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the_compton_effect [2021/02/03 18:41] – adminthe_compton_effect [2022/10/13 17:36] (current) – admin
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 \begin{align} \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} \end{align}
  
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 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 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 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 Squaring this equation gives
 \[E_0^2 + (p-p')^2c^2 + 2E_0(p-p')c = E_0^2 + P_e^2c^2,\] \[E_0^2 + (p-p')^2c^2 + 2E_0(p-p')c = E_0^2 + P_e^2c^2,\]
 which can be rearranged to which can be rearranged to
 \[P_e^2 = p^2 + p'^2 - 2pp' + \frac{2E_0(p-p')}{c}.\] \[P_e^2 = p^2 + p'^2 - 2pp' + \frac{2E_0(p-p')}{c}.\]
-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
 \[p^2 + p'^2 - 2 pp'\cos\theta = p^2 + p'^2 - 2pp' + \frac{2E_0(p-p')}{c}.\] \[p^2 + p'^2 - 2 pp'\cos\theta = p^2 + p'^2 - 2pp' + \frac{2E_0(p-p')}{c}.\]
 This can be rearranged to give This can be rearranged to give
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 {{:question-mark.png?direct&50|}} {{:question-mark.png?direct&50|}}
 ====== In Class Activity ====== ====== 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
   \[\frac{E_0(p-p')}{c} = pp'(1 - \cos \theta),\]   \[\frac{E_0(p-p')}{c} = pp'(1 - \cos \theta),\]
   show that the wavelength shift $\Delta \lambda = \lambda' - \lambda$ is   show that the wavelength shift $\Delta \lambda = \lambda' - \lambda$ is
   \[\Delta \lambda = \frac{h}{m_e c} \left ( 1 - \cos \theta \right ).\]   \[\Delta \lambda = \frac{h}{m_e c} \left ( 1 - \cos \theta \right ).\]