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Compton scattering 📂Quantum Mechanics

Compton scattering

Formula

Let λ\lambda be the wavelength of the incident light and λ′\lambda^{\prime} be the wavelength of the scattered photon. The following equation then holds true:

λ′−λ=hmec(1−cos⁡θ) \lambda^{\prime} -\lambda = \frac{h}{m_{e}c}(1-\cos\theta)

Here, hh is Planck’s constant, mem_{e} is the mass of the electron, cc is the speed of light, and θ\theta is the scattering angle. In terms of energy, we have:

cos⁡θ=1−mec2(E−E′)E′E \cos \theta=1-\frac{m_{e}c^{2}(E-E^{\prime})}{E^{\prime}E}

Explanation

Compton scattering1 refers to the phenomenon where X-rays scatter when they encounter electrons, resulting in both the X-rays and the electrons being deflected. When this happens, the wavelength of the scattered X-ray increases, which in terms of energy means that its energy decreases. This serves as evidence that X-rays, or light, possess particle-like properties.

Since λ′−λ=hmec(1−cos⁡θ)>0 \lambda ^{\prime}-\lambda=\dfrac{h}{m_{e} c}(1-\cos\theta) > 0 holds true, the wavelength of light increases after the collision. This matches well with experimental results and supports the particle nature of light.

Derivation

Strategy: Use the conservation of momentum and conservation of energy principles to derive the result.

Let pγ\mathbf{p}_\gamma be the momentum of the photon before the collision, pe\mathbf{p}_{e} be the momentum of the electron before the collision, pγ′\mathbf{p}_\gamma^{\prime} be the momentum of the photon after the collision, and pe′\mathbf{p}_{e}^{\prime} be the momentum of the electron after the collision. 5F4F54FA2.png

Part 1. Conservation of Momentum

Since there is no information about the electron after the collision, let’s solve for pe′\mathbf{p}_{e}^{\prime}.

pγ+pe=pγ′+pe′ \mathbf{p}_{\gamma}+\mathbf{p}_{e}=\mathbf{p}_{\gamma}^{\prime}+\mathbf{p}_{e}^{\prime}

Assuming the electron is initially at rest, we have pe=0\mathbf{p}_{e}=0.

pγ+pe−pγ′=pe′ \mathbf{p}_{\gamma}+\mathbf{p}_{e}-\mathbf{p}_{\gamma}^{\prime}=\mathbf{p}_{e}^{\prime}

Since the rest mass of the photon is 00, we have pγ=Ec=hνc p_\gamma=\dfrac{E}{c}=\dfrac{h\nu}{c}. Substituting this results in

h2ν2c2+h2ν′2c2−2h2νν′c2cos⁡θ=(pe′)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}

Part 2. Conservation of Energy

Now let EγE_\gamma be the energy of the photon before the collision, Eγ′E_{\gamma}^{\prime} be the energy of the photon after the collision, EeE_{e} be the energy of the electron before the collision, and Ee′E_{e}^{\prime} be the energy of the electron after the collision. Then,

Eγ′+Ee′=Eγ+Ee  ⟹  Ee′=Eγ+Ee−Eγ′  ⟹  (Ee′)2=(Eγ+Ee−Eγ′)2  ⟹  (Ee′)2=(Eγ)2+(Ee)2+(Eγ′)2+2EγEe−2EγEγ′−2EeEγ′ \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*}

The energy of the photon is E=hνE=h\nu, and the relativistic energy is E=(mc2)2+p2c2 E=\sqrt{(mc^2)^2+p^2c^2}. Reformulating this yields

h2ν2+me2c4+h2ν′2+2hνmec2−2h2νν′−2mec2hν′=mec4+(pe′)2c2 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

Rearranging for (pe′)2(p_{e}^{\prime})^2, we get

h2ν2c2+h2ν′2c2+2meh(ν−ν′)−2h2νν′c2=(pe′)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}

Part 3.

Using (1)(1) and (2)(2), we have

h2ν2c2+h2ν′2c2−2h2νν′c2cos⁡θ= h2ν2c2+h2ν′2c2+2meh(ν−ν′)−2h2νν′c2  ⟹  −2h2νν′c2cos⁡θ=2meh(ν−ν′)−2h2νν′c2  ⟹  2meh(ν−ν′)= 2h2νν′c2(1−cos⁡θ)  ⟹  (ν−ν′)= hmeνν′c2(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*}

Dividing both sides by νν′\nu\nu^{\prime} and multiplying by c,

cν′−cν=hme1c(1−cos⁡θ) \frac{c}{\nu^{\prime}}-\frac{c}{\nu}=\frac{h}{m_{e}}\frac{1}{c}(1-\cos\theta)

Since λ=cν\displaystyle \lambda=\frac{c}{\nu} holds, we get

λ′−λ=hmec(1−cos⁡θ) \lambda ^{\prime}-\lambda=\frac{h}{m_{e} c}(1-\cos\theta)

Due to E=hν=hcλE=h\nu=\dfrac{ hc }{ \lambda }, simplifying the above equation results in

cos⁡θ=1−mec2(E−E′)E′E \cos \theta=1-\frac{m_{e}c^{2}(E-E^{\prime})}{E^{\prime}E}

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  1. 컴프턴 효과(Compton Effect)라고도 한다. ↩︎