Consider the equivalent of Compton scattering, but the case in which a photon scatters off of a free proton. a) If \(140 .-\mathrm{keV} \mathrm{X}\) -rays bounce off of a proton at \(90.0^{\circ},\) what is their fractional change in energy \(\left(E_{0}-E\right) / E_{0} ?\) b) What energy of photon would be necessary to cause a \(1.00 \%\) change in energy at \(90.0^{\circ}\) scattering?

Short Answer

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b) What is the initial energy of the photon necessary to cause a 1% change in energy at a 90-degree scattering angle?

Step by step solution

01

Calculate the initial wavelength of the photon

Use the formula for the energy of the photon to calculate the initial wavelength of the photon: $$\lambda_0 = \dfrac{hc}{E_0} $$ where \(E_0 = 140 \, keV = 140 \times 10^3 eV\), \(h = 4.13567 \times 10^{-15} \, eV s\), and \(c = 2.998 \times 10^8 \, m/s\). Plug in the values and calculate \(\lambda_0\).
02

Calculate the change in wavelength of the photon

Use the Compton formula to calculate the change in wavelength \(\Delta \lambda\): $$\Delta \lambda = \dfrac{h}{m_p c}(1-\cos\theta) $$ where \(\theta = 90^\circ\), \(h = 4.13567 \times 10^{-15} \, eV s\), \(c = 2.998 \times 10^8 \, m/s\), and \(m_p = 1.67262 \times 10^{-27} \, kg\). Plug in the values and calculate \(\Delta \lambda\).
03

Calculate the final wavelength of the photon

The final wavelength of the photon \(\lambda\) would be the sum of the initial wavelength and the change in wavelength: $$\lambda = \lambda_0 + \Delta \lambda$$
04

Calculate the final energy of the photon

Use the formula for the energy of the photon to calculate the final energy of the photon: $$ E = \dfrac{hc}{\lambda} $$ where \(h = 4.13567 \times 10^{-15} \, eV s\), and \(c = 2.998 \times 10^8 \, m/s\). Plug in the values and calculate \(E\).
05

Calculate the fractional change in energy (a)

Calculate the fractional change in energy using the formula: $$ \dfrac{E_0 - E}{E_0} $$
06

Calculate the initial energy of the photon for a 1% change (b)

We need to find the energy \(E_0\) to cause a \(1.00 \%\) change in energy at \(90^\circ\) scattering. Let's use the given percentage and calculate the corresponding change in wavelength: $$ \Delta \lambda_{1\%} = \dfrac{E_0 - E_{1\%}}{E_0} \lambda_{1\%} $$ Using the Compton formula: $$ \Delta \lambda_{1\%} = \dfrac{h}{m_p c}(1-\cos\theta) $$ Solve the above equations for \(E_0\) and find the energy of the photon required to cause a 1% change in energy at a scattering angle of 90 degrees.

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Key Concepts

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

Photon Energy Calculation
Understanding photon energy is crucial when analyzing phenomena such as Compton scattering. Energy of a photon can be determined using the equation \(E = \frac{hc}{\lambda}\), where \(E\) is the photon energy, \(h\) is Planck's constant, \(c\) is the speed of light, and \(\lambda\) is the photon's wavelength.

For a photon with a given energy, such as the 140 keV X-rays in our exercise, converting this energy into the wavelength using the above formula gives insight into its characteristics prior to any interaction with particles like protons. It's akin to knowing the initial speed of a ball before it bounces off a wall; you can tell much about the post-collision scenario by understanding the pre-collision state.

The precision in photon energy calculation is paramount as it determines the subsequent steps and results, especially when dealing with fractional changes in energy after scattering events. Small inaccuracies could lead to significantly different interpretations of the phenomenon under study.
Compton Wavelength Shift
The Compton wavelength shift is a quantum mechanical phenomenon that illustrates the change in a photon's wavelength after it scatters off a target particle like a free proton. This shift can be quantified using the Compton formula \(\Delta \lambda = \frac{h}{m_p c}(1-\cos\theta)\) where \(\Delta \lambda\) is the change in the photon's wavelength, \(m_p\) is the proton's mass, \(\theta\) is the scattering angle, and other symbols have their usual meanings.

When a photon undergoes this energy exchange, the scattering angle is directly related to how much the wavelength shifts. In the case where the angle is 90 degrees, as in our example, the wavelength shift can be quite significant. This is akin to how the angle at which you throw a ball against a surface greatly affects how it bounces back.

Implications of the Shift

Understanding the Compton wavelength shift is not only about knowing how to perform the calculation; it is also about understanding the implications of this change. A larger shift means more energy has been transferred from the photon to the proton, which can have implications for the proton's kinetic energy and the photon's subsequent interactions.
Proton-Photon Interaction
When a photon scatters off a proton, it's an interaction that can be explained using quantum electrodynamics. This photon-proton interaction is less noticeable compared to electron-based scattering because protons, being much more massive than electrons, result in smaller energy transfers for given conditions.

In our exercise, the scattering is at a 90-degree angle involving 140 keV X-rays, requiring detailed calculations to determine the photon's fractional change in energy. The mass of the proton plays a significant role as it affects the Compton wavelength shift. The larger the target particle’s mass (like a proton, compared to an electron), the smaller the wavelength shift will be for a given energy of the incident photon.

While students need to be meticulous with the calculations, conceptual understanding of proton-photon interactions aids in grasping the subtleties of energy and momentum conservation in quantum scale interactions. This understanding is critical for progressing in fields such as particle physics and nuclear engineering where such interactions are the cornerstone of more complex processes.

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Most popular questions from this chapter

A photovoltaic device uses monochromatic light of wavelength 700 . \(\mathrm{nm}\) that is incident normally on a surface of area \(10.0 \mathrm{~cm}^{2}\). Calculate the photon flux rate if the light intensity is \(0.300 \mathrm{~W} / \mathrm{cm}^{2}\).

Consider an electron whose de Broglie wavelength is equal to the wavelength of green light (about \(550 \mathrm{nm}\) ). a) Treating the electron nonrelativistically, what is its speed? b) Does your calculation confirm that a nonrelativistic treatment is sufficient? c) Calculate the kinetic energy of the electron in \(\mathrm{eV}\).

Scintillation detectors for gamma rays transfer the energy of a gamma-ray photon to an electron within a crystal, via the photoelectric effect or Compton scattering. The electron transfers its energy to atoms in the crystal, which re-emit it as a light flash detected by a photomultiplier tube. The charge pulse produced by the photomultiplier tube is proportional to the energy originally deposited in the crystal; this can be measured so an energy spectrum can be displayed. Gamma rays absorbed by the photoelectric effect are recorded as a photopeak in the spectrum, at the full energy of the gammas. The Compton-scattered electrons are also recorded, at a range of lower energies known as the Compton plateau. The highest-energy of these form the Compton edge of the plateau. Gamma-ray photons scattered \(180 .^{\circ}\) by the Compton effect appear as a backscatter peak in the spectrum. For gamma-ray photons of energy \(511 \mathrm{KeV}\) calculate the energies of the Compton edge and the backscatter peak in the spectrum.

An accelerator boosts a proton's kinetic energy so that the de Broglie wavelength of the proton is \(3.5 \cdot 10^{-15} \mathrm{~m}\) What are the momentum and energy of the proton?

What is the wavelength of an electron that is accelerated from rest through a potential difference of \(1.00 \cdot 10^{-5} \mathrm{~V} ?\)

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