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Chapter 42: Problem 38

An X-ray beam with a wavelength of exactly \(5.00 \times 10^{-14} \mathrm{~m}\) strikes a proton that is at rest \(\left(m=1.67 \times 10^{-27} \mathrm{~kg}\right)\). If the X-rays are scattered through an angle of \(110^{\circ}\), what is the wavelength of the scattered X-rays?

Short Answer

Expert verified
The wavelength of the scattered X-rays is approximately \( 5.18 \times 10^{-14} \, \text{m} \).

Step by step solution

01

Identify and understand Compton's Equation

The problem discusses the scattering of X-rays, which is a typical situation for applying Compton's Effect. Compton's equation is given by \( \lambda' - \lambda = \frac{h}{m c} (1 - \cos \theta) \), where \( \lambda' \) is the wavelength after scattering, \( \lambda \) is the original wavelength, \( h \) is Planck's constant \((6.63 \times 10^{-34} \text{ Js})\), \( m \) is the mass of the proton, \( c \) is the speed of light \((3.00 \times 10^8 \text{ m/s})\), and \( \theta \) is the scattering angle. In this case, the angle \( \theta \) is \( 110^{\circ} \).
02

Convert angle to radians

Since trigonometric functions typically use radians, convert the angle from degrees to radians. The formula is \( \text{radians} = \frac{\pi}{180} \times \text{degrees} \). So, \( 110^{\circ} = \frac{\pi}{180} \times 110 \approx 1.91986 \text{ radians} \).
03

Calculate the change in wavelength using Compton's Equation

Plug the values into Compton's equation: \( \Delta \lambda = \frac{h}{m c} (1 - \cos(1.91986)) \). Calculate \( \cos(1.91986) \approx -0.342 \). Then, \( \Delta \lambda = \frac{6.63 \times 10^{-34}}{(1.67 \times 10^{-27})(3.00 \times 10^8)} \times (1 + 0.342) \).
04

Solve for the Change in Wavelength

Compute \( \Delta \lambda = \frac{6.63 \times 10^{-34}}{5.01 \times 10^{-19}} \times 1.342 \approx 1.77 \times 10^{-15} \text{ m} \). This is the increase in wavelength due to the scattering.
05

Find the new wavelength

Add the change in wavelength to the original wavelength: \( \lambda' = \lambda + \Delta \lambda = 5.00 \times 10^{-14} + 1.77 \times 10^{-15} \approx 5.18 \times 10^{-14} \text{ m} \).

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Key Concepts

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

X-ray diffraction
X-ray diffraction is a fascinating phenomenon that occurs when X-rays encounter a material and are scattered in various directions. This process can reveal insightful information about the material's internal structure, particularly at the atomic or molecular level. X-rays are a form of electromagnetic radiation with very short wavelengths, typically in the range of 0.01 to 10 nanometers. When these rays strike a material, they interact with the electrons within the atoms, causing the waves to scatter.
  • The scattering pattern is unique to each substance, allowing scientists to use X-ray diffraction to study and identify the arrangement of atoms in a crystal.
  • By analyzing these patterns, researchers can deduce the physical properties of materials.
In general, the application of X-ray diffraction extends to various fields, including chemistry, biology, and materials science, making it an invaluable tool for researchers and engineers alike.
Photon scattering
Photon scattering is a fundamental concept where photons, which are the basic units or quanta of light, interact with particles or matter and change direction. This process is crucial in a wide range of physical and scientific phenomena. When a photon collides with a particle, such as an electron or a proton, it may be deflected or absorbed and then re-emitted. This interaction can lead to changes in the energy or direction of the photon.
  • One famous type of photon scattering is the Compton Scattering, named after Arthur Compton, who first observed it in 1923.
  • Compton Scattering specifically involves X-ray photons and their interactions with target particles, like electrons, to shift their wavelengths.
Photon scattering is a key concept in understanding how light and electromagnetic radiation behave in various environments, influencing technologies like medical imaging and communication.
Wavelength shift
In the context of photon interactions, such as Compton Scattering, a wavelength shift indicates a change in the wavelength of a photon after it has collided with a particle. This shift is a form of energy transfer, reflecting changes in either momentum or energy between the interacting entities.The Compton Effect describes how the wavelength of X-rays increases after colliding with particles, like electrons, which was revolutionary in supporting the quantum theory of light. This shift provides vital evidence of the photon-electron interaction.
  • Using the Compton Equation: \( \lambda' - \lambda = \frac{h}{m c} (1 - \cos \theta) \), we can calculate the change in wavelength, where \( \lambda' \) is the scattered wavelength.
  • The equation shows that the shift depends on the angle of scattering \( \theta \) and the constants involved, linking physical properties to measurable outcomes.
Wavelength shift is an essential concept in modern physics, as it helps to explain phenomena that classical physics alone could not address, enhancing our understanding of light and matter interactions.

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Most popular questions from this chapter

What are the speed and momentum of a 500 -nm photon?

With what speed will the fastest photoelectrons be emitted from a surface whose threshold wavelength is \(600 \mathrm{~nm}\), when the surface is illuminated with light of wavelength \(4 \times 10^{-7} \mathrm{~m} ?\)

What is the de Broglie wavelength for a particle moving with speed \(2.0 \times 10^{6} \mathrm{~m} / \mathrm{s}\) if the particle is \((a)\) an electron, \((b)\) a proton, and \((c)\) a \(0.20\) -kg ball? We make use of the definition of the de Broglie wavelength: $$ \lambda=\frac{h}{m v}=\frac{6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}}{m\left(2.0 \times 10^{6} \mathrm{~m} / \mathrm{s}\right)}=\frac{3.31 \times 10^{-40} \mathrm{~m} \cdot \mathrm{kg}}{m} $$ Substituting the required values for \(m\), one finds that the wavelength is \(3.6 \times 10^{-10} \mathrm{~m}\) for the electron, \(2.0 \times 10^{-13} \mathrm{~m}\) for the proton, and \(1.7 \times 10^{-39} \mathrm{~m}\) for the \(0.20\) -kg ball.

What potential difference must be applied to stop the fastest photoelectrons emitted by a nickel surface under the action of ultraviolet light of wavelength \(200 \mathrm{~nm}\) ? The work function of nickel is \(5.01 \mathrm{eV}\). $$ \mathrm{E}=\frac{h \mathrm{c}}{\lambda}=\frac{\left(6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}\right)\left(2.998 \times 10^{8} \mathrm{~m} / \mathrm{s}\right)}{2000 \times 10^{-10} \mathrm{~m}}=9.95 \times 10^{-19} \mathrm{~J}=6.21 \mathrm{eV} $$ Then, from the photoelectric equation, the energy of the fastest emitted electron is $$ 6.21 \mathrm{eV}-5.01 \mathrm{eV}=1.20 \mathrm{eV} $$ Hence, a negative retarding potential of \(1.20 \mathrm{~V}\) is required. This is the stopping potential.

What must be the wavelength of a photon if it is to have the same momentum as an electron traveling at \(2.2 \mathrm{~km} / \mathrm{s}\) ?
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