## 21.3 Compton's Effect This section covers two key phenomena that demonstrate the particle nature of light: the Compton effect, where photons scatter off electrons, and pair production, where a photon's energy is converted into a matter-antimatter pair. ## Compton's Effect **Definition:** The phenomenon, observed by Arthur H. Compton in 1923, where X-rays scattered by electrons have a longer wavelength ($\lambda'$) than the incident X-rays ($\lambda$). This change in wavelength is known as the **Compton shift**. **Particle Model Explanation:** Compton explained this effect by treating the interaction as a collision between two particles: a photon and an electron (initially at rest). In this collision, both energy and momentum are conserved. - The incident photon transfers some of its energy and momentum to the electron. - Consequently, the scattered photon has less energy, a lower frequency, and thus a longer wavelength. **Conservation Laws:** - **Energy Conservation:** The initial energy of the photon equals the sum of the scattered photon's energy and the kinetic energy of the recoiling electron. $$ hf = K.E + hf' $$ - **Momentum Conservation:** The momentum of the system is conserved vectorially. $$ \mathbf{p} = \mathbf{p}_{\mathbf{o}} + \mathbf{p}' $$ **Photon Momentum:** The momentum of a photon is given by: $$ p = \frac{E}{c} = \frac{hf}{c} = \frac{h}{\lambda} $$ **Compton Shift Formula:** The change in wavelength, $\Delta \lambda = \lambda' - \lambda$, is given by: $$ \Delta \lambda = \frac{h}{m_{0} c}(1 - \cos \theta) $$ - $\theta$ is the scattering angle of the photon. - $m_0$ is the rest mass of the electron. - The term $\frac{h}{m_{0} c}$ is a constant called the **Compton wavelength**, with a value of $0.00243 \text{ nm}$. **Maximum Compton Shift:** The Compton shift is maximum when $\theta = 180^\circ$ (backward scattering), giving: $$ \Delta \lambda_{\text{max}} = \frac{2h}{m_{0} c} = 2 \times 0.00243 \text{ nm} = 0.00486 \text{ nm} $$ **Dependencies:** The Compton shift depends only on the scattering angle $\theta$ and is independent of the incident photon's wavelength and the nature of the scattering material. Since $(1 - \cos \theta)$ is always non-negative, the scattered wavelength $\lambda'$ is always greater than or equal to the incident wavelength $\lambda$. This effect provides evidence for the particle nature of electromagnetic radiation, supporting the concepts discussed in and . ### Pair Production **Definition:** The process where a high-energy photon, passing near a heavy nucleus, is absorbed and its energy is converted into a particle-antiparticle pair, typically an electron and a positron. **Significance:** This phenomenon is direct evidence of the conversion of energy into mass, as described by Einstein's mass-energy equivalence. **Conservation Laws:** 1. **Charge:** The net charge before (photon, 0) and after (electron -e, positron +e) is zero. 2. **Momentum:** The initial momentum of the photon is transferred to the resulting particles. The nearby nucleus is necessary to absorb some recoil momentum while conserving energy. 3. **Energy:** The photon's energy is converted into the rest mass energy and kinetic energy of the new particles. **Energy Threshold:** - The total energy of the created pair is the sum of their rest mass energies and kinetic energies: $$ hf = (K.E_{\text{electron}} + m_{0} c^{2}) + (K.E_{\text{positron}} + m_{0} c^{2}) $$ - Since the rest mass energy of an electron (or positron) is $m_0 c^2 = 0.511 \text{ MeV}$, the minimum photon energy required is: $$ hf_{\text{min}} = 2m_0 c^2 = 1.02 \text{ MeV} $$ - Any photon energy above this threshold appears as the kinetic energy of the electron and positron. ## Possible Questions/Answer **Q:** An X-ray of 0.20 nm is incident on a target. What is the Compton shift that can be expected at a $45^{\circ}$ scattering angle? **A:** The Compton shift is calculated using the formula $\Delta \lambda = \frac{h}{m_{0} c}(1-\cos \theta)$. Plugging in the constants and the angle: $\Delta \lambda = \frac{6.63 \times 10^{-34} \text{ J s}}{(9.11 \times 10^{-31} \text{ kg})(3 \times 10^{8} \text{ m/s})}(1-\cos 45^{\circ})$ $\Delta \lambda = (2.426 \times 10^{-12} \text{ m})(1 - 0.707) = 7.11 \times 10^{-13} \text{ m}$. ## Formula Table | Formula | Description | | :--- | :--- | | $\Delta \lambda = \frac{h}{m_{0} c}(1 - \cos \theta)$ | **Compton Shift:** The change in a photon's wavelength after scattering. | | $p = \frac{h}{\lambda}$ | **Photon Momentum:** The momentum of a photon. | | $hf_{\text{min}} = 2m_0 c^2 \approx 1.02 \text{ MeV}$ | **Pair Production Threshold:** The minimum energy a photon needs to create an electron-positron pair. | ## References (Derived from FBISE textbook)