Problem 42 - University Physics with Modern Physics

Question

An x-ray photon is scattered from a free electron (mass \(m )\) at rest. The wavelength of the scattered photon is \(\lambda^{\prime},\) and the final speed of the struck electron is \(v\) . (a) What was the initial wave-length \(\lambda\) of the photon? Express your answer in terms of \(\lambda^{\prime}, v\) and \(m .\) (Hint. Use the relativistic expression for the electron kinetic energy.) (b) Through what angle \(\phi\) is the photon scattered? Express your answer in terms of \(\lambda, \lambda^{\prime},\) and \(m\) . (c) Evaluate your results in parts (a) and (b) for a wavelength of \(5.10 \times 10^{-3}\) nm for the scattered photon and a final electron speed of \(1.80 \times 10^{8} \mathrm{m} / \mathrm{s}\) . Give \(\phi\) in degrees.

Step-by-Step Solution

Answer

(a) \( \lambda = \frac{hc}{\frac{hc}{\lambda'} - \frac{1}{2}mv^2} \). (b) \( \phi = \cos^{-1}(1 - \frac{(\lambda' - \lambda) mc}{h}) \). Evaluated: \( \phi \approx 15.5^\circ \).

Step 1: Understanding the Problem

We are dealing with the scattering of photons and electrons. To solve the exercise, we need to find the initial wavelength, \( \lambda \), and the scattering angle, \( \phi \). The change in wavelength can be analyzed using the Compton equation, and kinetic energy principles.

Step 2: Using Compton's Equation

The Compton shift formula is given by: \( \Delta \lambda = \lambda' - \lambda = \frac{h}{mc}(1 - \cos \phi) \). However, part (a) asks to find \( \lambda \) in terms of the given quantities, so we need to relate this to the given kinetic energy condition.

Step 3: Calculating Initial Wavelength \(\lambda\)

Use the relativistic kinetic energy formula: \( K = \frac{1}{2}mv^2 = \frac{h}{\lambda'}c - \frac{h}{\lambda}c \). Rearrange to solve for \( \lambda \): \( \frac{h}{\lambda} = \frac{h}{\lambda'} - \frac{1}{2}mv^2/c \). Thus, initial wavelength is \( \lambda = \frac{hc}{\frac{hc}{\lambda'} - \frac{1}{2}mv^2} \).

Step 4: Calculating the Scattering Angle \(\phi\)

Using \( \lambda \) from (a), find \( \cos \phi \) using Compton's formula: \( 1 - \cos \phi = \frac{\Delta \lambda mc}{h} \). Rearrange: \( \phi = \cos^{-1}(1 - \frac{\Delta \lambda mc}{h}) \) where \( \Delta \lambda = \lambda' - \lambda \).

Step 5: Evaluating Given Values

For \( \lambda' = 5.10 \times 10^{-3} \) nm and \( v = 1.80 \times 10^8 \text{ m/s} \), first find \( \lambda \) using \( \lambda = \frac{hc}{\frac{hc}{5.10 \times 10^{-12} \text{ m}} - \frac{1}{2}(9.11 \times 10^{-31} \text{ kg})(1.80 \times 10^8 \text{ m/s})^2} \). Then, calculate \( \phi \) with \( \phi = \cos^{-1}(1 - \frac{(\lambda' - \lambda) mc}{h}) \) to find the angle in degrees.

Key Concepts

Relativistic Kinetic Energy

When particles like electrons move at very high speeds, close to the speed of light, their kinetic energy cannot be calculated using the classical equation, which is \( K = \frac{1}{2}mv^2 \). Instead, we use relativistic kinetic energy to get more accurate results.
Relativistic kinetic energy is given by the formula:\[ K = (\gamma - 1)mc^2 \]where \( \gamma \) (the Lorentz factor) is determined by \[ \gamma = \frac{1}{\sqrt{1 - \left(\frac{v}{c}\right)^2}} \]This formula accounts for the fact that as the speed increases and approaches the speed of light \(c\), the effective mass of the electron becomes greater.
In the context of the Compton scattering problem, we use the relativistic expression to understand the energy transfer from the photon to the electron, which helps us find initial conditions like the wavelength of the photon.

Photon-Electron Interaction

In Compton scattering, a photon collides with an electron, causing the photon's path to change direction and its wavelength to shift. When the x-ray photon interacts with a free electron at rest, energy and momentum are transferred from the photon to the electron.
During this interaction:

• The photon provides energy, part of which is absorbed by the electron, causing it to move.
• The rest of the energy is carried away by the scattered photon, which loses a part of its initial energy, resulting in a longer wavelength.

This interaction is significant because it validates the particle-like behavior of light. The experiment conducted by Arthur Compton in 1923 demonstrated this phenomenon, showing that light interacts with matter not only as a wave but also as a particle.

Wavelength Shift

This phenomenon is crucial to understanding Compton scattering. The change in wavelength of the photon after it scatters off an electron is called the wavelength shift.
The shift is calculated using Compton's equation:\[ \Delta \lambda = \lambda' - \lambda = \frac{h}{mc}(1 - \cos \phi) \]where \( \lambda' \) is the scattered photon's wavelength, \( \lambda \) is the initial wavelength, and \( \phi \) is the scattering angle.
This shift occurs because some of the photon's energy is transferred to the electron, reducing its energy and hence increasing its wavelength. It is also an experimental observation supporting the photon theory of light.

Scattering Angle

The scattering angle \( \phi \) is the angle at which the photon is deflected after striking the electron.
Using the Compton equation, the scattering angle is derived from the change in the photon's wavelength:\[ 1 - \cos \phi = \frac{\Delta \lambda mc}{h} \]To find \( \phi \), we rearrange the equation:\[ \phi = \cos^{-1}\left(1 - \frac{\Delta \lambda mc}{h}\right) \]The angle provides insight into how much energy and momentum are transferred during the interaction. A larger scattering angle means more energy is transferred to the electron, leading to a larger shift in the wavelength of the photon. By measuring \( \phi \), we can deduce crucial details about the energy dynamics in the scattering process.