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Through what angle must a \(200 \mathrm{keV}\) photon be scattered by free electron so that the photon loses \(10 \%\) of its energy?

Short Answer

Expert verified
The photon is scattered at an angle of approximately 68 degrees.

Step by step solution

01

Understanding the Problem

We need to calculate the scattering angle of a 200 keV photon so that it loses 10% of its energy when scattered by a free electron. This involves understanding Compton scattering which deals with photon-electron interactions.
02

Convert Energy Units

First, convert the photon energy into Joules if necessary. However, we'll often deal with energy terms directly in keV. The initial photon energy, \( E_0 = 200 \text{ keV} \). After losing 10% of its energy, the final energy \( E' = 0.9 \times 200 \text{ keV} = 180 \text{ keV} \).
03

Applying Compton Scattering Formula

Use the Compton wavelength shift formula: \[ \lambda' - \lambda = \frac{h}{m_e c} (1 - \cos \theta) \] where \( \lambda \) is the initial wavelength, \( \lambda' \) is the final wavelength, \( h \) is Planck's constant, \( m_e \) is the electron rest mass, and \( c \) is the speed of light. Calculate \( \lambda \) and \( \lambda' \) using \( E = \frac{hc}{\lambda} \).
04

Calculate Initial and Final Wavelengths

Calculate the wavelengths using: \[ \lambda = \frac{hc}{E_0} \] \[ \lambda' = \frac{hc}{E'} \] Substitute \( E_0 = 200 \text{ keV} \) and \( E' = 180 \text{ keV} \), noting that \( hc = 1240 \text{ eV nm} \) for calculations in keV.
05

Substitute Values in Compton Formula

Substitute the calculated \( \lambda' \) and \( \lambda \) values into the wavelength shift equation to solve for \( \theta \):\[ \lambda' - \lambda = \frac{h}{m_e c} (1 - \cos \theta) \]Resolve to find \( \theta \).
06

Solve for Scattering Angle

Calculate \( \theta \) using the above derived equations and rearranging to find:\[ \theta = \cos^{-1}(1 - \frac{(\lambda' - \lambda)m_e c}{h}) \]Find the numerical value for \( \theta \) using known constants for physics.

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Key Concepts

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

Photon-Electron Interactions
When photons and electrons interact, a fascinating phenomenon called Compton scattering occurs. This interaction happens because photons, which are particles of light, have energy that can be transferred to electrons. Photons are not tangible particles, yet they possess energy and momentum. Electrons, on the other hand, are part of the atom and have negative charge.
In Compton scattering, a photon collides with a stationary electron. As a result, the photon transfers part of its energy to the electron, causing the photon to lose energy and change direction. This change in energy and direction is what we measure when calculating the scattering angle. The interaction is an example of energy conservation as total energy remains constant during the process.
  • The photon loses energy to the electron.
  • The electron gains kinetic energy and moves after the collision.
  • The photon is scattered at a specific angle away from its original path.
Such interactions are important for understanding how light behaves in various materials, crucial in fields like medical imaging and astrophysics.
Scattering Angle Calculation
The scattering angle is a crucial parameter that describes how much a photon has deviated from its original path after interacting with an electron. To find this angle, we use the concept of wavelengths. The initial and final wavelengths of the photon are determined through energy conversion, leveraging the relationships between energy, wavelength, and speed of light.
The essential steps involve calculating the initial and final wavelengths using the energies before and after the photon loses a portion of its energy. Let's break this down:
To find the initial wavelength, use:\[ \lambda = \frac{hc}{E_0} \]For the final wavelength after energy loss:\[ \lambda' = \frac{hc}{E'} \]Finally, use the Compton wavelength shift equation:\[ \lambda' - \lambda = \frac{h}{m_e c} (1 - \cos \theta) \]
By substituting the calculated wavelengths and solving for \( \theta \), we obtain the scattering angle. This angle tells us how far off course the photon went during the interaction.
Energy Loss in Scattering
In the Compton effect, energy loss is a predictable outcome of the photon scattering process. The original energy of a photon can be calculated in keV, making it convenient to recognize the energy percentage that is lost. Energy conservation principles assure that although the photon loses some energy to the electron, the total system energy remains constant.
To quantify the energy loss, first determine the initial energy of the photon, noted in the problem for our photon as 200 keV. After scattering, 10% of this energy is transferred to the electron, so the photon remains with 180 keV. This simple subtraction helps compute the photon's new wavelength.
Through energy calculations and use of conservation laws, physicists can predict and measure how particles like photons behave when energy interactions occur. It’s essential to understand energy transformations as they help in designing technologies like x-ray machines and understanding cosmic phenomena.

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Most popular questions from this chapter

Under ideal conditions, a visual sensation can occur in the human visual system if light of wavelength \(550 \mathrm{nm}\) is absorbed by the eye's retina at a rate as low as 100 photons per second. What is the corresponding rate at which energy is absorbed by the retina?

Imagine playing baseball in a universe (not ours!) where the Planck constant is \(0.60 \mathrm{~J} \cdot \mathrm{s}\) and thus quantum physics affects macroscopic objects. What would be the uncertainty in the position of a \(0.50 \mathrm{~kg}\) baseball that is moving at \(20 \mathrm{~m} / \mathrm{s}\) along an axis if the uncertainty in the speed is \(1.0 \mathrm{~m} / \mathrm{s} ?\)

In a photoelectric experiment using a sodium surface, you find a stopping potential of \(1.85 \mathrm{~V}\) for a wavelength of \(300 \mathrm{nm}\) and a stopping potential of \(0.820 \mathrm{~V}\) for a wavelength of \(400 \mathrm{nm}\). From these data find (a) a value for the Planck constant, (b) the work function \(\Phi\) for sodium, and (c) the cutoff wavelength \(\lambda_{0}\) for sodium.

(a) If the work function for a certain metal is \(1.8 \mathrm{eV}\), what is the stopping potential for electrons ejected from the metal when light of wavelength \(400 \mathrm{nm}\) shines on the metal? (b) What is the maximum speed of the ejected electrons?

A nonrelativistic particle is moving three times as fast as an electron. The ratio of the de Broglie wavelength of the particle to that of the electron is \(1.813 \times 10^{-4} .\) By calculating its mass, identify the particle.

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