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A photon of wavelength 4.50 pm scatters from a free electron that is initially at rest. (a) For \(\phi = 90.0^\circ\), what is the kinetic energy of the electron immediately after the collision with the photon? What is the ratio of this kinetic energy to the rest energy of the electron? (b) What is the speed of the electron immediately after the collision? (c) What is the magnitude of the momentum of the electron immediately after the collision? What is the ratio of this momentum value to the nonrelativistic expression \(mv\)?

Short Answer

Expert verified
(a) The kinetic energy is approximately 4.419 x 10^-14 J, with a ratio of 0.54 to rest energy. (b) Electron speed is 1.38 x 10^7 m/s. (c) Momentum is 1.26 x 10^-23 kg·m/s, with ratio 1 to nonrelativistic expression.

Step by step solution

01

Understand the Compton Scattering Formula

The Compton scattering formula describes the change in wavelength of a photon when it scatters from a particle, such as an electron. The formula is given by: \[ \Delta \lambda = \lambda' - \lambda = \frac{h}{m_e c}(1 - \cos \phi) \] where \(\lambda\) is the initial wavelength, \(\lambda'\) is the final wavelength, \(h\) is Planck's constant, \(m_e\) is the electron's rest mass, \(c\) is the speed of light, and \(\phi\) is the scattering angle.
02

Calculate Change in Wavelength (\(\Delta \lambda\))

Substitute the values into the Compton formula: \(\lambda = 4.50 \text{ pm} = 4.50 \times 10^{-12} \text{ m}\), \(\phi = 90.0^\circ\). The change in wavelength becomes \(\Delta \lambda = \frac{6.626 \times 10^{-34} \text{ J} \cdot \text{s}}{9.11 \times 10^{-31} \text{ kg} \cdot 3 \times 10^8 \text{ m/s}} \cdot (1 - \cos 90.0^\circ)\). Since \(\cos 90.0^\circ = 0\), \[\Delta \lambda = \frac{6.626 \times 10^{-34}}{9.11 \times 10^{-31} \cdot 3 \times 10^8} \approx 2.43 \times 10^{-12} \text{ m}\].
03

Find the Final Wavelength (\(\lambda'\))

The final wavelength is \(\lambda' = \lambda + \Delta \lambda\). So, \(\lambda' = 4.50 \times 10^{-12} \text{ m} + 2.43 \times 10^{-12} \text{ m} = 6.93 \times 10^{-12} \text{ m}\).
04

Calculate the Kinetic Energy of the Electron

The change in energy of the photon goes to the kinetic energy of the electron: \(E = \frac{hc}{\lambda} - \frac{hc}{\lambda'}\). Substituting, \[E = \frac{6.626 \times 10^{-34} \cdot 3 \times 10^8}{4.50 \times 10^{-12}} - \frac{6.626 \times 10^{-34} \cdot 3 \times 10^8}{6.93 \times 10^{-12}} \]. This gives \(E = 4.419 \times 10^{-14} \text{ J}\). The kinetic energy \(E_k\) of the electron is thus \(E_k = 4.419 \times 10^{-14} \text{ J}\).
05

Ratio of Kinetic Energy to Rest Energy of Electron

The rest energy \(E_{rest}\) of the electron is \(m_e c^2\). So \(E_{rest} = 9.11 \times 10^{-31} \times 9 \times 10^{16} = 8.19 \times 10^{-14} \text{ J}\). The ratio \(\frac{E_k}{E_{rest}} = \frac{4.419 \times 10^{-14}}{8.19 \times 10^{-14}} \approx 0.54\).
06

Calculate the Speed of the Electron

Using the kinetic energy \(E_k = \frac{1}{2}mv^2\), solve for \(v\): \[v = \sqrt{\frac{2E_k}{m_e}}\]. \(v = \sqrt{\frac{2 \times 4.419 \times 10^{-14}}{9.11 \times 10^{-31}}} \approx 1.38 \times 10^7 \text{ m/s}\).
07

Calculate the Momentum of the Electron

The momentum \(p\) of the electron is \(p = mv\), so \[p = 9.11 \times 10^{-31} \times 1.38 \times 10^7 \approx 1.26 \times 10^{-23} \text{ kg} \cdot \text{m/s}\].
08

Ratio of Momentum to Nonrelativistic Expression

The nonrelativistic expression for momentum is exactly \(mv\). Therefore, the ratio of the calculated momentum to \(mv\) is 1, since \(p = mv\) was used in both calculations.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Photon Wavelength
In the world of quantum mechanics, the concept of photon wavelength is crucial to understanding phenomena like Compton scattering. A photon's wavelength is the distance between successive peaks of its electromagnetic wave. It is typically measured in picometers (pm) for X-rays and gamma rays, where 1 picometer is equivalent to \(10^{-12}\) meters.

When a photon interacts with an electron, such as in Compton scattering, its wavelength changes. This change in wavelength, or "wavelength shift," provides insights into the energy and momentum exchange between the photon and the electron. Importantly, the initial and final wavelengths of a photon are directly related to its energy: the shorter the wavelength, the higher the energy of the photon.

During Compton scattering, the formula \(\Delta \lambda = \lambda' - \lambda = \frac{h}{m_e c}(1 - \cos \phi)\) predicts the change in wavelength. Here, \(\lambda\) is the original wavelength, \(\lambda'\) is the altered wavelength post-collision, \(h\) denotes Planck's constant, \(m_e\) represents the electron's rest mass, \(c\) is the speed of light, and \(\phi\) is the angle at which the scattered photon departs. This formula shows us that the scattering angle directly affects the amount of wavelength shift experienced by the photon.
Kinetic Energy of Electron
When a photon collides with an electron at rest, it transfers part of its energy to the electron, causing it to move. This transferred energy is known as the electron's kinetic energy. It is the energy an object possesses due to its motion.

For a given photon-electron interaction, the kinetic energy of the electron can be accounted for by the energy change in the photon. The formula \(E = \frac{hc}{\lambda} - \frac{hc}{\lambda'}\) represents this energy change, where \(\lambda\) and \(\lambda'\) are the initial and final wavelengths of the photon, respectively. Planck's constant \(h\) and the speed of light \(c\) are constants that bridge the gap between frequency and energy in light phenomena.

The kinetic energy \(E_k\) acquired by the electron is a fraction of its rest energy, which is calculated using Einstein’s equation \(E_{rest} = m_e c^2\). The relationship between kinetic and rest energy helps us understand how the motion energy of the electron compares with its intrinsic energy at rest. In practical scenarios, this ratio is useful for evaluating the extent of energy transfer during the interaction.
Momentum of Electron
Momentum, in physics, describes an object's motion quantity and is crucial when analyzing collisional interactions like Compton scattering. For an electron resulting from a photon collision, its momentum is directly tied to its mass \(m_e\) and velocity \(v\). The momentum \(p\) is calculated using the expression \(p = m_e v\).

In our situation, after the photon collides with an initially stationary electron, both the velocity and momentum of the electron increase. These changes are proportional to the amount of energy transferred to the electron from the photon.

The momentum of an electron relates directly to its kinetic energy through the relationship \(E_k = \frac{1}{2}m_e v^2\). By knowing either the kinetic energy or the momentum, you can derive the other characteristic to fully describe the electron’s post-collision state.
  • A critical aspect of momentum in these calculations is understanding how it conserves in isolated systems, like our photon-electron scenario.
  • This concept confirms that the momentum calculated for the electron stays consistent through non-relativistic frameworks, matching the values derived from typical kinetic energy calculations.
Understanding momentum in this context is instrumental for grasping the precise outcome of energy and movement predictions in quantum mechanics.

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