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A beam of photons with just barely enough individual photon energy to create electron-positron pairs undergoes Compton scattering from free electrons before any pair production occurs. (a) Can the Compton-scattered photons create electron-positron pairs? (b) Calculate the momentum magnitude of the photons found at a \(20.0^{\circ}\) angle from the direction of the original beam.

Short Answer

Expert verified
a) No, the Compton-scattered photons cannot create electron-positron pairs because they have less energy after scattering. b) The momentum magnitude of the photons depends on the original wavelength of the photons. After calculating the new wavelength using the Compton scattering equation, the momentum can be found using the formula \( p = \frac {h}{\lambda} \).

Step by step solution

01

Understanding pair production

In order to answer part a, it is important to understand that pair production (creation of an electron-positron pair) only happens if the photons have enough energy, typically more than 1.022 MeV, which is the rest mass energy of electron and positron. But in Compton scattering, photons lose energy. So if the photons had barely enough energy for pair production before Compton scattering, after scattering their energy will be less than required for pair production.
02

Compton Scattering equation

To answer part b, we need to use the formula for Compton scattering. The change in wavelength (\( \Delta \lambda \)) of the scattered photon can be given by the formula: \( \Delta \lambda = \frac {h}{m_e c}(1 - cos \Theta) \), where \( h \) is the Planck’s constant, \( m_e \) is the rest mass of the electron, \( c \) is the speed of light, and \( \Theta \) is the scattering angle.
03

Calculating change in photon wavelength

First, substituting the values in to the formula from Step 2, we find the change in photon's wavelength after the 20 degree scattering.
04

Calculating momentum

The momentum of a photon is given by \( p = \frac {h}{\lambda} \). We can use this formula to find the momentum of the scattered photons. Remembering that the new wavelength will be the original wavelength plus the change calculated in step 3.

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Key Concepts

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

Pair Production
Pair production is a fascinating process where energy turns into matter. Specifically, a photon with sufficiently high energy can transform into an electron-positron pair. The energy threshold for this occurrence is about 1.022 MeV, equivalent to the combined rest mass energy of an electron and a positron. This phenomenon illustrates the principle of energy-mass equivalence, emphasizing that energy can convert into mass and vice versa.
For a photon to undergo pair production, it must encounter a strong electric field, typically near a nucleus. The field provides additional stability needed for this conversion. However, if the photon doesn't meet or exceed the energy threshold, pair production will not occur.
In the context of Compton scattering, photons lose energy as they collide with electrons. If a photon has just enough energy for pair production and then undergoes Compton scattering, it loses energy, thus falling short of the threshold required for creating an electron-positron pair. Consequently, these scattered photons cannot produce pairs.
Photon Momentum
Photons, despite having no rest mass, do have momentum due to their energy and speed, which is always the speed of light. The momentum of a photon is mathematically related to its wavelength by the equation \( p = \frac{h}{\lambda} \), where \( p \) represents momentum, \( h \) is Planck’s constant, and \( \lambda \) is the photon's wavelength.
In any interaction, such as Compton scattering, the momentum of the photon changes along with its wavelength. The shift in momentum is a critical part of understanding how photons behave when they interact with matter. This shift also explains why they move differently after scattering, as their paths and speeds are altered while conserving energy and momentum.
Understanding photon momentum helps us grasp how energy transfers and changes form, painting a crucial part of the picture in the dynamics of subatomic particles.
Scattering Angle
The scattering angle in Compton scattering describes how much the direction of a photon changes after it interacts with a target, usually free electrons. When a photon strikes an electron, its path changes, and this change in direction is quantified as the scattering angle \( \Theta \).
The scattering angle is integral to calculations involving Compton scattering as it affects the wavelength shift of the photon. The relationship is given by the Compton wavelength shift equation: \( \Delta \lambda = \frac{h}{m_e c} (1 - \cos \Theta) \), where \( \Delta \lambda \) is the change in wavelength, \( m_e \) is the electron’s mass, and \( c \) is the speed of light.
For example, at a scattering angle of 20 degrees, we use this equation to find how the photon's wavelength changes. A larger scattering angle typically means a larger change in wavelength, illustrating how the degree of deflection impacts other properties of the photon.
Energy Conservation in Photon Interactions
Energy conservation is a foundational principle guiding photon interactions, including those occurring in phenomena like Compton scattering and pair production. In these processes, energy is neither created nor destroyed; rather, it is transformed from one form to another or transferred between particles.
In Compton scattering, when a photon collides with an electron, some of its energy transfers to the electron, changing the photon's wavelength and, consequently, its energy. The loss in photon energy corresponds to the gain in energy by the electron, maintaining the overall conservation of energy.
During pair production, if it occurs, a photon converts its energy into the mass and kinetic energy of the electron and positron pair. To maintain energy conservation, the total energy following the interaction equals the initial photon energy, allowing no surplus or deficit.
Through these interactions, energy transformation and transfer illustrate fundamental physics principles, showcasing the elegance of nature's finely-tuned systems.

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Most popular questions from this chapter

A photon with wavelength \(\lambda\) is incident on a stationary particle with mass \(M,\) as shown in Fig. \(\mathbf{P 3 8 . 3 9 .}\) The photon is annihilated while an electron-positron pair is produced. The target particle moves off in the original direction of the photon with speed \(V_{M}\). The electron travels with speed \(v\) at angle \(\phi\) with respect to that direction. Owing to momentum conservation, the positron has the same speed as the electron. (a) Using the notation \(\gamma=\left(1-v^{2} / c^{2}\right)^{-1 / 2}\) and \(\gamma_{M}=\left(1-V_{M}^{2} / c^{2}\right)^{-1 / 2},\) write a relativistic expression for energy conservation in this process. Use the symbol \(m\) for the electron mass. (b) Write an analogous expression for momentum conservation. (c) Eliminate the ratio \(h / \lambda\) between your energy and momentum equations to derive a relationship between \(v / c, V_{M} / c,\) and \(\phi\) in terms of the ratio \(m / M\), keeping \(\gamma\) and \(\gamma_{M}\) as a useful shorthand notation. (d) Consider the case where \(V_{M} \ll c\), and use a binomial expansion to derive an expression for \(\gamma_{M}\) to the first order in \(V_{M}\). Use that result to rewrite your previous result as an expression for \(V_{M}\) in terms of \(c, v, m / M,\) and \(\phi .\) (e) Are there choices of \(v\) and \(\phi\) for which \(V_{M}=0 ?\) (f) Suppose the target particle is a proton. If the electron and positron remain stationary, so that \(v=0\), then with what speed does the proton move, in \(\mathrm{km} / \mathrm{s} ?\) (g) If the electron and positron each have total energy \(5.00 \mathrm{MeV}\) and move with \(\phi=60^{\circ}\), then what is the speed of the proton? (Hint: First solve for \(\gamma\) and \(v .)\) (h) What is the energy of the incident photon in this case?

Nuclear fusion reactions at the center of the sun produce gamma-ray photons with energies of about \(1 \mathrm{MeV}\left(10^{6} \mathrm{eV}\right) .\) By contrast, what we see emanating from the sun's surface are visiblelight photons with wavelengths of about \(500 \mathrm{nm}\). A simple model that explains this difference in wavelength is that a photon undergoes Compton scattering many times-in fact, about \(10^{26}\) times, as suggested by models of the solar interior - as it travels from the center of the sun to its surface. (a) Estimate the increase in wavelength of a photon in an average Compton-scattering event. (b) Find the angle in degrees through which the photon is scattered in the scattering event described in part (a). (Hint: A useful approximation is \(\cos \phi \approx 1-\phi^{2} / 2,\) which is valid for \(\phi \ll 1 .\) Note that \(\phi\) is in radians in this expression.) (c) It is estimated that a photon takes about \(10^{6}\) years to travel from the core to the surface of the sun. Find the average distance that light can travel within the interior of the sun without being scattered. (This distance is roughly equivalent to how far you could see if you were inside the sun and could survive the extreme temperatures there. As your answer shows, the interior of the sun is very opaque.)

If a photon of wavelength \(0.04250 \mathrm{nm}\) strikes a free electron and is scattered at an angle of \(35.0^{\circ}\) from its original direction, find (a) the change in the wavelength of this photon; (b) the wavelength of the scattered light; (c) the change in energy of the photon (is it a loss or a gain?); (d) the energy gained by the electron.

An electron accelerates through a potential difference of \(50.0 \mathrm{kV}\) in an \(\mathrm{x}\) -ray tube. When the electron strikes the target, \(70.0 \%\) of its kinetic energy is imparted to a single photon. Find the photon's frequency, wavelength, and magnitude of momentum.

In developing night-vision equipment, you need to measure the work function for a metal surface, so you perform a photoelectric-effect experiment. You measure the stopping potential \(V_{0}\) as a function of the wavelength \(\lambda\) of the light that is incident on the surface. You get the results in the table. $$ \begin{array}{l|llllll} \boldsymbol{\lambda}(\mathbf{n m}) & 100 & 120 & 140 & 160 & 180 & 200 \\ \hline \boldsymbol{V}_{0}(\mathbf{V}) & 7.53 & 5.59 & 3.98 & 2.92 & 2.06 & 1.43 \end{array} $$ In your analysis, you use \(c=2.998 \times 10^{8} \mathrm{~m} / \mathrm{s}\) and \(e=1.602 \times 10^{-19} \mathrm{C}\) which are values obtained in other experiments. (a) Select a way to plot your results so that the data points fall close to a straight line. Using that plot, find the slope and \(y\) -intercept of the best-fit straight line to the data. (b) Use the results of part (a) to calculate Planck's constant \(h\) (as a test of your data) and the work function (in eV) of the surface. (c) What is the longest wavelength of light that will produce photoelectrons from this surface? (d) What wavelength of light is required to produce photoelectrons with kinetic energy \(10.0 \mathrm{eV} ?\)

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