Problem 21 - University Physics with Modern Physics

Question

If a photon of wavelength 0.04250 nm strikes a free electron and is scattered at an angle of \(35.0^{\circ}\) from its original direction, find (a) the change in the wavelength of this photon; (b) the wavelength of the scattered light; (c) the change in energy of the photon (is it a loss or a gain?); (d) the energy gained by the electron.

Step-by-Step Solution

Answer

The wavelength shift is 0.01499 nm, scattered wavelength is 0.05749 nm, photon energy decreases by 1.22 × 10⁻¹⁵ J, and the electron gains this energy.

Step 1: Identify the formula for Compton wavelength shift

The change in wavelength due to Compton scattering is given by the Compton wavelength shift formula: \[ \Delta \lambda = \frac{h}{m_e c} (1 - \cos \theta) \] where \( h \) is Planck's constant \( (6.626 \times 10^{-34} \, \text{m}^2 \cdot \text{kg/s}) \), \( m_e \) is the electron mass \( (9.109 \times 10^{-31} \, \text{kg}) \), \( c \) is the speed of light \( (3.00 \times 10^8 \, \text{m/s}) \), and \( \theta \) is the scattering angle in radians.

Step 2: Calculate the change in wavelength

First, calculate the Compton wavelength constant: \[ \frac{h}{m_e c} = \frac{6.626 \times 10^{-34}}{9.109 \times 10^{-31} \times 3.00 \times 10^8} \approx 2.43 \times 10^{-12} \, \text{m} \] Then, convert the angle to radians: \( 35.0^{\circ} = 0.611 \text{ rad} \).Now compute the change in wavelength: \[ \Delta \lambda = (2.43 \times 10^{-12}) (1 - \cos 0.611) \approx 1.499 \times 10^{-13} \, \text{m} = 0.01499 \, \text{nm} \]

Step 3: Calculate the wavelength of the scattered light

The initial wavelength of the photon is 0.04250 nm. Add the shift to this initial wavelength to find the new wavelength:\[ \lambda_{new} = \lambda_{old} + \Delta \lambda = 0.04250 \text{ nm} + 0.01499 \text{ nm} = 0.05749 \text{ nm} \]

Step 4: Determine the change in energy of the photon

Energy is related to wavelength by the equation: \[ E = \frac{hc}{\lambda} \]Calculate the energies before and after the shift:- Initial energy: \[ E_{old} = \frac{6.626 \times 10^{-34} \times 3.00 \times 10^8}{0.04250 \times 10^{-9}} \approx 4.69 \times 10^{-15} \text{ J} \]- Final energy: \[ E_{new} = \frac{6.626 \times 10^{-34} \times 3.00 \times 10^8}{0.05749 \times 10^{-9}} \approx 3.47 \times 10^{-15} \text{ J} \]Therefore, the change in energy is:\[ \Delta E_{photon} = E_{new} - E_{old} \approx 3.47 \times 10^{-15} - 4.69 \times 10^{-15} = -1.22 \times 10^{-15} \text{ J} \]The negative sign indicates a loss in energy.

Step 5: Calculate the energy gained by the electron

Since energy is conserved, the energy lost by the photon is gained by the electron. Therefore, the energy gained by the electron is: \[ \Delta E_{electron} = 1.22 \times 10^{-15} \text{ J} \]

Key Concepts

Understanding Wavelength Shift in Compton Scattering

Compton scattering occurs when a photon, which is a particle of light, collides with a free electron, causing the photon to scatter at a different angle and wavelength. The change in wavelength, known as the wavelength shift, is a key aspect of Compton scattering. It is calculated using the formula:
\[ \Delta \lambda = \frac{h}{m_e c} (1 - \cos \theta) \]
Here, \( h \) is Planck's constant, \( m_e \) is the mass of the electron, \( c \) is the speed of light, and \( \theta \) is the scattering angle.

• Planck's constant: \( 6.626 \times 10^{-34} \, \text{m}^2 \, \text{kg/s} \)
• Electron mass: \( 9.109 \times 10^{-31} \, \text{kg} \)
• Speed of light: \( 3.00 \times 10^8 \, \text{m/s} \)

The wavelength shift, \( \Delta \lambda \), depends on the scattering angle \( \theta \). As the angle increases, the change in wavelength also becomes more significant.
Through calculations, the shift in this case is approximately \( 0.01499 \, \text{nm} \). This shift represents how far the photon's wavelength has moved from its original value due to the scattering process.

Photon Energy Before and After Scattering

Photon energy is intrinsically linked to its wavelength. A photon's energy can be determined using the equation:

• \( E = \frac{hc}{\lambda} \)

Where \( E \) is the energy, \( h \) is Planck's constant, \( c \) is the speed of light, and \( \lambda \) is the wavelength.
In this problem, we calculate the photon energy before and after it experiences a wavelength shift due to scattering.

• Initial energy of the photon: Around \( 4.69 \times 10^{-15} \text{ J} \)
• Final energy after scattering: Around \( 3.47 \times 10^{-15} \text{ J} \)

This demonstrates that the photon loses energy as it changes wavelength. The negative energy change indicates this loss, reflecting the energy transfer to another particle—in this case, the electron.

Electron Energy Gain in Compton Scattering

When a photon undergoes Compton scattering, it's not just the photon that is affected—the scattering also impacts the electron it collides with. According to the conservation of energy, the total energy remains constant, meaning any energy lost by the photon must be gained by the electron.
In the presented situation, the photon loses \( 1.22 \times 10^{-15} \text{ J} \) of energy through scattering.
Hence, this energy is transferred to the electron, increasing its kinetic energy by the same amount.

• Energy gained by the electron: \( 1.22 \times 10^{-15} \text{ J} \)

This gain is crucial for understanding how photons and particles like electrons interact, showcasing a foundational principle of physics: energy conservation.
Each interaction between a photon and an electron, such as Compton scattering, provides insight into quantum mechanics and the dual nature of light. Understanding these processes is essential for fields like material science and medical imaging.