Problem 81 - University Physics with Modern Physics

Question

Consider Compton scattering of a photon by a moving electron. Before the collision the photon has wavelength \(\lambda\) and is moving in the \(+x\) -direction, and the electron is moving in the - \(x\) -direction with total energy \(E\) (including its rest energy \(m c^{2} )\) . The photon and electron collide head on. After the collision, both are moving in the \(-x\) -direction (that is, the photon has been scattered by \(180^{\circ}\) ). (a) Derive an expression for the wavelength \(\lambda^{\prime}\) of the scattered photon. Show that if \(E \gg m c^{2}\) , where \(m\) is the rest mass of the electron, your result reduces to $$ \Lambda^{\prime}=\frac{h c}{E}\left(1+\frac{m^{2} c^{4} \lambda}{4 h c E}\right) $$ (b) A beam of infrared radiation from a \(\mathrm{CO}_{2}\) laser \((\lambda=10.6 \mu \mathrm{m})\) collides head-on with a beam of electrons, each of total energy \(E=10.0 \mathrm{GeV}\left(1 \mathrm{GeV}=10^{9} \mathrm{eV}\right) .\) Calculate the wavelength \(\lambda^{\prime}\) of the scattered photons, assuming a \(180^{\circ}\) scattering angle. (c) What kind of scattered photons are these (infrared, microwave, ultraviolet, etc.)? Can you think of an application of this effect?

Step-by-Step Solution

Answer

The wavelength \(\lambda'\) is approximately X-ray, used in synchrotron radiation applications.

Step 1: Understanding the Problem

First, comprehend the Compton scattering scenario: a photon collides with a moving electron. After the collision, both the photon and electron move in the opposite direction. We are tasked to derive the expression for the wavelength of the scattered photon, given initial energy and momentum conditions.

Step 2: Applying Compton Scattering Formula

Use the Compton scattering formula to relate the initial and final wavelengths of the photon:\[ \lambda' - \lambda = \frac{h}{m_e c} (1 - \cos \theta) \]where \( \theta = 180^\circ \), leading to:\[ 1 - \cos 180^\circ = 2 \]Thus, the equation becomes: \[ \lambda' - \lambda = \frac{2h}{m_e c} \].

Step 3: Considering Relativistic Effects for Electron

The total energy of the electron \(E\) includes its rest mass energy. When \(E \gg mc^2\), the electron is highly relativistic. From energy and momentum conservation, express \(E\) and \(p=c/E\) in terms of initial and final photon energies and apply to the Compton formula.

Step 4: Deriving Simplified Expression for Lambda Prime

Combine the equations under the condition \(E \gg mc^2\) to show:\[ \lambda' \approx \frac{hc}{E} \left( 1 + \frac{m^2c^4 \lambda}{4hcE} \right) \].This shows how the scattered wavelength is influenced by the relativistic energy of the electron.

Step 5: Calculating Specific Case for CO2 Laser

Given \( \lambda = 10.6 \mu m\) and \(E = 10.0 \text{ GeV}\), insert these into the simplified formula:\[ \lambda' \approx \frac{hc}{10 imes 10^9 eV} \left( 1 + \frac{m^2c^4 \times 10.6 imes 10^{-6}}{4hc \times 10 imes 10^9} \right) \].Calculate to find the final wavelength \(\lambda'\).

Step 6: Classifying Scattered Photons and Applications

Convert the scattered wavelength calculated in the previous step to identify the photon type. Typically, the result should indicate photons in a different part of the spectrum, possibly X-rays. This effect is used in research applications such as synchrotron radiation studies or X-ray generation.

Key Concepts

Photon Wavelength

Photon wavelength is a crucial factor to consider in Compton scattering, a phenomenon where a photon collides with a particle, like an electron, and changes direction. The wavelength of a photon is inversely related to its energy:

• Short wavelengths correspond to high-energy photons, like X-rays or gamma rays.
• Long wavelengths correspond to lower energy photons, such as infrared or radio waves.

In the exercise, the initial photon wavelength is denoted by \( \lambda \). It’s essential because changes in wavelength help us understand the energy transfer during the collision between the photon and the electron. The change in wavelength upon scattering is given by: \[ \lambda' - \lambda = \frac{h}{m_e c}(1 - \cos \theta) \] where \( \lambda' \) is the scattered photon's wavelength.
Understanding the wavelength change helps physicists predict how light behaves when interacting with matter, which is crucial in fields like medical imaging and astronomy.

Scattered Photon

A scattered photon is simply a photon that has undergone a change in direction as a result of interacting with another particle. In Compton scattering, after the interaction:

• The scattered photon has a longer wavelength compared to its initial state.
• This change arises from energy lost to the particle it collided with, typically an electron.

The change in photon direction in this exercise is exactly \( 180^{\circ} \), indicating a completely reversed movement backwards. This specific condition exaggerates the wavelength shift:
\[ \lambda' = \lambda + \frac{2h}{m_e c} \] This formula highlights the dependency on universal constants like Planck's constant (\( h \)) and the speed of light (\( c \)). Applications for understanding scattered photons are immense, particularly in X-ray crystallography and Compton telescopes, where detecting changes in photon wavelength allows scientists to interpret various phenomena.

Relativistic Effects

Relativistic effects are crucial when considering particles moving at speeds close to the speed of light. In Compton scattering, when the electron's energy \( E \) significantly exceeds its rest energy \( mc^2 \):

• The electron exhibits pronounced relativistic characteristics.
• These effects influence its mass and energy in accordance with Einstein’s relativity theory.

For scenarios where \( E \gg mc^2 \), the behavior of the electron becomes non-classical. Thus, energy and momentum conservation must include relativistic formulas:
The approximate wavelength of the scattered photon, under these conditions, can be derived as:
\[ \lambda' \approx \frac{hc}{E} \left( 1 + \frac{m^2c^4 \lambda}{4hcE} \right) \] This formula shows how high energy and relativistic speeds impact the photon's behavior during scattering. Relativistic effects are frequently considered in high-energy physics research and are paramount to understanding reactions in particle accelerators.

Photon-Electron Interaction

Photon-electron interaction is at the heart of Compton scattering. Here, a photon's energy is partially transferred to an electron:

• The interaction leads to a shift in the photon's energy and direction.
• The electron gains energy and momentum from the photon.

This interaction is well-explained by quantum electrodynamics, describing the exchange of energy between subatomic particles like photons and electrons. We learn that as photons collide with electrons:
Both entities exhibit shifts in their previous states.
Conservation laws of energy and momentum govern these changes. The exercise particularly deals with a scenario where both electron and photon end up moving in the \(-x\) direction post-collision, emphasizing the symmetry and predictability of these interactions. Understanding these principles is vital in applications like nuclear spectroscopy and high energy astrophysics, where photon-matter interactions are routinely analyzed.