Problem 35 - University Physics with Modern Physics

Question

A photon with wavelength 0.1100 nm collides with a free electron that is initially at rest. After the collision the wavelength is 0.1132 nm. (a) What is the kinetic energy of the electron after the collision? What is its speed? (b) If the electron is suddenly stopped (for example, in a solid target), all of its kinetic energy is used to create a photon. What is the wavelength of this photon?

Step-by-Step Solution

Answer

(a) The electron's kinetic energy is approximately \( 4.11 \times 10^{-16} \text{ J} \) and its speed is \( 9.50 \times 10^6 \text{ m/s} \). (b) The wavelength of the photon produced is approximately \( 0.485 \text{ nm} \).

Step 1: Determine the change in wavelength

First, calculate the change in wavelength using the formula \( \Delta \lambda = \lambda' - \lambda \), where \( \lambda = 0.1100 \) nm and \( \lambda' = 0.1132 \) nm. Substitute these values to get:\[ \Delta \lambda = 0.1132 \text{ nm} - 0.1100 \text{ nm} = 0.0032 \text{ nm} \]

Step 2: Use the Compton wavelength shift equation

The Compton wavelength shift equation is given by:\[ \Delta \lambda = \frac{h}{m_e c} (1 - \cos \theta) \]where \( h \) is Planck's constant \( 6.626 \times 10^{-34} \text{ J s} \), \( m_e \) is the electron mass \( 9.109 \times 10^{-31} \text{ kg} \), and \( c \) is the speed of light \( 3.00 \times 10^{8} \text{ m/s} \). Solving for \( \cos \theta \) gives:\[ 1 - \cos \theta = \frac{\Delta \lambda m_e c}{h} \]Substitute the known values into the equation to find:\[ 1 - \cos \theta = \frac{0.0032 \times 10^{-9} \times 9.109 \times 10^{-31} \times 3.00 \times 10^8}{6.626 \times 10^{-34}} \approx 0.0483 \]

Step 3: Calculate the kinetic energy of the electron

The kinetic energy \( K \) of the electron is related to the change in energy of the photon by the equation:\[ K = \Delta E = h c \left( \frac{1}{\lambda} - \frac{1}{\lambda'} \right) \]Substitute \( h = 6.626 \times 10^{-34} \text{ J s} \), \( c = 3.00 \times 10^8 \text{ m/s} \), \( \lambda = 0.1100 \times 10^{-9} \text{ m} \), and \( \lambda' = 0.1132 \times 10^{-9} \text{ m} \) into the equation:\[ K = 6.626 \times 10^{-34} \times 3.00 \times 10^8 \left( \frac{1}{0.1100 \times 10^{-9}} - \frac{1}{0.1132 \times 10^{-9}} \right) \approx 4.11 \times 10^{-16} \text{ J} \]

Step 4: Convert electron kinetic energy to electron speed

Use the kinetic energy formula \( K = \frac{1}{2} m v^2 \) to find the speed \( v \) of the electron.Rearranging gives \( v = \sqrt{\frac{2K}{m}} \).Substitute \( K = 4.11 \times 10^{-16} \text{ J} \) and \( m = 9.109 \times 10^{-31} \text{ kg} \):\[ v = \sqrt{\frac{2 \times 4.11 \times 10^{-16}}{9.109 \times 10^{-31}}} \approx 9.50 \times 10^6 \text{ m/s} \]

Step 5: Find the wavelength of the photon produced

The energy of the photon that could be generated by stopping the electron is \( K \) because all kinetic energy is transferred to the photon. The energy of a photon \( E \) is given by \( E = \frac{hc}{\lambda_f} \), where \( \lambda_f \) is the final wavelength.Rearranging gives \( \lambda_f = \frac{hc}{E} \).Substitute the values \( E = 4.11 \times 10^{-16} \text{ J} \):\[ \lambda_f = \frac{6.626 \times 10^{-34} \times 3.00 \times 10^8}{4.11 \times 10^{-16}} \approx 4.85 \times 10^{-10} \text{ m} \]

Key Concepts

Photon-Electron Collision

In the realm of quantum physics, when a high-energy photon collides with a free electron, an intriguing phenomenon occurs. This interaction is known as a photon-electron collision and is part of a broader concept named Compton scattering. During a photon-electron collision, the photon of light—though without mass—carries energy and momentum. When it impacts an electron, some of its energy is transferred to the electron, causing the electron to bounce or move while altering the path and energy of the photon.
Much like a game of billiards, this collision changes the direction and speed of the involved particles. It's essential because it showcases the particle nature of light and contributes to our understanding of quantum mechanics. In situations where the photon hits the electron, the photon releases some of its energy, which then increases the kinetic energy of the electron.
Understanding this interaction is crucial when analyzing certain properties, such as scattering angles and energy distribution, which are derived from how photons and electrons behave during collisions. Thus, a photon-electron collision is foundational for exploring how light and matter interact on a very tiny scale.

Kinetic Energy

Kinetic energy is the energy possessed by an object due to its motion. In the context of Compton scattering, kinetic energy becomes particularly relevant for the electron post-collision. Before the photon hits the electron, the electron might be at rest or quiescent. However, once the collision occurs, the electron gains kinetic energy as it moves.
To calculate this kinetic energy, the principle that energy is transferred from the photon to the electron comes into play. The equation is given by \( K = \Delta E = h c \left( \frac{1}{\lambda} - \frac{1}{\lambda'} \right) \), where \( K \) represents the kinetic energy, \( h \) is Planck's constant, \( c \) is the speed of light, and \( \lambda \) and \( \lambda' \) are the initial and final wavelengths, respectively.
This energy is significant because it dictates how fast the electron moves after the collision. By understanding the amount of kinetic energy transferred to an electron, we can infer its speed and how much of the energy from the incident photon was used to alter the electron's state of motion.

Wavelength Shift

One of the most fascinating outcomes of a photon-electron collision is the shift in wavelengths of the involved photon. This shift is known as the Compton wavelength shift. Initially, the photon has a specific wavelength before the interaction. Post-collision, a noticeable change in its wavelength is observed, due to the energy transferred during the impact.
Mathematically, this shift is expressed using the formula \( \Delta \lambda = \lambda' - \lambda \). For instance, if a photon's initial wavelength is 0.1100 nm and it becomes 0.1132 nm post-collision, the wavelength shift \( \Delta \lambda \) is 0.0032 nm.
Such shifts are crucial for physicists because they offer deeper insight into the wave-particle duality of light. The Compton wavelength shift confirms that light behaves as both a particle and a wave, thereby providing evidence for the dual nature of light. This phenomenon not only supported early quantum theories but also demonstrated that the wavelength shift could be directly linked to the angle and energy change during the scattering event of the photon and electron.