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Chapter 38: Problem 18

A photon with wavelength \(\lambda\) = 0.1385 nm scatters from an electron that is initially at rest. What must be the angle between the direction of propagation of the incident and scattered photons if the speed of the electron immediately after the collision is 8.90 \(\times\) 10\(^6\) m/s?

Short Answer

Expert verified
The scattering angle \(\theta\) is determined via the Compton shift and kinetic energy of the electron. Calculate \(\theta\) using the described steps.

Step by step solution

01

Compton Wavelength Shift Equation

The Compton scattering equation relates the change in wavelength (\(\Delta \lambda\)) to the scattering angle (\(\theta\)):\[\Delta \lambda = \lambda' - \lambda = \frac{h}{m_e c} (1 - \cos \theta)\]where \(h\) is Planck's constant \(6.626 \times 10^{-34} \; m^2 \cdot kg/s\), \(m_e\) is the electron rest mass \(9.11 \times 10^{-31} \; kg\), and \(c\) is the speed of light \(3 \times 10^8 \; m/s\).For Compton scattering, the initial wavelength \(\lambda\) is given as 0.1385 nm.
02

Calculate Kinetic Energy of Electron

The kinetic energy \(KE\) of the scattered electron is calculated using:\[KE = \frac{1}{2} m_e v^2\]Given the speed \(v = 8.90 \times 10^6 \; m/s\), substitute in to find:\[KE = \frac{1}{2} \times 9.11 \times 10^{-31} \times (8.90 \times 10^6)^2 \approx 3.59 \times 10^{-18} \; J\]
03

Relate Energy to Wavelength Shift

The energy of the photon before scattering is:\[E_{initial} = \frac{hc}{\lambda}\]The energy of the photon after scattering is the initial energy minus the electron's kinetic energy:\[E_{final} = E_{initial} - KE\]Using these expressions, calculate the change in wavelength \(\Delta \lambda\).
04

Solve for Scattering Angle

Using the change in wavelength from Step 3, substitute \(\Delta \lambda\) into the Compton equation from Step 1:\[\Delta \lambda = \frac{h}{m_e c} (1 - \cos \theta)\]Rearrange to solve for \(\cos \theta\):\[\cos \theta = 1 - \frac{m_e c \Delta \lambda}{h}\]Finally, calculate \(\theta\) by using: \[\theta = \cos^{-1}(\cdot)\]

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Key Concepts

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

Wavelength Shift
In Compton scattering, understanding the wavelength shift is crucial to grasp how scattered photons interact with electrons. Let's break down the fundamental elements:
  • The shift in wavelength (\(\Delta \lambda\)) is the difference between the initial wavelength (\(\lambda\)) of the photon and the wavelength after scattering (\(\lambda'\)).
  • This shift occurs because photons lose energy to electrons during scattering, leading to an increase in wavelengths.
  • The Compton equation is foundational in calculating this shift: \[ \Delta \lambda = \lambda' - \lambda = \frac{h}{m_e c} (1 - \cos \theta) \]
  • Here, \(h\) is Planck's constant, \(m_e\) is the electron's rest mass, and \(c\) is the speed of light.
By substituting known values into this equation, we can determine how much the photon's wavelength changes after interaction with the electron.
Kinetic Energy of Electron
After a photon collides with an electron, part of its energy is transferred, giving kinetic energy to the electron. This energy transfer is pivotal in understanding the aftermath of photon-electron collisions.
  • The kinetic energy (\(KE\)) of the electron can be computed using the formula \[ KE = \frac{1}{2} m_e v^2 \] where \(v\) is the speed of the electron post-collision.
  • In the given problem, the electron's speed is 8.90 \(\times\) 10\(^6\) m/s, leading to its calculated \(KE\) of approximately 3.59 \(\times\) 10\(^{-18}\) J.
  • This kinetic energy effectively represents the portion of energy the photon loses during its interaction with the electron.
Understanding the kinetic energy is key to linking energy dynamics involved in Compton scattering, aiding in calculations like those for the wavelength shift.
Photon Scattering Angle
The angle at which a photon scatters, known as the scattering angle, is a fundamental factor in analyzing Compton scattering. This angle helps determine how the photon's path changes post-collision.
  • The scattering angle (\(\theta\)) is directly involved in the calculation of the wavelength shift through the Compton equation.
  • It is determined via the wavelength change: \[ \cos \theta = 1 - \frac{m_e c \Delta \lambda}{h} \]
  • After calculating the cosine of the angle, you can find the actual scattering angle \(\theta\) using the inverse cosine function, \(\theta = \cos^{-1} (\cdot)\).
  • This computation reveals how much the photon veered off its initial path.
Understanding the scattering angle is vital because it helps you establish the correlation between energy lost by the photon and its direction change, thereby explaining much of the behavior observed during Compton scattering.

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Most popular questions from this chapter

A laser produces light of wavelength 625 nm in an ultrashort pulse. What is the minimum duration of the pulse if the minimum uncertainty in the energy of the photons is 1.0\(\%\)?

A photon with wavelength 0.1100 nm collides with a free electron that is initially at rest. After the collision the wavelength is 0.1132 nm. (a) What is the kinetic energy of the electron after the collision? What is its speed? (b) If the electron is suddenly stopped (for example, in a solid target), all of its kinetic energy is used to create a photon. What is the wavelength of this photon?

The photoelectric work function of potassium is 2.3 eV. If light that has a wavelength of 190 nm falls on potassium, find (a) the stopping potential in volts; (b) the kinetic energy, in electron volts, of the most energetic electrons ejected; (c) the speed of these electrons.

The cathode-ray tubes that generated the picture in early color televisions were sources of x rays. If the acceleration voltage in a television tube is 15.0 kV, what are the shortest-wavelength x rays produced by the television?

Consider Compton scattering of a photon by a \(moving\) electron. Before the collision the photon has wavelength \(\lambda\) and is moving in the +\(x\)-direction, and the electron is moving in the -\(x\)-direction with total energy \(E\) (including its rest energy \(mc^2\)). The photon and electron collide head-on. After the collision, both are moving in the -\(x\)-direction (that is, the photon has been scattered by 180\(^\circ\)). (a) Derive an expression for the wavelength \(\lambda'\) of the scattered photon. Show that if \(E \gg mc^2\), where m is the rest mass of the electron, your result reduces to $$\lambda' = {hc \over E} (1 + {m^2c^4\lambda \over 4hcE}) $$ (b) A beam of infrared radiation from a CO\(_2\) laser (\(\lambda = 10.6 \mu{m}\)) collides head-on with a beam of electrons, each of total energy \(E\) = 10.0 GeV (1 GeV = 10\(^9\) eV). Calculate the wavelength \(\lambda'\) of the scattered photons, assuming a 180\(^\circ\) scattering angle. (c) What kind of scattered photons are these (infrared, microwave, ultraviolet, etc.)? Can you think of an application of this effect?
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