Problem 41 - 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 \(\mathrm{nm}\) (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) KE = \( 3.73 \times 10^{-16} \) J, speed = \( 9.04 \times 10^6 \) m/s; (b) Wavelength = \( 5.32 \) pm.

Step 1: Understanding the Problem

In this problem, we are analyzing a Compton scattering event where a photon collides with an electron at rest, causing a change in the photon's wavelength. We need to determine the kinetic energy and speed of the electron after the collision and the wavelength of a photon that could be produced if the electron's kinetic energy is converted into the energy of a photon.

Step 2: Calculate Change in Wavelength

The change in the photon's wavelength is given by the Compton shift formula: \[\Delta \lambda = \lambda' - \lambda = \frac{h}{m_e c} (1 - \cos \theta)\]where \( h = 6.63 \times 10^{-34} \) J·s is Planck’s constant, \( m_e = 9.11 \times 10^{-31} \) kg is the electron mass, and \( c = 3.00 \times 10^8 \) m/s is the speed of light. In this case, \( \Delta \lambda = 0.1132 \text{ nm} - 0.1100 \text{ nm} = 0.0032 \text{ nm} = 3.2 \times 10^{-12} \text{ m} \). We will use this change to find the angle of scattering or other needed values.

Step 3: Calculate Electron's Kinetic Energy

The energy gained by the electron can be determined using the change in energy of the photon:\( E_i = \frac{hc}{\lambda} \) and \( E_f = \frac{hc}{\lambda'} \) are the initial and final energies of the photon. The kinetic energy \( KE \) of the electron is then given by:\[ KE = E_i - E_f = hc \left( \frac{1}{\lambda} - \frac{1}{\lambda'} \right) \]Substitute the given values (\( \lambda = 0.1100 \) nm, \( \lambda' = 0.1132 \) nm, \( h = 6.63 \times 10^{-34} \) J·s, \( c = 3 \times 10^8 \) m/s):\[ KE = 6.63 \times 10^{-34} \times 3 \times 10^8 \left( \frac{1}{0.1100 \times 10^{-9}} - \frac{1}{0.1132 \times 10^{-9}} \right) \approx 3.73 \times 10^{-16} \text{ J} \]

Step 4: Calculate Electron's Speed

Using the kinetic energy formula \( KE = \frac{1}{2} m_e v^2 \), we solve for velocity \( v \):\[ v = \sqrt{\frac{2 \cdot KE}{m_e}} \]Substituting the values for kinetic energy and the mass of the electron:\[ v = \sqrt{\frac{2 \cdot 3.73 \times 10^{-16}}{9.11 \times 10^{-31}}} \approx 9.04 \times 10^6 \text{ m/s} \]

Step 5: Calculate Wavelength of Photon Created from Kinetic Energy

If the electron is stopped and its kinetic energy is converted to a photon, we can use the energy-wavelength relationship: \[ E = \frac{hc}{\lambda} \Rightarrow \lambda = \frac{hc}{E} \]We know that \( E = KE = 3.73 \times 10^{-16} \text{ J} \), so:\[ \lambda = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{3.73 \times 10^{-16}} \approx 5.32 \times 10^{-12} \text{ m} = 5.32 \text{ pm} \]

Step 6: Conclusion and Final Results

The kinetic energy of the electron after the collision is approximately \( 3.73 \times 10^{-16} \) J, and its speed is about \( 9.04 \times 10^6 \) m/s. If this kinetic energy is fully converted into a photon, the resulting photon's wavelength would be approximately \( 5.32 \) pm.

Key Concepts

Photon Wavelength

When we dive into the concept of Compton scattering, the idea of a photon's wavelength is essential. Wavelength refers to the distance between successive peaks of a wave. For a photon, the initial wavelength \( \lambda \) plays a critical role when it collides with a stationary electron. In this case, the photon's initial wavelength was 0.1100 nm and changed to 0.1132 nm after the collision.

This change in wavelength, known as the Compton shift, is measurable and key to understanding the energy transfer that occurs during the collision. The formula to calculate the change in wavelength is:

• \( \Delta \lambda = \lambda' - \lambda \)

where \( \lambda' \) is the wavelength post-collision. This change helps us calculate other factors, like the scattered angle and the electron's kinetic energy post-collision.

Kinetic Energy Calculation

Calculating the kinetic energy of an electron after a collision involves understanding how energy is transferred from the photon to the electron. Initially, the photon's energy is calculated using the formula:

• \( E = \frac{hc}{\lambda} \)

where \( h \) is Planck's constant \( (6.63 \times 10^{-34} \text{ J·s}) \), and \( c \) is the speed of light \( (3 \times 10^8 \text{ m/s}) \).

The kinetic energy \( KE \) gained by the electron is based on the change in energy of the photon from initial \( E_i \) to final energy \( E_f \):

• \( KE = E_i - E_f \)

Using the given wavelength values, this results in a kinetic energy of approximately \( 3.73 \times 10^{-16} \text{ J} \). This energy reflects how much the electron accelerated after interacting with the photon.

Electron Speed

The speed of the electron post-collision is directly related to its kinetic energy. For this calculation, we use the kinetic energy formula:

• \( KE = \frac{1}{2} m_e v^2 \)

where \( m_e \) is the electron's mass \( (9.11 \times 10^{-31} \text{ kg}) \).

By rearranging the formula to solve for velocity \( v \), we get:

• \( v = \sqrt{\frac{2 \cdot KE}{m_e}} \)

After substituting the previously calculated kinetic energy \( 3.73 \times 10^{-16} \text{ J} \), the speed of the electron comes out to approximately \( 9.04 \times 10^6 \text{ m/s} \). This high speed indicates just how much energy was transferred during the collision.

Energy Conversion

In scenarios where an electron is abruptly stopped, such as by colliding with a solid, its kinetic energy can be wholly converted into a photon. The concept of energy conversion in this context involves using the kinetic energy to produce new photon energy.

The relationship between photon energy \( E \) and wavelength \( \lambda \) is expressed by the formula:

• \( E = \frac{hc}{\lambda} \Rightarrow \lambda = \frac{hc}{E} \)

Substituting the kinetic energy \( E = 3.73 \times 10^{-16} \text{ J} \) from the electron, we find the wavelength of the resulting photon is approximately \( 5.32 \text{ pm} \) (picometers).

This transformation showcases how energy conservation governs particle interactions and conversions into electromagnetic waves, maintaining the total energy throughout the process.