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An X-ray photon is scattered at an angle of \(\theta=180.0^{\circ}\) from an electron that is initially at rest. After scattering, the electron has a speed of \(4.67 \times 10^{6} \mathrm{~m} / \mathrm{s}\). Find the wavelength of the incident X-ray photon.

Short Answer

Expert verified
The wavelength of the incident X-ray photon is approximately \( 2.43 \times 10^{-11} \; m \).

Step by step solution

01

Compton Wavelength Shift Formula

The Compton effect describes how X-ray photons scatter off electrons. The formula we use to find the change in wavelength is given by: \ \[ \Delta \lambda = \lambda' - \lambda = \frac{h}{m_ec} (1 - \cos \theta) \] \ where \( \lambda' \) is the wavelength after scattering, \( \lambda \) is the initial wavelength, \( h \) is the Planck's constant, \( m_e \) is the electron mass, and \( c \) is the speed of light. For \( \theta = 180^{\circ}, \; (1 - \cos \theta) = 2. \)
02

Calculate Electron's Kinetic Energy

The kinetic energy \( KE \) of the electron after scattering is given by \( KE = \frac{1}{2} m_e v^2 \). Substitute \( m_e = 9.11 \times 10^{-31} \; \text{kg} \) and \( v = 4.67 \times 10^6 \; \text{m/s} \): \ \[ KE = \frac{1}{2} \times 9.11 \times 10^{-31} \times (4.67 \times 10^6)^2 = 9.94 \times 10^{-18} \; \text{J} \]
03

Energy-Wavelength Relation for Photon

The energy of the incident photon \( E_i \) is related to its wavelength by the equation \( E_i = \frac{hc}{\lambda} \), and for the scattered photon, the energy \( E_f = \frac{hc}{\lambda'} \). Use Planck's constant \( h = 6.626 \times 10^{-34} \; J \cdot s \) and \( c = 3 \times 10^8 \; m/s \).
04

Energy Conservation Equation

Using energy conservation, the initial energy of the system is equal to the final energy: \ \[ \frac{hc}{\lambda} = \frac{hc}{\lambda'} + KE \] \ Rearrange to find \( \frac{hc}{\lambda} - \frac{hc}{\lambda'} = KE \). Substitute \( KE = 9.94 \times 10^{-18} \; J \) from Step 2.
05

Solve for Initial Wavelength \( \lambda \)

From Step 1 we know \( \Delta \lambda = \frac{h}{m_e c} (1 - \cos 180^{\circ}) = 2 \times \frac{h}{m_e c} \). Solve for \( \lambda \) using \ \[ \frac{hc}{\lambda} = KE + \frac{hc}{\lambda + \Delta \lambda} \], \ Substitute \( \Delta \lambda = 2 \times \frac{6.626 \times 10^{-34}}{9.11 \times 10^{-31} \times 3 \times 10^8} = 4.85 \times 10^{-12} \; m \). Substitute back to find \( \lambda \).
06

Calculate the Numerical Answer

Substitute the calculated values to solve for \( \lambda \). Use the given values and rearrange the relation: \ \[ \lambda = \frac{hc}{\frac{hc}{\lambda + 4.85 \times 10^{-12}} + 9.94 \times 10^{-18}} \] \ After calculation, \( \lambda \approx 2.43 \times 10^{-11} \; \text{m} \).

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Key Concepts

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

X-ray Photon
An X-ray photon is a form of high-energy electromagnetic radiation. These photons have significantly shorter wavelengths than visible light, making them capable of penetrating most substances, which is why they are widely used in medical imaging and materials science. X-ray photons are energy carriers that travel at the speed of light. They are generated by accelerating electrons striking a metal target within X-ray tubes. When discussing X-ray photons in the context of Compton Scattering, we pay close attention to their energy and wavelength, which play pivotal roles in understanding how these photons interact with matter.
In the specific case of Compton Scattering, an X-ray photon is scattered from an electron, changing its wavelength and energy in the process. This concept is essential for comprehending how X-rays are affected when they collide with electrons, revealing important atomic and molecular information.
Wavelength Shift
The concept of wavelength shift is at the heart of Compton Scattering. When an X-ray photon collides with an electron, its path is deflected, leading to a change in its wavelength. This change is referred to as the wavelength shift. The Compton effect quantitatively describes this shift using the formula:
  • \( \Delta \lambda = \lambda' - \lambda = \frac{h}{m_ec} (1 - \cos \theta) \)
This equation helps us determine how much the wavelength of the photon changes after scattering. Here, \( \lambda \) is the original wavelength, \( \lambda' \) is the new wavelength, and \( \theta \) is the scattering angle.
For a maximum shift, as seen in our exercise where \( \theta = 180^{\circ} \), the formula simplifies to \( \Delta \lambda = 2 \times \frac{h}{m_e c} \). This specific case exemplifies a scenario where the photon undergoes a full directional reversal, showcasing the largest possible change in wavelength.
Energy Conservation
Energy conservation is a fundamental principle in physics. In Compton Scattering, this principle helps us understand how energy is distributed between the photon and the electron after interaction. The energy of the incident photon is initially given by the equation:
  • \( E_i = \frac{hc}{\lambda} \)
After scattering, the energy of the photon becomes \( E_f = \frac{hc}{\lambda'} \), and the kinetic energy of the electron is expressed as:
  • \( KE = \frac{1}{2} m_e v^2 \)
Energy conservation states that the total initial energy of the system (only the photon’s energy) is equal to the sum of the scattered photon's energy and the kinetic energy of the electron:
  • \( \frac{hc}{\lambda} = \frac{hc}{\lambda'} + KE \)
By applying this principle, we can deduce the relationships required to solve for unknown quantities such as the initial wavelength or energy, ensuring energy has been accurately accounted for during scattering.
Photon-Electron Interaction
Photon-electron interaction is a crucial phenomenon in Compton Scattering. When an X-ray photon encounters an electron, it transfers some of its energy to the electron. This causes the electron to gain kinetic energy and move, changing the photon's direction and wavelength.
Understanding this interaction involves several key concepts:
  • **Momentum Transfer:** The photon carries momentum, which it transfers to the electron during collision.
  • **Energy Distribution:** As energy is passed to the electron, the photon's energy decreases, hence the shift to a longer wavelength.
Quantum mechanics details this interaction, helping predict the outcomes of such scattering events. The ability of the photon to impart energy and momentum results in a kinetic boost to the electron, evidencing the wave-particle duality nature of electromagnetic radiation.

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Most popular questions from this chapter

A photon of known wavelength strikes a free electron that is initially at rest, and the photon is scattered straight backward. (a) How is the kinetic energy KE of the recoil electron related to the energies \(E\) and \(E_{\prime}\) of the incident and scattered photons, respectively? (b) If the wavelength \(\lambda\) of the incident photon is known, how can the photon's energy \(E\) be determined? (c) How can the wavelength \(\mathcal{N}\) of the scattered photon and, hence, its energy \(E^{\prime}\) be found if we know the wavelength of the incident photon and the fact that the photon is scattered straight backward? A photon of wavelength \(0.45000 \mathrm{~nm}\) strikes a free electron that is initially at rest. The photon is scattered straight backward. What is the speed of the recoil electron after the collision?

Heat is a form of energy. One way to supply the heat needed to warm an object is to irradiate the object with electromagnetic waves. (a) How is the heat \(Q\) needed to raise the temperature of an object by \(\Delta T\) degrees related to its specific heat capacity \(c_{\text {specific heat }}\) and mass \(m ?\) (b) Is the energy carried by an infrared photon greater than, smaller than, or the same as the energy carried by a visible photon of light? Explain. (c) Assuming that all the photons are absorbed by the object, is the number of infrared photons needed to supply a given amount of heat greater or smaller than the number of visible photons? Explain. A glass plate has a mass of \(0.50 \mathrm{~kg}\) and a specific heat capacity of \(840 \mathrm{~J} /(\mathrm{kg} \cdot \mathrm{C}\) \({ }^{\circ}\) ). The wavelength of infrared light is \(6.0 \times 10^{-5} \mathrm{~m}\), while the wavelength of blue visible light is \(4.7 \times 10^{-7} \mathrm{~m}\). Find the number of infrared photons and the number of visible photons needed to raise the temperature of the glass plate by \(2.0 \mathrm{C}^{\circ}\), assuming that all photons are absorbed by the glass. Verify that your answers are consistent with your answers to the Concept Questions.

Consult Interactive LearningWare 29.1 at for background material relating to this problem. An owl has good night vision because its eyes can detect a light inten sity as small as \(5.0 \times 10^{-13} \mathrm{~W} / \mathrm{m}^{2}\). What is the minimum number of photons per second that an owl eye can detect if its pupil has a diameter of \(8.5 \mathrm{~mm}\) and the light has a wavelength of \(510 \mathrm{nm} ?\)

Concept Questions Consider a Young's double-slit experiment that uses electrons. In versions \(A\) and \(B\) of the experiment the first-order bright fringes are located at angles \(\theta_{\mathrm{A}}\) and \(\theta_{\mathrm{B}},\) where \(\theta_{\mathrm{B}}\) is greater than \(\theta_{\mathrm{A}} .\) In these two versions, the separation between the slits is the same but the electrons have different momenta. (a) In which version is the wavelength of an electron greater? (b) In which version is the momentum of an electron greater? Explain your answers. Problem In a Young's double-slit experiment the angle that locates the first- order bright fringes is \(\theta_{A}=1.6 \times 10^{-4}\) degrees when the magnitude of the electron momentum is \(p_{\mathrm{A}}=1.2 \times 10^{-22} \mathrm{~kg} \cdot \mathrm{m} / \mathrm{s}\). With the same double slit, what momentum magnitude \(p_{\mathrm{B}}\) is necessary, so that an angle of \(\theta_{\mathrm{B}}=4.0 \times 10^{-4}\) degrees locates the first-order bright fringes?

A laser emits \(1.30 \times 10^{18}\) photons per second in a beam of light that has a diameter of \(2.00 \mathrm{~mm}\) and a wavelength of \(514.5 \mathrm{nm} .\) Determine (a) the average electric field strength and (b) the average magnetic field strength for the electromagnetic wave that constitutes the beam.

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