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An incident x-ray photon of wavelength 0.0900 nm is scattered in the backward direction from a free electron that is initially at rest. (a) What is the magnitude of the momentum of the scattered photon? (b) What is the kinetic energy of the electron after the photon is scattered?

Short Answer

Expert verified
(a) Calculate momentum of the photon with new wavelength; (b) Electron's KE is change in photon's energy.

Step by step solution

01

Determine initial momentum of the photon

The momentum of a photon is given by the formula \( p = \frac{h}{\lambda} \), where \( h \) is Planck's constant \( 6.626 \times 10^{-34} \text{ m}^2 \text{kg/s} \) and \( \lambda = 0.0900 \text{ nm} = 0.0900 \times 10^{-9} \text{ m} \). Calculate the initial momentum \( p_i \) of the incident photon.
02

Determine wavelength of scattered photon

For backward scattering, the change in wavelength \( \Delta \lambda \) is given by \( \Delta \lambda = \frac{2h}{m_e c} (1 - \cos \theta) \) with \( \theta = 180^\circ \). For backward direction, \( \cos 180^\circ = -1 \). Thus, \( \Delta \lambda = \frac{2h}{m_e c} \). Here \( m_e = 9.11 \times 10^{-31} \text{ kg} \) is the electron mass and \( c = 3 \times 10^8 \text{ m/s} \). Calculate \( \Delta \lambda \) and determine \( \lambda' = \lambda + \Delta \lambda \), the wavelength of the scattered photon.
03

Calculate momentum of the scattered photon

Using the new wavelength \( \lambda' \), calculate the momentum of the scattered photon with \( p' = \frac{h}{\lambda'} \).
04

Determine kinetic energy of the electron

The kinetic energy of the electron can be calculated using the energy conservation principle. The change in energy of the photon goes into the electron's kinetic energy. Initial energy \( E_i = \frac{hc}{\lambda} \) and final energy \( E_f = \frac{hc}{\lambda'} \). The kinetic energy \( KE = E_i - E_f \).

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Key Concepts

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

Photon Momentum
In the fascinating world of physics, photon momentum plays a crucial role, especially when discussing phenomena like Compton scattering. Unlike traditional concepts of momentum, which involve mass, photon momentum does not depend on mass since photons are massless particles. Instead, the momentum of a photon is determined by its wavelength, using the formula \[ p = \frac{h}{\lambda} \]Here, \( h \) is Planck's constant \( 6.626 \times 10^{-34} \text{ m}^2 \text{kg/s} \), and \( \lambda \) represents the wavelength of the photon.In Compton scattering, calculating the initial momentum of the incident photon is the first step in understanding how the photon interacts with an electron. The formula indicates that as the wavelength decreases, the momentum of the photon increases, highlighting the photon’s wave-particle duality. This behavior explains why high-energy photons, like X-ray or gamma-ray photons, can effectively transfer momentum to electrons, even though they lack mass. To truly grasp photon momentum, it is helpful to remember:
  • Photons have no rest mass, yet carry momentum based on their wavelength.
  • Shorter wavelengths equate to higher momentum for the photon.
These principles are key when analyzing scenarios like the Compton effect, where photons impart momentum to electrons, resulting in motion.
Wavelength Shift
The concept of wavelength shift is pivotal in understanding Compton scattering and is what results in the observable phenomena of scattered photons having a different wavelength than the incident photons. This shift happens due to the interaction between the photon and a free electron, leading to a change in the photon's direction and energy. The change in wavelength \( \Delta \lambda \) is given by \[ \Delta \lambda = \frac{2h}{m_e c} (1 - \cos \theta) \]where \( m_e \) is the electron mass (\( 9.11 \times 10^{-31} \text{ kg} \)), \( c \) is the speed of light (\( 3 \times 10^8 \text{ m/s} \)), and \( \theta \) is the angle of scattering.In backward scattering, particularly when \( \theta = 180^\circ \), the cosine term becomes \(-1\), simplifying the equation significantly, leading to the maximum possible change in wavelength.
  • The shift in wavelength depends on the angle of scattering.
  • Larger angles correspond to greater changes in wavelength.
Understanding the wavelength shift is crucial when exploring how photon energies change after interacting with electrons, a feature essential in medical imaging and other high-energy applications.
Electron Kinetic Energy
One of the most fascinating outcomes of Compton scattering is the increase in the kinetic energy of the electron that interacts with the photon. As the photon imparts energy and momentum to the electron, it shifts the dynamics of the system significantly.The conservation of energy principle governs this interaction, where the energy lost by the photon is gained by the electron as kinetic energy. Initially, the photon has energy \[ E_i = \frac{hc}{\lambda} \]and the energy of the photon after scattering is\[ E_f = \frac{hc}{\lambda'} \].The difference between these energies gives us the kinetic energy \( KE \) of the electron:\[ KE = E_i - E_f \].This equation shows how even massless photons can elevate the energy state of electrons, serving as a vital process in various scientific fields. Some points to note include:
  • The kinetic energy of the electron depends on the change in photon energy.
  • Conservation of energy ensures the energy lost by the photon turns into the kinetic energy of the electron.
Analyzing electron kinetic energy in Compton scattering provides insights into energy transfer at atomic levels, underpinning many technological advances and research areas. Understanding this energy transfer allows for medical diagnostics and materials analysis, capitalizing on the physics of scattering.

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Most popular questions from this chapter

A 2.50-W beam of light of wavelength 124 nm falls on a metal surface. You observe that the maximum kinetic energy of the ejected electrons is 4.16 eV. Assume that each photon in the beam ejects a photoelectron. (a) What is the work function (in electron volts) of this metal? (b) How many photoelectrons are ejected each second from this metal? (c) If the power of the light beam, but not its wavelength, were reduced by half, what would be the answer to part (b)? (d) If the wavelength of the beam, but not its power, were reduced by half, what would be the answer to part (b)?

An x ray with a wavelength of 0.100 nm collides with an electron that is initially at rest. The x ray's final wavelength is 0.110 nm. What is the final kinetic energy of the electron?

A photon has momentum of magnitude 8.24 \(\times\) 10\(^{-28}\) kg \(\bullet\) m/s. (a) What is the energy of this photon? Give your answer in joules and in electron volts. (b) What is the wavelength of this photon? In what region of the electromagnetic spectrum does it lie?

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.

(a) If the average frequency emitted by a 120-W light bulb is 5.00 \(\times\) 10\(^{14}\) Hz and 10.0\(\%\) of the input power is emitted as visible light, approximately how many visible-light photons are emitted per second? (b) At what distance would this correspond to 1.00 \(\times\) 10\(^{11}\) visible-light photons per cm\(^2\) per second if the light is emitted uniformly in all directions?

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