Problem 76 - University Physics with Modern Physics

Question

(a) Calculate the maximum increase in photon wavelength that can occur during Compton scattering. (b) What is the energy (in electron volts) of the lowest- energy \(x\) -ray photon for which Compton scattering could result in doubling the original wave- length?

Step-by-Step Solution

Answer

(a) Maximum wavelength increase: \( 4.86 \times 10^{-12} \text{m} \). (b) Energy of lowest-energy photon: 511 keV.

Step 1: Understand Compton Scattering

Compton scattering is a phenomenon where a photon scatters off a free electron, resulting in a change in the wavelength of the photon. The change in wavelength \( \Delta \lambda \) is given by the equation \( \Delta \lambda = \frac{h}{m_e c} (1 - \cos \theta) \), where \( h \) is Planck's constant, \( m_e \) is the electron mass, \( c \) is the speed of light, and \( \theta \) is the scattering angle.

Step 2: Calculate Maximum Wavelength Increase

To find the maximum increase in wavelength, set the scattering angle \( \theta = 180^\circ \). This results in \( 1 - \cos \theta = 2 \). Substituting these values into the equation gives \( \Delta \lambda = \frac{h}{m_e c} \times 2 \).

Step 3: Calculate Values

Substitute the known values into the equation: Planck's constant \( h = 6.626 \times 10^{-34} \, \text{Js} \), mass of electron \( m_e = 9.109 \times 10^{-31} \, \text{kg} \), and speed of light \( c = 3 \times 10^8 \, \text{m/s} \). This results in \[ \Delta \lambda = \frac{2 \times 6.626 \times 10^{-34}}{9.109 \times 10^{-31} \times 3 \times 10^8} = 4.86 \times 10^{-12} \text{m} \].

Step 4: Condition for Doubling Wavelength

For the wavelength to double, \( \Delta \lambda = \lambda \). From the Compton formula \( \Delta \lambda = \frac{h}{m_e c} (1 - \cos \theta) \), set \( \Delta \lambda = \lambda \). Thus, \( \lambda = \frac{h}{m_e c} \) for the smallest photon wavelength where this condition is met.

Step 5: Calculate Energy of Lowest-energy Photon

Use the relation \( \lambda = \frac{hc}{E} \) to find the energy. Substituting \( E = \frac{hc}{\lambda} = \frac{h \times c}{\frac{h}{m_e c}} = m_e c^2 \). Calculate using \( m_e = 9.109 \times 10^{-31} \, \text{kg} \) and \( c = 3 \times 10^8 \, \text{m/s} \) to get \( E = 8.19 \times 10^{-14} \text{J} \). Convert this energy to electron volts: \[ E = \frac{8.19 \times 10^{-14}\, \text{J}}{1.602 \times 10^{-19} \text{J/eV}} = 511 \text{keV} \].

Key Concepts

Photon Wavelength Change

Compton scattering is a fascinating process in physics where the interaction between a photon and a free electron causes the photon's wavelength to change. The change in wavelength, often denoted as \( \Delta \lambda \), can be calculated using the formula:

• \( \Delta \lambda = \frac{h}{m_e c} (1 - \cos \theta) \)

Here:

• \( h \) is Planck's constant \( (6.626 \times 10^{-34} \text{ Js}) \)
• \( m_e \) is the electron mass \( (9.109 \times 10^{-31} \text{ kg}) \)
• \( c \) is the speed of light \( (3 \times 10^8 \text{ m/s}) \)
• \( \theta \) is the scattering angle

To find the maximum possible increase in wavelength, the scattering angle \( \theta \) is set to \( 180^\circ \). This means the photon and electron move in opposite directions after the collision, maximizing the energy transfer. In this scenario, \( 1 - \cos \theta = 2 \), giving us \( \Delta \lambda = \frac{2h}{m_e c} \).Substituting the known values into this equation gives you the maximum \( \Delta \lambda \), approximately \( 4.86 \times 10^{-12} \text{ m} \). This comprehensive formula allows us to understand how varying the scattering angle impacts the wavelength change.

Scattering Angle in Physics

The scattering angle, \( \theta \), plays a crucial role in determining how much a photon's direction and energy change during Compton scattering. It is the angle between the original and the scattered photon paths. Here is how it works:

• If \( \theta \) is small, meaning the angle is close to 0, the change in wavelength is minimal because the photon’s path only slightly changes direction.
• As \( \theta \) approaches \( 180^\circ \), the photon's path and the electron's path differ more radically, maximizing the change in wavelength and energy transfer.

In physics problems involving scattering, such as this one, it's common to calculate how variables like the scattering angle affect outcomes. The formula \( (1 - \cos \theta) \) is important because it defines the extent of the interaction's effect on the photon's wavelength change.Understanding the scattering angle can also help predict the photon's new direction post-collision. This kind of insight is valuable in both theoretical and applied physics, such as material analysis or understanding cosmic phenomena.

Photon Energy Calculation

The energy of a photon is closely linked to its wavelength. The equation that captures this relationship is:

• \( E = \frac{hc}{\lambda} \)

Where \( E \) is the energy, \( h \) is Planck's constant, \( c \) is the speed of light, and \( \lambda \) is the wavelength. In the context of the Compton effect, if we are looking for the energy of a photon which can double its wavelength after scattering, we set \( \lambda = \Delta \lambda \). Continued calculations show that the energy of the photon which doubles its wavelength is equal to \( m_e c^2 \). Substituting in the known values gives:

• \( m_e c^2 = 8.19 \times 10^{-14} \text{ J} \)

To convert this energy from joules to electron volts, use the conversion factor \( 1 \, \text{eV} = 1.602 \times 10^{-19} \text{ J} \), resulting in about \( 511 \text{ keV} \). This calculation shows that a low-energy x-ray photon must have this much energy to potentially double its wavelength due to Compton scattering. Understanding this transformation is useful for analyzing photon interactions in quantum physics and technology.