/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 38 A beam of \(x\) rays with wavele... [FREE SOLUTION] | 91Ó°ÊÓ

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A beam of \(x\) rays with wavelength 0.0500 \(\mathrm{nm}\) is Compton- scattered by the electrons in a sample. At what angle from the incident beam should you look to find \(x\) rays with a wavelength of (a) \(0.0542 \mathrm{nm} ;\) (b) \(0.0521 \mathrm{nm} ;(\mathrm{c}) 0.0500 \mathrm{nm} ?\)

Short Answer

Expert verified
(a) \(137.55^\circ\); (b) \(82.2^\circ\); (c) \(0^\circ\).

Step by step solution

01

Understanding Compton Scattering

Compton scattering describes the scattering of X-rays when they collide with electrons. The shift in wavelength, \( \Delta \lambda \), is given by the formula:\[\Delta \lambda = \lambda' - \lambda = \frac{h}{mc}(1 - \cos \theta)\]where \( \lambda \) is the initial wavelength, \( \lambda' \) is the scattered wavelength, \( h \) is Planck's constant (\(6.626 \times 10^{-34} \ \text{J} \cdot \text{sec}\)), \( m \) 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.
02

Calculation for Scenario (a)

Given: \( \lambda = 0.0500 \ \text{nm} \), \( \lambda' = 0.0542 \ \text{nm} \). Convert to meters (\(1 \ \text{nm} = 10^{-9} \ \text{m}\)).\[\Delta \lambda = \lambda' - \lambda = 0.0542 \times 10^{-9} \ \text{m} - 0.0500 \times 10^{-9} \ \text{m}\]\[\Delta \lambda = 4.2 \times 10^{-12} \ \text{m}\]Solve for \( \theta \) using \[ \Delta \lambda = \frac{h}{mc}(1 - \cos \theta)\].\[ 4.2 \times 10^{-12} = \frac{6.626 \times 10^{-34}}{9.109 \times 10^{-31} \, \times 3 \times 10^{8}}(1 - \cos \theta)\]Calculate:\[ 4.2 \times 10^{-12} = 2.43 \times 10^{-12} \, (1 - \cos \theta)\]\[\frac{4.2}{2.43} = 1 - \cos \theta\]\[\cos \theta = 1 - 1.73 \approx -0.73\]Find \( \theta \):\[\theta = \arccos(-0.73) \approx 137.55^\circ\].
03

Calculation for Scenario (b)

Given: \( \lambda = 0.0500 \ \text{nm} \), \( \lambda' = 0.0521 \ \text{nm} \).\[\Delta \lambda = 2.1 \times 10^{-12} \ \text{m}\]Using the same formula:\[\Delta \lambda = \frac{h}{mc}(1 - \cos \theta)\]\[ 2.1 \times 10^{-12} = 2.43 \times 10^{-12} \, (1 - \cos \theta)\]\[\frac{2.1}{2.43} = 1 - \cos \theta\]\[\cos \theta = 1 - 0.865 = 0.135\]Find \(\theta\):\[\theta = \arccos(0.135) \approx 82.2^\circ\].
04

Calculation for Scenario (c)

Given: \(\lambda = \lambda' = 0.0500 \ \text{nm}\).Thus, \(\Delta \lambda = 0\).From the formula:\[0 = \frac{h}{mc}(1 - \cos \theta)\]This implies:\(1 - \cos \theta = 0\)\(\cos \theta = 1\)\(\theta = \arccos(1) = 0^\circ\).

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Key Concepts

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

X-ray scattering
X-ray scattering is a fundamental process where X-ray photons are deflected from their original path after interacting with particles. When X-rays encounter electrons, they scatter, and this scattering is a key concept in various scientific fields, such as material science and physics.
When an X-ray hits an electron, it can be deflected, or scattered, in different directions. This scattering phenomena is influenced by the energy and angle at which the X-rays interact with electrons. This concept is known as Compton scattering, named after Arthur Compton, who discovered that X-rays scattered by electrons resulted in a change in wavelength. Compton's experiments were groundbreaking, showcasing that light doesn't just behave like a wave, but also exhibits particle-like properties. To dive deeper, when X-ray photons scatter, the energy transferred typically causes a longer wavelength. This scattering helps us derive information about the structure of materials on an atomic scale. Changing the angles and wavelengths during experiments allows researchers to infer rich details about electron arrangements and material composition.
wavelength shift
In Compton scattering, the term 'wavelength shift' refers to the change in wavelength as photons scatter off electrons. Originally, X-rays have a certain initial wavelength. However, upon scattering, this wavelength increases. This change can be calculated using the equation: \[\Delta \lambda = \lambda' - \lambda\]where \( \lambda \) is the initial wavelength, and \( \lambda' \) is the scattered wavelength. The shift is due to the energy loss that photons experience when they collide with electrons. This alteration in energy leads to an increase in wavelength, typically measured in nanometers. A longer wavelength implies that the photon lost some of its energy to the electron during the collision. Understanding this shift is essential for accurately interpreting X-ray scattering data in practical applications. It also provides insights into the energy and momentum of the electrons in the sample. By knowing both initial and scattered wavelengths, scientists can determine crucial interactions within the subject material.
electron collision
Electron collision is a central aspect of Compton scattering. When an X-ray photon collides with an electron, the interaction results in the electron gaining energy and the photon losing energy. This is a type of inelastic collision, where energy transfer occurs between the colliding particles.
Such collisions are key in understanding the dynamics at the atomic level, offering precise insights into electron behavior. During the collision, the X-ray photon imparts some of its energy to the electron, thereby reducing its own energy, which manifests as an increase in the photon's wavelength. Post-collision, the electron is typically ejected from its atom and may move in a new direction with higher kinetic energy. This process is crucial in pure physics research and materials science because it enables the study of electronic and atomic structure. The amount of energy transferred in these collisions can be calculated using considerations of conservation of momentum and energy equations.
scattering angle calculation
The scattering angle calculation is integral to understanding Compton scattering. By determining the angle at which the X-ray scatters after hitting an electron, researchers can deduce fundamental characteristics of the scattering event. This angle, represented as \( \theta \), is significant because it dictates the degree of wavelength shift during scattering.
Calculating \( \theta \) requires using the Compton wavelength shift formula:\[\Delta \lambda = \frac{h}{mc}(1 - \cos \theta)\]Here, \( h \) is Planck's constant, \( m \) is the mass of the electron, and \( c \) is the speed of light. Solving this formula for \( \theta \) involves finding the inverse cosine, using the known values of the initial and final wavelengths.Understanding the scattering angle is critical for multiple technical applications. By measuring this angle, scientists can gain insights into the forces and collisions happening on an atomic scale. It also helps determine the spatial arrangement of atoms within a sample, aiding in the analysis of the material's internal structure.

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Most popular questions from this chapter

Protons are accelerated from rest by a potential difference of 4.00 \(\mathrm{kV}\) and strike a metal target. If a proton produces one photon on impact, what is the minimum wavelength of the resulting x rays? How does your answer compare to the minimum wave-length if \(4.00-\mathrm{keV}\) electrons are used instead? Why do x-ray tubes use electrons rather than protons to produce x rays? use electrons rather than protons to produce x rays?

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