e^2 c^2},\]
+and rearranging gives
+\[E_0 + (p-p')c = \sqrt{E_0^2 + P_e^2 c^2}.\]
+Squaring this equation gives
+\[E_0^2 + (p-p')^2c^2 + 2E_0(p-p')c = E_0^2 + P_e^2c^2,\]
+which can be rearranged to
+\[P_e^2 = p^2 + p'^2 - 2pp' + \frac{2E_0(p-p')}{c}.\]
+We can use this together with equation (1) to eliminate $P_e^2$ and obtain
+\[p^2 + p'^2 - 2 pp'\cos\theta = p^2 + p'^2 - 2pp' + \frac{2E_0(p-p')}{c}.\]
+This can be rearranged to give
+\[\frac{E_0(p-p')}{c} = pp'(1 - \cos \theta).\]
+From this point, it is an in-class activity to show that the shift in wavelength between the incident and scattered radiation is
+\[\boxed{\Delta \lambda = \frac{h}{m_e c} \left ( 1 - \cos \theta \right ).}\]
+Using $\cos \theta = \left ( 1 - 2\sin^2 \frac{\theta}{2}\right )$, we can also write this as
+\[\boxed{\Delta \lambda = \frac{2 h }{m_e c} \sin^2 \frac{\theta}{2} = 2\lambda_C \sin^2 \frac{\theta}{2},}\]
+where
+\[\boxed{\lambda_C = \frac{h}{m_ec},}\]
+is called the //**Compton wavelength**// of the electron.
+The Compton wavelength of an electron is $2.426 \times 10^{-12} \,\text{m}$ and typical X-rays have wavelengths between $10^{-12}\, \text{m}$ and $10^{-9} \, m$, so the shift in wavelength due to Compton scattering would be noticeable for small wavelength X-rays.
+{{:question-mark.png?direct&50|}}
+====== In Class Activity ======
+  - By the quantum postulate, we have $E = h\nu$ and $E' = h\nu'$, where $\nu$ and $\nu'$ are the frequencies of the incident and scattered radiation respectively.  Using these, together with $E_0 = m_e c^2$, $E = pc$, $E'=p'c$, and
+  \[\frac{E_0(p-p')}{c} = pp'(1 - \cos \theta),\]
+  show that the wavelength shift $\Delta \lambda = \lambda' - \lambda$ is
+  \[\Delta \lambda = \frac{h}{m_e c} \left ( 1 - \cos \theta \right ).\]