컴프턴 산란
📂양자역학 컴프턴 산란 공식 λ \lambda λ λ ′ \lambda^{\prime} λ ′
λ ′ − λ = h m e c ( 1 − cos θ )
\lambda^{\prime} -\lambda = \frac{h}{m_{e}c}(1-\cos\theta)
λ ′ − λ = m e c h ( 1 − cos θ )
이때 h h h m e m_{e} m e c c c θ \theta θ
cos θ = 1 − m e c 2 ( E − E ′ ) E ′ E
\cos \theta=1-\frac{m_{e}c^{2}(E-E^{\prime})}{E^{\prime}E}
cos θ = 1 − E ′ E m e c 2 ( E − E ′ )
설명 컴프턴 산란
λ ′ − λ = h m e c ( 1 − cos θ ) > 0 \lambda ^{\prime}-\lambda=\dfrac{h}{m_{e} c}(1-\cos\theta) > 0 λ ′ − λ = m e c h ( 1 − cos θ ) > 0
유도 전략: 운동량 보존 법칙과 에너지 보존 법칙을 사용하여 결과를 이끌어낸다.
p γ \mathbf{p}_\gamma p γ p e \mathbf{p}_{e} p e p γ ′ \mathbf{p}_\gamma^{\prime} p γ ′ p e ′ \mathbf{p}_{e}^{\prime} p e ′
Part 1. 운동량 보존 법칙
p γ + p e = p γ ′ + p e ′
\mathbf{p}_{\gamma}+\mathbf{p}_{e}=\mathbf{p}_{\gamma}^{\prime}+\mathbf{p}_{e}^{\prime}
p γ + p e = p γ ′ + p e ′
충돌 후의 전자에 대해서는 정보가 없으므로 p e ′ \mathbf{p}_{e}^{\prime} p e ′
p γ + p e − p γ ′ = p e ′
\mathbf{p}_{\gamma}+\mathbf{p}_{e}-\mathbf{p}_{\gamma}^{\prime}=\mathbf{p}_{e}^{\prime}
p γ + p e − p γ ′ = p e ′
충돌 전 전자는 정지상태 이므로 p e = 0 \mathbf{p}_{e}=0 p e = 0
( p γ − p γ ′ ) 2 = ( p e ′ ) 2 ⟹ ( p γ ) 2 + ( p γ ′ ) 2 − 2 p γ ⋅ p γ ′ = ( p e ′ ) 2
\begin{align*}
&&(\mathbf{p}_{\gamma}-\mathbf{p}_{\gamma}^{\prime})^2&=(\mathbf{p}_{e}^{\prime})^2
\\ \implies &&(p_\gamma)^2 +(p_\gamma^{\prime})^2-2 \mathbf{p}_{\gamma} \cdot \mathbf{p}_{\gamma}^{\prime} &=(p_{e}^{\prime})^2
\end{align*}
⟹ ( p γ − p γ ′ ) 2 ( p γ ) 2 + ( p γ ′ ) 2 − 2 p γ ⋅ p γ ′ = ( p e ′ ) 2 = ( p e ′ ) 2
광자의 정지질량은 0 0 0 p γ = E c = h ν c p_\gamma=\dfrac{E}{c}=\dfrac{h\nu}{c} p γ = c E = c h ν
h 2 ν 2 c 2 + h 2 ν ′ 2 c 2 − 2 h 2 ν ν ′ c 2 cos θ = ( p e ′ ) 2 (1)
\frac{h^2\nu^2}{c^2}+\frac{h^2{\nu^{\prime}}^{2}}{c^2}-\frac{2h^2\nu\nu^{\prime}}{c^2}\cos\theta=(p_{e}^{\prime})^2\tag{1}
c 2 h 2 ν 2 + c 2 h 2 ν ′ 2 − c 2 2 h 2 ν ν ′ cos θ = ( p e ′ ) 2 ( 1 )
Part 2. 에너지 보존 법칙
이제 E γ E_\gamma E γ E γ ′ E_{\gamma}^{\prime} E γ ′ E e E_{e} E e E e ′ E_{e}^{\prime} E e ′
E γ ′ + E e ′ = E γ + E e ⟹ E e ′ = E γ + E e − E γ ′ ⟹ ( E e ′ ) 2 = ( E γ + E e − E γ ′ ) 2 ⟹ ( E e ′ ) 2 = ( E γ ) 2 + ( E e ) 2 + ( E γ ′ ) 2 + 2 E γ E e − 2 E γ E γ ′ − 2 E e E γ ′
\begin{align*}
&& E_{\gamma}^{\prime}+E_{e}^{\prime} &= E_\gamma+E_{e} \\
\implies && E_{e}^{\prime} &= E_\gamma+E_{e}-E_{\gamma}^{\prime} \\
\implies && (E_{e}^{\prime})^2 &= (E_\gamma+E_{e}-E_{\gamma}^{\prime})^2 \\
\implies && (E_{e}^{\prime})^2 &= (E_\gamma)^2+(E_{e})^2+(E_{\gamma}^{\prime})^2+2E_\gamma E_{e}-2E_\gamma E_{\gamma}^{\prime}-2E_{e}E_{\gamma}^{\prime}
\end{align*}
⟹ ⟹ ⟹ E γ ′ + E e ′ E e ′ ( E e ′ ) 2 ( E e ′ ) 2 = E γ + E e = E γ + E e − E γ ′ = ( E γ + E e − E γ ′ ) 2 = ( E γ ) 2 + ( E e ) 2 + ( E γ ′ ) 2 + 2 E γ E e − 2 E γ E γ ′ − 2 E e E γ ′ E = h ν E=h\nu E = h ν E = ( m c 2 ) 2 + p 2 c 2 E=\sqrt{(mc^2)^2+p^2c^2} E = ( m c 2 ) 2 + p 2 c 2 h 2 ν 2 + m e 2 c 4 + h 2 ν ′ 2 + 2 h ν m e c 2 − 2 h 2 ν ν ′ − 2 m e c 2 h ν ′ = m e c 4 + ( p e ′ ) 2 c 2
h^2\nu^2+m_{e}^2c^4+h^2{\nu^{\prime}}^{2}+2h\nu m_{e}c^2-2h^2\nu\nu^{\prime}-2m_{e}c^2h\nu^{\prime}=m_{e}c^4+(p_{e}^{\prime})^2c^2
h 2 ν 2 + m e 2 c 4 + h 2 ν ′ 2 + 2 h ν m e c 2 − 2 h 2 ν ν ′ − 2 m e c 2 h ν ′ = m e c 4 + ( p e ′ ) 2 c 2 ( p e ′ ) 2 (p_{e}^{\prime})^2 ( p e ′ ) 2 h 2 ν 2 c 2 + h 2 ν ′ 2 c 2 + 2 m e h ( ν − ν ′ ) − 2 h 2 ν ν ′ c 2 = ( p e ′ ) 2 (2)
\frac{h^2\nu^2}{c^2} +\frac{h^2{\nu^{\prime}}^{2}}{c^2} +2m_{e} h(\nu-\nu^{\prime}) -\frac{2h^2\nu\nu^{\prime}}{c^2}=(p_{e}^{\prime})^2 \tag{2}
c 2 h 2 ν 2 + c 2 h 2 ν ′ 2 + 2 m e h ( ν − ν ′ ) − c 2 2 h 2 ν ν ′ = ( p e ′ ) 2 ( 2 )
Part 3.
( 1 ) (1) ( 1 ) ( 2 ) (2) ( 2 )
h 2 ν 2 c 2 + h 2 ν ′ 2 c 2 − 2 h 2 ν ν ′ c 2 cos θ = h 2 ν 2 c 2 + h 2 ν ′ 2 c 2 + 2 m e h ( ν − ν ′ ) − 2 h 2 ν ν ′ c 2 ⟹ − 2 h 2 ν ν ′ c 2 cos θ = 2 m e h ( ν − ν ′ ) − 2 h 2 ν ν ′ c 2 ⟹ 2 m e h ( ν − ν ′ ) = 2 h 2 ν ν ′ c 2 ( 1 − cos θ ) ⟹ ( ν − ν ′ ) = h m e ν ν ′ c 2 ( 1 − cos θ )
\begin{align*}
&& \frac{h^2\nu^2}{c^2}+\frac{h^2{\nu^{\prime}}^{2}}{c^2}-\frac{2h^2\nu\nu^{\prime}}{c^2}\cos\theta&=\ \frac{h^2\nu^2}{c^2} +\frac{h^2{\nu^{\prime}}^{2}}{c^2} +2m_{e} h(\nu-\nu^{\prime}) -\frac{2h^2\nu\nu^{\prime}}{c^2}
\\ \implies && -\frac{2h^2\nu\nu^{\prime}}{c^2}\cos\theta& =2m_{e} h(\nu-\nu^{\prime}) -\frac{2h^2\nu\nu^{\prime}}{c^2}
\\ \implies && 2m_{e} h(\nu-\nu^{\prime})&=\ \frac{2h^2\nu\nu^{\prime}}{c^2}(1-\cos\theta)
\\ \implies && (\nu-\nu^{\prime})&=\ \frac{h}{m_{e}}\frac{\nu\nu^{\prime}}{c^2}(1-\cos\theta)
\end{align*}
⟹ ⟹ ⟹ c 2 h 2 ν 2 + c 2 h 2 ν ′ 2 − c 2 2 h 2 ν ν ′ cos θ − c 2 2 h 2 ν ν ′ cos θ 2 m e h ( ν − ν ′ ) ( ν − ν ′ ) = c 2 h 2 ν 2 + c 2 h 2 ν ′ 2 + 2 m e h ( ν − ν ′ ) − c 2 2 h 2 ν ν ′ = 2 m e h ( ν − ν ′ ) − c 2 2 h 2 ν ν ′ = c 2 2 h 2 ν ν ′ ( 1 − cos θ ) = m e h c 2 ν ν ′ ( 1 − cos θ )
양변에 ν ν ′ \nu\nu^{\prime} ν ν ′
c ν ′ − c ν = h m e 1 c ( 1 − cos θ )
\frac{c}{\nu^{\prime}}-\frac{c}{\nu}=\frac{h}{m_{e}}\frac{1}{c}(1-\cos\theta)
ν ′ c − ν c = m e h c 1 ( 1 − cos θ )
λ = c ν \displaystyle \lambda=\frac{c}{\nu} λ = ν c
λ ′ − λ = h m e c ( 1 − cos θ )
\lambda ^{\prime}-\lambda=\frac{h}{m_{e} c}(1-\cos\theta)
λ ′ − λ = m e c h ( 1 − cos θ )
E = h ν = h c λ E=h\nu=\dfrac{ hc }{ \lambda } E = h ν = λ h c
cos θ = 1 − m e c 2 ( E − E ′ ) E ′ E
\cos \theta=1-\frac{m_{e}c^{2}(E-E^{\prime})}{E^{\prime}E}
cos θ = 1 − E ′ E m e c 2 ( E − E ′ )
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