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--- the_compton_effect [2021/02/03 07:27] – admin
+++ the_compton_effect [2022/10/13 17:36] (current) – admin
@@ Line 12: / Line 12: @@
 \begin{align}
-  P_e^2 & = p^2 + {p'}^2 - 2  \vec{p} \cdot \vec{p}' \nonumber\\
+  P_e^2  & = p^2 + {p'}^2 - 2  \vec{p} \cdot \vec{p}' \\
-  & = p^2 + {p'}^2 - 2pp' \cos\theta. \label{momeq}
+  & = p^2 + {p'}^2 - 2pp' \cos\theta. \qquad\qquad\qquad (1)
 \end{align}
@@ Line 20: / Line 20: @@
 \[E + E_0 = E' + E_e.\]
-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 $E_e = \sqrt{m_e c^4 + P_e^2 c^2}$ or, equivalently $E_e = \sqrt{E_0^2 + P_e^2 c^2}$.  Substituting all of this into the conservation of energy formula gives
+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
-\[pc + E_0 = p'c + \sqrt{E^2 + P_e^2 c^2},\]
+\[pc + E_0 = p'c + \sqrt{E_0^2 + P_e^2 c^2},\]
 and rearranging gives
-\[E_0 + (p-p')c = \sqrt{E^2 + P_e^2 c^2}.\]
+\[E_0 + (p-p')c = \sqrt{E_0^2 + P_e^2 c^2}.\]
 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 \eqref{momeq} to eliminate $P_e^2$ and obtain
+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
@@ Line 44: / Line 44: @@
 is called the //**Compton wavelength**// of the electron.
-The Compton wavelength of an electron is $2.426 \times 10^{-12} \,m$ and typical X-rays have wavelengths between $10^{-12}\, m$ and $10^{-9} \, m$, so the shift in wavelength due to Compton scattering would be noticeable for small wavelength X-rays.
+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
+  - 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
   \[\Delta \lambda = \frac{h}{m_e c} \left ( 1 - \cos \theta \right ).\]