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Chapter 38: Problem 19

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.

Short Answer

Expert verified
The photon's wavelength increases by about \(0.0000624 \text{ nm}\), scattered to \(0.0425684 \text{ nm}\). Photon loses energy, which the electron gains.

Step by step solution

01

Understand the Compton Effect

The problem involves the Compton effect, which describes the change in wavelength of X-ray or gamma-ray photons when they collide with electrons. This is crucial to calculate the change in wavelength and energy.
02

Compton Wavelength Shift Formula

The formula for the shift in wavelength is given by \( \Delta \lambda = \frac{h}{m_e c} (1 - \cos \theta) \), where \( h \) 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 \(3 \times 10^8 \text{ m/s}\), and \( \theta \) is the scattering angle \(35.0^\circ\). Convert the angle to radians and substitute the values to find \( \Delta \lambda \).
03

Calculate Change in Wavelength

Convert the angle to radians: \( \theta = 35.0^\circ \times \frac{\pi}{180} = 0.6109 \text{ radians} \). Then use the formula: \( \Delta \lambda = \frac{6.626 \times 10^{-34}}{9.109 \times 10^{-31} \times 3 \times 10^8} (1 - \cos(0.6109)) \). Calculate \( \Delta \lambda \).
04

Calculate Wavelength of Scattered Photon

The wavelength of the scattered photon is the original wavelength plus the change: \( \lambda' = \lambda + \Delta \lambda \). Substituting \( \lambda = 0.04250 \text{ nm} = 0.04250 \times 10^{-9} \text{ m} \) and \( \Delta \lambda \) from the previous step. Obtain \( \lambda' \).
05

Calculate Energy Change of Photon

The energy change is found by \( \Delta E = E_f - E_i \), where \( E = \frac{hc}{\lambda} \). Calculate initial energy \( E_i \) using initial wavelength \( \lambda \) and final energy \( E_f \) using \( \lambda' \). Subtract to find \( \Delta E \) and determine if the energy is lost or gained by the photon.
06

Calculate Energy Gained by Electron

Energy gained by the electron equals the energy lost by the photon, i.e., \( \Delta E \). Find \( \Delta K_e = |\Delta E| \) where \( \Delta K_e \) is the kinetic energy gained by the electron.

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Key Concepts

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

Photon Scattering
When a photon encounters a free electron, it can interact with it in a process known as photon scattering. This means that the photon's path changes direction upon hitting the electron. The scattering alters both the direction and energy characteristics of the photon. For a clearer understanding:
  • A photon is a particle of light that carries energy.
  • An electron is a subatomic particle present in atoms.
  • During scattering, photons impact free electrons, causing them to move.
In our specific scenario, the photon initially directed along a certain path encounters an electron and is deflected by an angle of 35.0掳. This interaction is central to understanding changes in the photon's wavelength and energy.
Wavelength Shift
Wavelength shift, in the context of the Compton effect, refers to the change in the wavelength of a photon as a result of scattering. This shift occurs because of the energy exchange between the photon and electron. The amount of shift is calculated using the Compton wavelength shift formula:\[ \Delta \lambda = \frac{h}{m_e c} (1 - \cos \theta) \]Where:
  • \( \Delta \lambda \) is the change in wavelength.
  • \( h \) is Planck's constant.
  • \( m_e \) is the mass of the electron.
  • \( c \) represents the speed of light.
  • \( \theta \) is the scattering angle, converted to radians before calculation.
The shift in wavelength is essential as it indicates how much energy has been transferred during the interaction. It's calculated by substituting known values into the formula. The result tells us whether the photon lost or gained energy.
Energy Conservation
In physics, energy conservation is a fundamental principle stating that the total energy in a closed system remains constant. This is deeply relevant in the explanation of the Compton effect during photon scattering. Before scattering, the photon possesses a certain energy determined by its wavelength. Post-scattering, energy is still conserved, but some of the photon's energy is transferred to the electron, altering the photon's wavelength.The energy of the photon can be expressed through the formula:\[ E = \frac{hc}{\lambda} \]Where:
  • \( E \) is the energy of the photon.
  • \( \lambda \) is its wavelength.
When calculated for initial and final wavelengths, the photon's energy shows a shift demonstrating energy transfer. The gained energy by the electron will equal the photon's energy loss, thus upholding energy conservation.
Electron-Photon Interaction
This interaction is the heart of the Compton effect, illustrating the dance between photons and electrons. Here鈥檚 how the interaction unfolds:
  • The photon approaches a free electron with its travel path and energy intact.
  • Upon collision, it imparts some momentum and energy to the electron, causing the photon to scatter.
  • This scattering changes the photon's path angle and alters its wavelength, reflecting energy transfer to the electron.
The electron gains kinetic energy from this energy shift. Through this process, the Compton effect gives solid evidence of light behaving as particles. Understanding these interactions helps explain many phenomena in quantum physics and is pivotal in comprehending the dual nature of light and electrons.

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Most popular questions from this chapter

A laser used to weld detached retinas emits light with a wavelength of 652 nm in pulses that are 20.0 ms in duration. The average power during each pulse is 0.600 W. (a) How much energy is in each pulse in joules? In electron volts? (b) What is the energy of one photon in joules? In electron volts? (c) How many photons are in each pulse?

A horizontal beam of laser light of wavelength 585 nm passes through a narrow slit that has width 0.0620 mm. The intensity of the light is measured on a vertical screen that is 2.00 m from the slit. (a) What is the minimum uncertainty in the vertical component of the momentum of each photon in the beam after the photon has passed through the slit? (b) Use the result of part (a) to estimate the width of the central diffraction maximum that is observed on the screen.

Nuclear fusion reactions at the center of the sun produce gamma-ray photons with energies of about 1 MeV (10\(^6\) eV). By contrast, what we see emanating from the sun's surface are visiblelight photons with wavelengths of about 500 nm. A simple model that explains this difference in wavelength is that a photon undergoes Compton scattering many times-in fact, about 10\(^{26}\) times, as suggested by models of the solar interior-as it travels from the center of the sun to its surface. (a) Estimate the increase in wavelength of a photon in an average Compton-scattering event. (b) Find the angle in degrees through which the photon is scattered in the scattering event described in part (a). (\(Hint\): A useful approximation is cos \(\phi \approx 1 - \phi^2 /2\), which is valid for \(\phi \ll\) 1. Note that \(\phi\) is in radians in this expression.) (c) It is estimated that a photon takes about 10\(^6\) years to travel from the core to the surface of the sun. Find the average distance that light can travel within the interior of the sun without being scattered. (This distance is roughly equivalent to how far you could see if you were inside the sun and could survive the extreme temperatures there. As your answer shows, the interior of the sun is \(very\) opaque.)

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 \(mv\)?

When ultraviolet light with a wavelength of 400.0 nm falls on a certain metal surface, the maximum kinetic energy of the emitted photoelectrons is measured to be 1.10 eV. What is the maximum kinetic energy of the photoelectrons when light of wavelength 300.0 nm falls on the same surface?
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