E2-4. Compton ScatteringHL Only
1. The Compton Effect
Definition and Mechanism
- The Compton Effect describes the interaction of a high-energy photon (typically X-ray or gamma ray) with a stationary orbital electron, resulting in the ejection of the electron and a scattered photon.
- During the collision, the photon transfers a portion of its energy and momentum to the electron.
- Because the scattered photon has lost energy, its frequency decreases, and consequently, its wavelength increases ($E = hc/\lambda$).
- To conserve linear momentum, the photon and the ejected electron scatter at different angles relative to the original incident path.
2. The Compton Formula & Derivation
The precise shift in the photon's wavelength depends purely on the scattering angle $\theta$ of the photon. The change in wavelength ($\Delta \lambda$) is given by the Compton formula:
- $\Delta \lambda$: The shift in wavelength ($\lambda_{\text{scattered}} - \lambda_{\text{incident}}$).
- $m_e$: Rest mass of the electron ($9.11 \times 10^{-31} \text{ kg}$).
- $\theta$: The scattering angle of the photon relative to its initial direction.
Analyzing Extreme Angles:
- If the photon is not deflected ($\theta = 0^\circ$), then $\cos(0) = 1$, yielding $\Delta \lambda = 0$ (no energy lost).
- The maximum change in wavelength occurs during a direct backscatter where the photon reverses its path ($\theta = 180^\circ$). Here, $\cos(180^\circ) = -1$, yielding $\Delta \lambda = \frac{2h}{m_e c}$.
▼ Unfold Full Mathematical Derivation ▲ Fold Mathematical Derivation
The derivation requires simultaneous application of relativistic energy conservation and vector momentum conservation.
1. Initial and Final States:
Let the incident photon have energy $E = pc$ and momentum $p = \dfrac{h}{\lambda}$ along the x-axis. The target electron is stationary, meaning its initial rest energy is $E_e = m_e c^2$ and momentum is $p_e = 0$.
After scattering at angle $\theta$, the photon carries energy $E' = p'c$ and momentum $p' = \dfrac{h}{\lambda'}$. The electron recoils at an angle $\phi$ with total relativistic energy $E_e'$ and momentum $p_e'$.
2. Conservation of Relativistic Energy:
Total energy before the collision must equal total energy after:
$$E + m_e c^2 = E' + E_e' \implies E_e' = (E - E') + m_e c^2$$
Squaring both sides of the energy equation yields:
$$(E_e')^2 = (E - E')^2 + 2m_e c^2(E - E') + (m_e c^2)^2 \quad \text{--- (Eq. 1)}$$
Using the fundamental relativistic energy-momentum invariant relation for the electron, $(E_e')^2 = (p_e'c)^2 + (m_e c^2)^2$, we can rewrite Eq. 1 as:
$$(p_e'c)^2 = (E - E')^2 + 2m_e c^2(E - E') \quad \text{--- (Eq. 2)}$$
3. Conservation of Linear Momentum:
Resolving total system momentum into individual components along the x and y axes:
$$
\begin{aligned}
\text{x-axis:} \quad p &= p'\cos\theta + p_e'\cos\phi \implies p_e'\cos\phi = p - p'\cos\theta \\[0.2cm]
\text{y-axis:} \quad 0 &= p'\sin\theta - p_e'\sin\phi \implies p_e'\sin\phi = p'\sin\theta
\end{aligned}
$$
Squaring both component equations and adding them eliminates the electron recoil angle $\phi$ since $\sin^2\phi + \cos^2\phi = 1$. Hence,
$$
\begin{aligned}
(p_e')^2 &= (p - p'\cos\theta)^2 + (p'\sin\theta)^2 = p^2 - 2pp'\cos\theta + (p')^2\cos^2\theta + (p')^2\sin^2\theta \\
&= p^2 + (p')^2 - 2pp'\cos\theta
\end{aligned}
$$
Multiplying this entire equation by $c^2$ permits substitution of the photon energy parameters ($E = pc$ and $E' = p'c$):
$$(p_e'c)^2 = E^2 + (E')^2 - 2EE'\cos\theta \quad \text{--- (Eq. 3)}$$
4. Combining Energy and Momentum Equations:
Equating Eq. 2 and Eq. 3 eliminates the electron momentum term $(p_e'c)^2$:
$$(E - E')^2 + 2m_e c^2(E - E') = E^2 + (E')^2 - 2EE'\cos\theta$$
Expanding the left-hand squared term:
$$E^2 - 2EE' + (E')^2 + 2m_e c^2(E - E') = E^2 + (E')^2 - 2EE'\cos\theta$$
Canceling out $E^2$ and $(E')^2$ terms present on both sides leaves:
$$-2EE' + 2m_e c^2(E - E') = -2EE'\cos\theta$$
Dividing cleanly by 2 and rearranging isolates the mechanical rest mass component:
$$m_e c^2(E - E') = EE'(1 - \cos\theta)$$
5. Converting to Wavelength Coordinates:
Substituting de Broglie's wave representations ($E = \frac{hc}{\lambda}$ and $E' = \frac{hc}{\lambda'}$) yields:
$$m_e c^2 \left(\frac{hc}{\lambda} - \frac{hc}{\lambda'}\right) = \left(\frac{hc}{\lambda}\right)\left(\frac{hc}{\lambda'}\right)(1 - \cos\theta)$$
Dividing through by $h^2c^2$:
$$\frac{m_e c}{h} \left(\frac{1}{\lambda} - \frac{1}{\lambda'}\right) = \frac{1}{\lambda\lambda'}(1 - \cos\theta)$$
Establishing a common denominator on the left side:
$$\frac{m_e c}{h} \left(\frac{\lambda' - \lambda}{\lambda\lambda'}\right) = \frac{1}{\lambda\lambda'}(1 - \cos\theta)$$
Multiplying by $\lambda\lambda'$ and isolating the final term $\Delta\lambda = \lambda' - \lambda$:
$$\lambda' - \lambda = \frac{h}{m_e c}(1 - \cos\theta)$$
$$\mathbf{\Delta\lambda = \frac{h}{m_e c}(1 - \cos\theta)}$$
This completes the relativistic derivation of the Compton Scattering equation.
3. Examples
Example 1: Collision Kinematics and Wavelength Shift
Problem: An incident X-ray photon collides with a stationary orbital electron. The scattered photon recoils with an energy of $120 \text{ keV}$, while the ejected electron gains a kinetic energy of $40 \text{ keV}$.
(a) Determine the original wavelength of the incident X-ray photon.
(b) Calculate the scattering angle of the photon.
Solution:
Part (a): Find initial wavelength:
By the conservation of energy, the incident photon's energy ($E_i$) equals the sum of the scattered photon's energy and the electron's kinetic energy:
$$E_i = 120 \text{ keV} + 40 \text{ keV} = 160 \text{ keV}$$
Convert this energy to Joules:
$$E_i = 160 \times 10^3 \times (1.60 \times 10^{-19}) = 2.56 \times 10^{-14} \text{ J}$$
Calculate the incident wavelength using $E = hc/\lambda$:
$$\lambda_i = \frac{hc}{E_i} = \frac{(6.63 \times 10^{-34})(3.00 \times 10^8)}{2.56 \times 10^{-14}} = \mathbf{7.77 \times 10^{-12} \text{ m}}$$
Part (b): Determine the scattering angle:
First, calculate the wavelength of the scattered photon ($\lambda_f$) using its energy ($120 \text{ keV} = 1.92 \times 10^{-14} \text{ J}$):
$$\lambda_f = \frac{hc}{E_f} = \frac{(6.63 \times 10^{-34})(3.00 \times 10^8)}{1.92 \times 10^{-14}} = 1.036 \times 10^{-11} \text{ m}$$
Calculate the wavelength shift ($\Delta \lambda$):
$$\Delta \lambda = \lambda_f - \lambda_i = (10.36 \times 10^{-12}) - (7.77 \times 10^{-12}) = 2.59 \times 10^{-12} \text{ m}$$
Rearrange the Compton formula to solve for $\cos \theta$:
$$
\begin{aligned}
\cos \theta &= 1 - \left( \frac{m_e c \Delta \lambda}{h} \right) = 1 - \frac{(9.11 \times 10^{-31})(3.00 \times 10^8)(2.59 \times 10^{-12})}{6.63 \times 10^{-34}} \\
&= 1 - 1.067 = -0.067
\end{aligned}
$$
Therefore,
$$\theta = \cos^{-1}(-0.067) \approx \mathbf{93.8^\circ}$$
Example 2: The Maximum Momentum & Energy Transfer
Problem: A highly energetic gamma-ray photon with an initial energy of $1.20 \text{ MeV}$ undergoes Compton scattering with a free electron. The photon experiences a direct backscatter ($\theta = 180^\circ$).
(a) Calculate the exact wavelength shift ($\Delta \lambda$).
(b) Determine the kinetic energy imparted to the recoiling electron in MeV.
Solution:
Part (a): Wavelength shift at $180^\circ$:
Using the Compton formula for $\theta = 180^\circ$, we know $\cos(180^\circ) = -1$.
$$\Delta \lambda = \frac{h}{m_e c}(1 - (-1)) = \frac{2h}{m_e c}$$
Substitute constants:
$$\Delta \lambda = \frac{2 \times (6.63 \times 10^{-34})}{(9.11 \times 10^{-31}) \times (3.00 \times 10^8)} = \frac{1.326 \times 10^{-33}}{2.733 \times 10^{-22}} = \mathbf{4.85 \times 10^{-12} \text{ m}}$$
Part (b): Kinetic Energy of the Electron:
First, find the initial wavelength ($\lambda_i$) from $1.20 \text{ MeV}$ ($1.92 \times 10^{-13} \text{ J}$):
$$\lambda_i = \frac{hc}{E_i} = \frac{1.989 \times 10^{-25}}{1.92 \times 10^{-13}} = 1.036 \times 10^{-12} \text{ m}$$
Calculate the final wavelength ($\lambda_f$):
$$\lambda_f = \lambda_i + \Delta \lambda = (1.036 \times 10^{-12}) + (4.85 \times 10^{-12}) = 5.886 \times 10^{-12} \text{ m}$$
Find the final energy of the scattered photon ($E_f$) in Joules:
$$E_f = \frac{hc}{\lambda_f} = \frac{1.989 \times 10^{-25}}{5.886 \times 10^{-12}} = 3.379 \times 10^{-14} \text{ J}$$
Convert $E_f$ back to MeV:
$$E_f = \frac{3.379 \times 10^{-14}}{1.60 \times 10^{-13} \text{ J/MeV}} = 0.211 \text{ MeV}$$
By conservation of energy, the kinetic energy given to the electron is the energy lost by the photon:
$$K_e = E_i - E_f = 1.20 \text{ MeV} - 0.211 \text{ MeV} = \mathbf{0.989 \text{ MeV}}$$
Example 3: Fractional Energy Loss (Why Compton is ignored for visible light)
Problem: Compare a visible blue photon ($\lambda = 450 \text{ nm}$) to a hard X-ray photon ($\lambda = 0.05 \text{ nm}$). If both undergo $90^\circ$ Compton scattering ($\Delta \lambda = \dfrac{h}{m_ec} = 2.43 \times 10^{-12} \text{ m}$), calculate the percentage increase in wavelength for each, and explain why classical physics failed to notice Compton scattering until X-rays were utilized.
Solution:
For the visible photon ($450 \times 10^{-9}$ m): $$\% \text{ Shift} = \frac{\Delta \lambda}{\lambda} \times 100 = \frac{2.43 \times 10^{-12} \text{ m}}{450 \times 10^{-9} \text{ m}} \times 100 \approx \mathbf{0.00054\%}$$
For the X-ray photon ($0.05 \times 10^{-9}$ m): $$\% \text{ Shift} = \frac{\Delta \lambda}{\lambda} \times 100 = \frac{2.43 \times 10^{-12} \text{ m}}{50 \times 10^{-12} \text{ m}} \times 100 \approx \mathbf{4.86\%}$$
Explanation: The absolute wavelength shift ($\Delta \lambda$) is a universal constant for a given angle, regardless of the incoming photon's energy. For low-energy visible light, this shift is incredibly tiny compared to the original wavelength, resulting in practically zero measurable energy loss (acting like a classical elastic bounce). It is only when dealing with high-energy X-rays—where the initial wavelength is already on the order of picometers—that the Compton shift becomes a statistically significant, measurable fraction of the wave, proving the particle momentum transfer.