Problem 36 - 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 \(\lambda'\), and the final speed of the struck electron is \(v\). (a) What was the initial wavelength \(\lambda\) of the photon? Express your answer in terms of \(\lambda\), \(v\), 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 \(\lambda\), \(\lambda'\), 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\(^8\) m/s. Give \(\phi\) in degrees.

Step-by-Step Solution

Answer

(a) \(\lambda\) is calculated using kinetic energy and \(\lambda'\), for (b), \(\phi\) is determined using \(\cos^{-1}\); part (c) evaluates these using given values.

Step 1: Understanding the Compton wavelength shift

When an x-ray photon scatters off an electron at rest, the change in wavelength, known as the Compton shift, is given by \[ \Delta \lambda = \lambda' - \lambda = \frac{h}{m c} (1 - \cos \phi) \]where \(h\) is Planck's constant, \(m\) is the mass of the electron, \(c\) is the speed of light, \(\lambda\) is the initial wavelength, and \(\phi\) is the scattering angle.

Step 2: Relating electron kinetic energy to wavelength

The kinetic energy \(K\) of the electron after scattering is given by the relativistic formula:\[ K = \frac{1}{2}mv^2 = m_e c^2( \gamma - 1) \]where \(\gamma = \frac{1}{\sqrt{1 - (\frac{v}{c})^2}}\) is the Lorentz factor. Since the energy difference of the photon is equal to the kinetic energy of the electron:\[ h c (\frac{1}{\lambda} - \frac{1}{\lambda'}) = \frac{1}{2}mv^2 \]

Step 3: Solve for initial wavelength λ

Combine the formula from Step 1 and Step 2:\[ \frac{1}{\lambda} = \frac{1}{\lambda'} + \frac{mv^2}{2hc} \]Rearranging gives the initial wavelength \(\lambda\) as:\[ \lambda = \left( \frac{1}{\lambda'} + \frac{mv^2}{2hc}\right)^{-1} \]

Step 4: Find scattering angle φ

Using the Compton shift equation from Step 1 with the given \(\lambda\) and \(\lambda'\), solve for \(\phi\):\[ 1 - \cos \phi = \frac{\Delta \lambda m c}{h} = \frac{(\lambda' - \lambda) m c}{h} \]This gives the angle as:\[ \phi = \cos^{-1} \left( 1 - \frac{(\lambda' - \lambda) m c}{h} \right) \]

Step 5: Evaluate results for given values

Using the provided values \(\lambda' = 5.10 \times 10^{-3}\) nm and electron speed \(v = 1.80 \times 10^8\) m/s, we calculate \(\lambda\):First, convert \(\lambda'\) to meters (\(5.10 \times 10^{-3} \times 10^{-9}\) m). Then solve:\[ \lambda = \left( \frac{1}{5.10 \times 10^{-3} \times 10^{-9}} + \frac{(9.11 \times 10^{-31} \times 1.80^2 \times 10^{16})}{2 \times 6.63 \times 10^{-34}\times 3 \times 10^8} \right)^{-1} \]Then, calculate angle \(\phi\) using the values from Steps 1 and 4.

Key Concepts

Relativistic Kinetic Energy

In the realm of physics, it's important to consider relativistic effects when dealing with high-speed particles, like electrons that are struck by X-ray photons. When a photon collides with a stationary electron and transfers energy to it, the electron will gain kinetic energy and begin to move. For such scenarios, we must use the relativistic kinetic energy formula instead of the classical one. The relativistic kinetic energy (K) is given by the equation:
\[ K = m_e c^2 ( \gamma - 1) \]
where \gamma, the Lorentz factor, is \gamma = \frac{1}{\sqrt{1 - (\frac{v}{c})^2}}. This formula accounts for the fact that as the electron approaches the speed of light \(c\), its mass appears to increase, and so the translational kinetic energy diverges from the simpler classical formula.

• Mass (\(m_e\)): Refers to the rest mass of the electron, approximately \(9.11 \times 10^{-31} kg\).
• Speed (\(v\)): The velocity of the electron after the collision.
• Light Speed (\(c\)): Speed of light, approximately \(3 \times 10^8 m/s\).

By setting the change in the photon's energy equal to the electron’s kinetic energy, we derived the initial wavelength of the photon for part of this problem.

Wavelength Shift

When we talk about the wavelength shift in context of photon-electron scattering, we're diving into Compton scattering territory. This phenomenon occurs when a photon interacts with an electron and causes the photon's wavelength to change—a fundamental find by Arthur Compton that confirmed light has particle-like properties. The change in wavelength is represented by the Compton equation:
\[ \Delta \lambda = \lambda' - \lambda = \frac{h}{m_e c} (1 - \cos \phi) \]
Here:

• \( \lambda' \): the wavelength of the scattered photon.
• \( \lambda \): the initial wavelength of the photon.
• \( \phi \): the scattering angle, which is the angle from the initial path that the photon has been scattered.

This shift in wavelength is essential in analyzing the energy transformation during the collision. The difference in energy before and after the collision can be calculated using the relationship between energy and wavelength, yielding insights into the subsequent motion of the electron.

Scattering Angle Calculation

Determining the scattering angle (\(\phi\)) is crucial for understanding the deviation in direction a photon undergoes after interacting with an electron. The relationship between the scattering angle and the wavelength shift is established via the rearranged Compton equation:
\[ 1 - \cos \phi = \frac{(\lambda' - \lambda) m_e c}{h} \]
This can be solved to give:
\[ \phi = \cos^{-1} \left( 1 - \frac{(\lambda' - \lambda) m_e c}{h} \right) \]
The angle is calculable when knowing the initial and final wavelengths of the photon. Relevant constants and knowns include:

• Planck's Constant (\(h\)): \(6.63 \times 10^{-34} Js\).
• Electron Mass (\(m_e\)): \(9.11 \times 10^{-31} kg\).
• \( c \): The speed of light, \(3 \times 10^8 m/s\).

This calculation provides a vital piece of the puzzle, indicating at what angle relative to its original path the photon emerges post-collision. Interpreting this result helps physicists sketch the paths of particles in scattering experiments, playing a key role in particle physics and quantum mechanics.