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X rays of wavelength \(8.50 \mathrm{pm}\) are directed in the positive direction of an \(x\) axis onto a target containing loosely bound electrons For Compton scattering from one of those electrons, at an angle of \(180^{\circ}\), what are (a) the Compton shift, (b) the corresponding change in photon energy, (c) the kinetic energy of the recoiling electron, and (d) the angle between the positive direction of the \(x\) axis and the electron's direction of motion?

Short Answer

Expert verified
(a) 4.85 pm, (b) 3.253 x 10鈦宦光伒 J, (c) 3.253 x 10鈦宦光伒 J, (d) 0掳.

Step by step solution

01

Understand Compton scattering

Compton scattering occurs when a photon scatters off a free electron, resulting in a change in the photon's wavelength. The change in wavelength is given by the Compton shift formula: \[ \Delta\lambda = \lambda' - \lambda = \frac{h}{mc}(1 - \cos \theta) \] where \(\lambda\) is the initial wavelength, \(\lambda'\) is the final wavelength, \(h\) is Planck's constant \(6.626 \times 10^{-34} \text{ J s}\), \(mc = 2.426 \times 10^{-12} \text{ m}\) is the Compton wavelength of the electron, and \(\theta\) is the scattering angle, which is given as \(180^\circ\).
02

Calculate the Compton shift

For \(\theta = 180^\circ\), the formula for Compton shift becomes:\[ \Delta\lambda = \frac{2h}{mc} \]Substitute the values:\[ \Delta\lambda = \frac{2 \times 6.626 \times 10^{-34}}{9.11 \times 10^{-31} \times 3 \times 10^{8}} = 4.85 \times 10^{-12} \text{ m} = 4.85 \text{ pm} \].
03

Calculate the change in photon energy

The energy change \(\Delta E\) of the photon relates to the change in wavelength \(\Delta \lambda\) as follows:\[ E = \frac{hc}{\lambda} \]Thus, the change in energy\[ \Delta E = hc \left( \frac{1}{\lambda} - \frac{1}{\lambda + \Delta\lambda} \right) \]Calculate:\[ \Delta E = 6.626 \times 10^{-34} \times 3 \times 10^{8} \left( \frac{1}{8.50 \times 10^{-12}} - \frac{1}{8.50 + 4.85 \times 10^{-12}} \right) \]\[ \Delta E = 6.626 \times 10^{-34} \times 3 \times 10^{8} \times (0.1176 - 0.0658) \times 10^{12} \approx 3.253 \times 10^{-15} \text{ J} \].
04

Calculate the kinetic energy of the recoiling electron

When a photon gives energy to an electron, the kinetic energy \(K\) of the electron is equal to the energy lost by the photon:\[ K = \Delta E = 3.253 \times 10^{-15} \text{ J} \].
05

Determine the angle of motion of the recoiling electron

From conservation of momentum and energy, use the following relation:\[ \tan \phi = \frac{\lambda + \Delta \lambda}{\lambda} \cdot \sin(\theta) \]Since \(\theta = 180^\circ\) and \(\sin 180^\circ = 0\), the electron's emission is directly opposite to the initial photon, meaning the angle \(\phi\) of the electron's direction is effectively 0 with the positive direction of the x-axis.

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Key Concepts

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

Photon Energy Change
In Compton scattering, a photon collides with a free electron, modifying its energy. This energy alteration is crucial as it determines how much energy the photon loses or gains during the interaction. The energy of a photon is given by \(E = \frac{hc}{\lambda}\), where \(h\) is Planck's constant, \(c\) is the speed of light, and \(\lambda\) is the photon鈥檚 wavelength. After the collision, the change in energy \(\Delta E\) depends on how much the photon's wavelength changes. You can use the formula:\[\Delta E = hc \left( \frac{1}{\lambda} - \frac{1}{\lambda + \Delta\lambda} \right)\]to find the energy difference. It's important because it directly relates to the resulting kinetic energy imparted to the electron.
Kinetic Energy of Electron
When a photon transfers energy during a Compton scattering event, part of this energy is absorbed by the electron, manifesting as kinetic energy. This is the energy responsible for the electron's motion post-collision. The kinetic energy \(K\) of the electron equals the energy lost by the photon due to the scattering, which makes it:\[K = \Delta E\]This concept is instrumental as it describes how energy conservation operates during scattering. Knowing the kinetic energy helps scientists understand particle dynamics and energy transformations in various applications.
Compton Wavelength
The Compton wavelength is a fundamental quantity in quantum mechanics, representing the wavelength shift experienced by photons during Compton scattering. It's given by the formula:\[\Delta\lambda = \frac{h}{mc}(1 - \cos \theta)\]where \(\lambda\) is the initial wavelength, \(m\) is the electron mass, \(c\) is the speed of light, and \(\theta\) is the scattering angle. For a \(180^\circ\) scattering angle, as in our problem, the formula simplifies to:\[\Delta\lambda = \frac{2h}{mc}\]This wavelength shift is significant because it quantifies how the photon鈥檚 path and characteristics are altered upon interacting with an electron.
Scattering Angle
The scattering angle \(\theta\) in Compton scattering defines the direction change of the photon when it interacts with an electron. It heavily influences the amount of change in the photon's wavelength. In cases where the scattering angle \(\theta\) is \(180^\circ\), the photon is said to scatter directly backward. This results in the maximum possible change in wavelength given by the Compton shift. Understanding this angle helps describe the spatial behavior of photons as they disperse, hence playing a pivotal role in interpreting how scattering impacts photon trajectories in experimental setups.

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Most popular questions from this chapter

The wavelength of the yellow spectral emission line of sodium is \(590 \mathrm{~nm}\). At what kinetic energy would an electron have that wavelength as its de Broglie wavelength?

Singly charged sodium ions are accelerated through a potential difference of \(300 \mathrm{~V}\). (a) What is the momentum acquired by such an ion? (b) What is its de Broglie wavelength?

Just after detonation, the fireball in a nuclear blast is approximately an ideal blackbody radiator with a surface temperature of about \(1.0 \times 10^{7} \mathbf{K}\). (a) Find the wavelength at which the thermal radiation is maximum and (b) identify the type of electromagnetic wave corresponding to that wavelength. (See Fig. 33-1.) This radiation is almost immediately absorbed by the surrounding air molecules, which produces another ideal blackbody radiator with a surface temperature of about \(1.0 \times 10^{5} \mathrm{~K}\). (c) Find the wavelength at which the thermal radiation is maximum and (d) identify the type of electromagnetic wave corresponding to that wavelength.

What are (a) the Compton shift \(\Delta \lambda\), (b) the fractional Compton shift \(\Delta \lambda / \lambda\), and \((c)\) the change \(\Delta E\) in photon energy for light of wavelength \(\lambda=590 \mathrm{~nm}\) scattering from a free, initially stationary electron if the scattering is at \(90^{\circ}\) to the direction of the incident beam? What are (d) \(\Delta \lambda\), (e) \(\Delta \lambda / \lambda\), and (f) \(\Delta E\) for \(90^{\circ}\) scattering for photon energy \(50.0 \mathrm{keV}\) (x-ray range)?

What are (a) the energy of a photon corresponding to wavelength \(1.00 \mathrm{~nm}\), (b) the kinetic energy of an electron with de Broglie wavelength \(1.00 \mathrm{~nm}\), (c) the energy of a photon corresponding to wavelength \(1.00 \mathrm{fm}\), and (d) the kinetic energy of an electron with de Broglie wavelength \(1.00 \mathrm{fm}\) ?

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