/*! 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 23 \(X\) rays with an initial wavel... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

\(X\) rays with an initial wavelength of \(0.900 \times 10^{-10} \mathrm{m}\) undergo Compton scattering. For what scattering angle is the wavelength of the scattered x rays greater by 1.0\(\%\) than that of the incident \(x\) rays?

Short Answer

Expert verified
The scattering angle \(\theta\) where the wavelength is 1.0% greater is approximately 19.4 degrees.

Step by step solution

01

Understanding the Problem

In this problem, we're examining the change in wavelength of X-rays due to Compton scattering. Initially, the wavelength of the X-rays is \(0.900 \times 10^{-10} \text{ m}\). We need to find the scattering angle where the wavelength of the scattered rays is greater by 1.0\(\%\). This implies that \(\Delta \lambda = 0.01 \times 0.900 \times 10^{-10} \text{ m}\).
02

Compton Wavelength Shift Equation

The Compton effect equation gives the change in wavelength as \( \Delta \lambda = \lambda' - \lambda = \frac{h}{mc}(1 - \cos\theta) \), where \(\lambda\) is the initial wavelength, \(\lambda'\) is the scattered wavelength, \(h\) is Planck's constant, \(m\) is the electron mass, \(c\) is the speed of light, and \(\theta\) is the scattering angle.
03

Calculate Wavelength Change

Calculate \(\Delta \lambda\) from the 1.0\(\%\) increase: \(\Delta \lambda = 0.01 \times 0.900 \times 10^{-10} \text{ m} = 0.009 \times 10^{-10} \text{ m} \).
04

Substitute into the Compton Equation

Substitute \(\Delta \lambda\) into the Compton equation: \[ 0.009 \times 10^{-10} = \frac{h}{mc}(1 - \cos\theta) \] where \(h = 6.626 \times 10^{-34} \text{ Js}\), \(m = 9.109 \times 10^{-31} \text{ kg} \), and \(c = 3 \times 10^{8} \text{ m/s} \).
05

Solve for Scattering Angle \(\theta\)

From the equation \[ 0.009 \times 10^{-10} = \frac{6.626 \times 10^{-34}}{9.109 \times 10^{-31} \times 3 \times 10^{8}}(1 - \cos\theta) \], rearrange to find \(\cos\theta\). Calculate to find \(\theta\), ensuring all constants are correctly inserted for computation.
06

Calculate Numerical Value for \(\theta\)

Solving the equation, we get \(\cos\theta = 1 - \frac{0.009 \times 10^{-10}}{2.426 \times 10^{-12}} \). This simplifies to \(\theta\). Calculate \(\theta\) using the inverse cosine to find the angle corresponding to this value.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

X-ray Scattering
X-ray scattering is a pivotal concept in physics, particularly when discussing the interaction between X-rays and matter. When X-rays strike matter, they can scatter, which means the direction of their propagation changes. This process is crucial in many scientific applications, like material analysis and medical imaging. When X-rays scatter, they may also experience a change in wavelength, known as Compton scattering, named after the American physicist Arthur H. Compton. He discovered that electromagnetic radiation, such as X-rays, scatters in a manner that can only be explained if light behaves as particles, called photons.
  • Compton scattering supports the particle theory of light.
  • It involves the collision of X-ray photons with electrons.
  • As a result, part of the photon's energy is transferred to the electron, resulting in a change in the photon's direction and wavelength.
Understanding X-ray scattering provides insight into both particle physics and practical applications related to material properties.
Wavelength Shift
In the process of Compton scattering, a crucial outcome is the change in wavelength of the scattered X-ray, termed as wavelength shift. This shift is essential for understanding how X-rays behave when they interact with electrons in a material. The Compton wavelength shift equation is formulated as:\[ \Delta \lambda = \lambda' - \lambda = \frac{h}{mc}(1 - \cos\theta) \]where:
  • \(\Delta \lambda\) is the wavelength shift.
  • \(\lambda'\) is the scattered wavelength.
  • \(\lambda\) is the initial wavelength.
  • \(h\) is Planck's constant.
  • \(m\) is the electron mass.
  • \(c\) is the speed of light.
  • \(\theta\) is the scattering angle.
The equation reveals that the wavelength shift depends on the angle at which the X-ray is scattered. This dependency is significant in studying the energy distribution of scattered X-rays, helping to reveal information about a material's electronic structure.
Scattering Angle
The scattering angle, denoted as \(\theta\), is a critical factor in Compton scattering, affecting how much the wavelength of the incident X-ray will shift. This angle is measured between the original direction of the X-ray and its new direction after scattering. In mathematical terms, the magnitude of the wavelength shift is directly linked to the cosine of this scattering angle:\[ \Delta \lambda = \frac{h}{mc}(1 - \cos\theta) \]Key points to understand about the scattering angle:
  • The greater the scattering angle, the more energy is transferred from the X-ray to the electron, resulting in a larger wavelength shift.
  • If \(\theta = 0\), there is no change in the direction, thus no wavelength shift.
  • If \(\theta = 180\) degrees, the X-ray is scattered backwards, which results in the maximum possible wavelength shift.
Calculating the scattering angle helps physicists determine structural properties of materials and the energetic interactions between particles within them.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

CP An x-ray tube is operating at voltage \(V\) and current \(I\) (a) If only a fraction \(p\) of the electric power supplied is converted into \(x\) rays, at what rate is energy being delivered to the target? (b) If the target has mass \(m\) and specific heat \(c\) (in \(J / k g \cdot K ),\) at what average rate would its temperature rise if there were no thermal losses? (c) Evaluate your results from parts (a) and (b) for an x-ray tube operating at 18.0 \(\mathrm{kV}\) and 60.0 \(\mathrm{mA}\) that converts 1.0\(\%\) of the electric power into \(\mathrm{x}\) rays. Assume that the \(0.250-\mathrm{kg}\) target is made of lead \((c=130 \mathrm{J} / \mathrm{kg} \cdot \mathrm{K})\) . (d) What must the physical properties of a practical target material be? What would be some suitable target elements?

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?

An x-ray photon is scattered from a free electron (mass \(m )\) at rest. The wavelength of the scattered photon is \(\lambda^{\prime},\) and the final speed of the struck electron is \(v\) . (a) What was the initial wave-length \(\lambda\) of the photon? Express your answer in terms of \(\lambda^{\prime}, v\) and \(m .\) (Hint. Use the relativistic expression for the electron kinetic energy.) (b) Through what angle \(\phi\) is the photon scattered? Express your answer in terms of \(\lambda, \lambda^{\prime},\) and \(m\) . (c) Evaluate your results in parts (a) and (b) for a wavelength of \(5.10 \times 10^{-3}\) nm for the scattered photon and a final electron speed of \(1.80 \times 10^{8} \mathrm{m} / \mathrm{s}\) . Give \(\phi\) in degrees.

CP A photon with wavelength \(\lambda=0.1050 \mathrm{nm}\) is incident on an electron that is initially at rest. If the photon scatters at an angle of \(60.0^{\circ}\) from its original direction, what are the magnitude and direction of the linear momentum of the electron just after the collision with the photon?

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 \(\mathrm{cV}\) . 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 eachsecond 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)?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.