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In the Compton effect, an \(X\) -ray photon scatters from a free electron. Find the change in the photon's wavelength if it scatters at an angle of \((a) \theta=30.0^{\circ}\), (b) \(\theta=90.0^{\circ},\) and (c) \(\theta=180.0^{\circ}\) relative to the incident direction.

Short Answer

Expert verified
For \(\theta=30.0^{\circ}\), \(\Delta\lambda=\frac{h}{m_e c} (1-\frac{\sqrt{3}}{2})\); for \(\theta=90.0^{\circ}\), \(\Delta\lambda=\frac{h}{m_e c}\); for \(\theta=180.0^{\circ}\), \(\Delta\lambda=\frac{2h}{m_e c}\).

Step by step solution

01

Understanding the Compton Effect Formula

The change in wavelength, known as the Compton wavelength shift, is determined using the Compton formula: \[ \Delta \lambda = \lambda' - \lambda = \frac{h}{m_e c} (1 - \cos\theta) \]where \(h\) is Planck's constant, \(m_e\) is the electron rest mass, \(c\) is the speed of light, and \(\theta\) is the angle of scattering.
02

Insert Given Angles into the Formula

For part (a), set \(\theta = 30.0^{\circ}\). Convert it to radians as \[\theta = \frac{30 \times \pi}{180} = \frac{\pi}{6}\]
03

Calculate the Wavelength Shift for 30 Degrees

Using the converted angle, compute the change in wavelength:\[ \Delta \lambda = \frac{h}{m_e c} (1 - \cos 30^{\circ}) \]\(\cos 30^{\circ} = \frac{\sqrt{3}}{2}\), resulting in:\[ \Delta \lambda = \frac{h}{m_e c} \left(1 - \frac{\sqrt{3}}{2}\right) \]
04

Insert 90 Degree Angle into the Formula

For part (b), set \(\theta = 90.0^{\circ}\). Convert this to radians as \[\theta = \frac{90 \times \pi}{180} = \frac{\pi}{2}\]
05

Calculate the Wavelength Shift for 90 Degrees

Using the assigned value, calculate the change in wavelength with the new angle:\[ \Delta \lambda = \frac{h}{m_e c} (1 - \cos 90^{\circ}) \] \(\cos 90^{\circ} = 0\), hence \[ \Delta \lambda = \frac{h}{m_e c} \]
06

Insert 180 Degree Angle into the Formula

For part (c), we have \(\theta = 180.0^{\circ}\), which converts to radians as \[\theta = \frac{180 \times \pi}{180} = \pi\]
07

Calculate the Wavelength Shift for 180 Degrees

Utilize the value to find the change for 180 degrees:\[ \Delta \lambda = \frac{h}{m_e c} (1 - \cos 180^{\circ}) \]\(\cos 180^{\circ} = -1\), so\[ \Delta \lambda = \frac{h}{m_e c} (1 - (-1)) = \frac{2h}{m_e c} \]

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Key Concepts

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

Wavelength Shift
The term 'wavelength shift' in the context of the Compton effect refers to the change in the wavelength of an X-ray photon after it has scattered off a free electron. The scattering of X-ray photons causes a measurable shift in the wavelength of the photons, an effect discovered by Arthur Compton in 1923. This shift is expressed mathematically by the formula for the Compton effect: \[ \Delta \lambda = \frac{h}{m_e c} (1 - \cos \theta) \] where:
  • \( \Delta \lambda \) represents the wavelength shift.
  • \( h \) is Planck's constant.
  • \( m_e \) is the electron's rest mass.
  • \( c \) is the speed of light.
  • \( \theta \) is the scattering angle.
The shift depends solely on the angle \( \theta \) of scattering, making it pivotal for understanding how energy and momentum are transferred during such interactions.
X-ray Photon
X-ray photons are high-energy particles of light, or quanta, that fall within the X-ray portion of the electromagnetic spectrum. They have the ability to penetrate various materials, which is why they are frequently used in medical imaging and analyzing the structure of materials. In the scenario of the Compton effect, X-ray photons are incident on a target electron. As they strike the electron, their energy and momentum are partially transferred, which results in a change - or shift - in their original wavelength. The Compton formula quantifies this interaction, providing a detailed understanding of how photon properties change as a consequence of their interactions with matter. It's crucial for students to grasp that although energy is transferred from the photon to the electron, the total energy before and after the interaction must be conserved according to the law of conservation of energy.
Scattering Angle
When analyzing the Compton effect, the scattering angle \( \theta \) is a critical factor. This angle is defined as the angle between the path of the incoming X-ray photon and its direction after hitting the electron. In practice, the scattering angle determines the extent of the wavelength shift. As evidenced by the Compton formula: \[ \Delta \lambda = \frac{h}{m_e c} (1 - \cos \theta) \] The mathematical expression \( \cos \theta \) plays a vital role. As the scattering angle \( \theta \) increases, the value of \( \cos \theta \) decreases, and thus the wavelength shift \( \Delta \lambda \) increases. Specific angles, such as \( \theta = 90^{\circ} \) and \( \theta = 180^{\circ} \), show significant shifts as illustrated in textbook exercises. The larger the angle, the greater the expected change in wavelength.
Planck's constant
Planck's constant \( h \) is a fundamental value in the field of quantum mechanics. Its precise value is approximately \( 6.62607015 \times 10^{-34} \) Js. Within the formula for the Compton effect: \[ \Delta \lambda = \frac{h}{m_e c} (1 - \cos \theta) \] Planck's constant serves as a multiplier factor, determining the scale of the wavelength shift. It represents the proportionate constant in the relationship between the energy of a photon and the frequency of its associated electromagnetic wave. In the context of the Compton effect, it highlights the quantization of electromagnetic energy interactions, showing that changes in energy (and thus wavelength) occur in discrete amounts, intimately linking the particle-like and wave-like behaviors of electromagnetic radiation.
Electron Rest Mass
The electron rest mass \( m_e \) is another essential constant in the Compton effect equation, approximately \( 9.10938356 \times 10^{-31} \) kg. This constant represents the mass of an electron when it is not moving relative to an observer.Within the Compton wavelength shift equation: \[ \Delta \lambda = \frac{h}{m_e c} (1 - \cos \theta) \] The inverse relationship between \( m_e \) and \( \Delta \lambda \) suggests that the electron's rest mass moderates the extent of the wavelength shift. Its inclusion in the formula indicates the dependence of the shift on the electron's intrinsic properties. The value of the rest mass is crucial for calculations, helping to demonstrate how matter's intrinsic characteristics impact photon-electron interactions. Understanding \( m_e \) aids in comprehending how energy and momentum conservation principles dictate the results of scattering events.

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Most popular questions from this chapter

Consider two different photons that scatter through an angle of \(180^{\circ}\) from a free electron. One is a visible-light photon with \(\lambda=520 \mathrm{nm},\) the other is an \(\mathrm{X}\) -ray photon with \(\lambda=0.030 \mathrm{nm}\) (a) Which (if either) photon experiences the greater change in wavelength as a result of the scattering? Explain. (b) Which photon experiences the greater percentage change in wavelength? Explain. (c) Calculate the percentage change in wavelength of each photon.

A blue-green photon \((\lambda=486 \mathrm{nm})\) is absorbed by a free hydrogen atom, initially at rest. What is the recoil speed of the hydrogen atom after absorbing the photon?

The measurement of an electron's energy requires a time interval of \(1.0 \times 10^{-8} \mathrm{s} .\) What is the smallest possible uncertainty in the electron's energy?

A hydrogen atom, initially at rest, emits an ultraviolet photon with a wavelength of \(\lambda=122 \mathrm{nm}\). What is the recoil speed of the atom after emitting the photon?

(a) As you accelerate your car away from a stoplight, does the de Broglie wavelength of the car increase, decrease, or stay the same? (b) Choose the best explanation from among the following: I. The de Broglie wavelength will increase because the momentum of the car has increased. II. The momentum of the car increases. It follows that the de Broglie wavelength will decrease, because it is inversely proportional to the wavelength. III. The de Broglie wavelength of the car depends only on its mass, which doesn't change by pulling away from the stoplight. Therefore, the de Broglie wavelength stays the same.

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