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Consider a collision between an x-ray photon of initial energy \(50.0 \mathrm{keV}\) and an electron at rest, in which the photon is scattered backward and the electron is knocked forward. (a) What is the energy of the backscattered photon? (b) What is the kinetic energy of the electron?

Short Answer

Expert verified
(a) Backscattered photon energy: \(\approx 20.0 \; \text{keV}\). (b) Electron's kinetic energy: \(\approx 30.0 \; \text{keV}\).

Step by step solution

01

Define the Energy Conservation Principle

In a photon-electron collision, energy conservation states that the initial energy (photon energy) is equal to the sum of the scattered photon's energy and the kinetic energy of the electron after collision.
02

Apply Compton Scattering Formula

The change in wavelength, also known as Compton shift, is given by:\[\Delta \lambda = \lambda' - \lambda = \frac{h}{m_e c} (1 - \cos \theta)\]where \(\lambda'\) is the wavelength of the scattered photon, \(\lambda\) is the initial wavelength, \(h\) is Planck's constant (\(6.626 \times 10^{-34} \; \text{m}^2\text{kg/s}\)), \(m_e\) is the electron mass (\(9.109 \times 10^{-31} \; \text{kg}\)), and \(c\) is the speed of light (\(3 \times 10^8 \; \text{m/s}\)). \(\theta = 180^\circ\) for backward scattering.
03

Convert Initial Energy to Wavelength

The energy of the photon is related to its wavelength by \[E = \frac{hc}{\lambda}\]Thus, for the initial photon energy \(E_0 = 50.0 \; \text{keV} = 50.0 \times 10^3 \times 1.602 \times 10^{-19} \; \text{J}\),\[\lambda_0 = \frac{hc}{E_0}\]
04

Calculate Scattered Photon Wavelength

Using the Compton scattering formula, calculate \(\lambda'\):\[\lambda' = \lambda_0 + 2\frac{h}{m_e c}\]Substitute \(\Delta \lambda = 2\frac{h}{m_e c}\) as \(\theta = 180^\circ\).
05

Calculate Backscattered Photon Energy

The energy of the backscattered photon is given by:\[E' = \frac{hc}{\lambda'}\]Substitute \(\lambda'\) from Step 4 to find \(E'\).
06

Apply Energy Conservation to Find Electron's Kinetic Energy

From energy conservation:\[E_0 = E' + KE_e\]Solve for the electron's kinetic energy (\(KE_e\)): \[KE_e = E_0 - E'\].

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Key Concepts

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

Photon-Electron Collision
In a photon-electron collision, an x-ray photon crashes into an electron. The electron is initially at rest. When the photon impacts the electron, it is scattered in a direction different from which it traveled initially. This scattering process is called Compton scattering. The electron, likewise, gains energy and moves forward in the direction away from the colliding photon.
The collision follows the law of conservation of energy and momentum, meaning the total energy and total momentum are the same before and after the collision. Understanding this link helps us identify how the energy of the photon and the electron change after the impact. Photons and electrons are very different. Photons have no mass and always travel at the speed of light. Electrons have mass and gain energy that becomes kinetic energy when they move.
Energy Conservation
Energy conservation is a fundamental principle in physics. It tells us that in an isolated system, energy cannot be created or destroyed. Instead, it can only be transformed from one form to another. In the case of Compton scattering, the system includes the photon and the electron. The initial total energy is the sum of the x-ray photon's energy and the electron's rest energy (which can be ignored since the electron is at rest).
After the collision, the total energy splits into the energy of the scattered photon and the kinetic energy of the electron. Mathematically, this can be expressed as:
  • Initial photon energy (\(E_0\))
  • Equal to the sum of the scattered photon energy (\(E'\))
  • And the kinetic energy of the electron (\(KE_e\))
In formula terms: \[E_0 = E' + KE_e\]. So, if you know any two of these energies, you can easily find the third.
Photon Energy
Photon energy is the energy carried by a single photon. It is directly related to the frequency or wavelength of the photon. This relationship is given by the equation:\[E = \frac{hc}{\lambda}\]where \(E\) is the energy, \(h\) is Planck's constant, \(c\) is the speed of light, and \(\lambda\) is the wavelength of the photon.
For our collision problem, the initial photon energy \(E_0\) is given as 50.0 keV, which can be converted to joules for calculation purposes. In Compton scattering, the photon's energy changes due to its change in wavelength. A longer wavelength means less energy.
After the collision, the photon that scatters backward has less energy than before because it gave some energy to the electron. Using the energy-wavelength relationship, we can find the new energy of the scattered photon.
Kinetic Energy of Electron
When the photon collides with the electron and transfers energy, this energy becomes the electron's kinetic energy. Kinetic energy is the energy an object possesses due to its motion. In Compton scattering, the initial stationary electron gains kinetic energy and starts moving after its interaction with the photon.
The kinetic energy of the electron can be deduced from the energy conservation formula:\[KE_e = E_0 - E'\]where \(KE_e\) is the electron's kinetic energy, \(E_0\) is the initial photon energy before the collision, and \(E'\) is the energy of the photon after being scattered.
This equation shows that the energy lost by the photon is equal to the kinetic energy gained by the electron. Even though electrons, when compared to photons, are substantially larger and slower, these processes happen on a very fast timescale in subatomic environments.

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Most popular questions from this chapter

Electrons accelerated to an energy of \(50 \mathrm{GeV}\) have a de Broglie wavelength \(\lambda\) small enough for them to probe the structure within a target nucleus by scattering from the structure. Assume that the energy is so large that the extreme relativistic relation \(p=E / c\) between momentum magnitude \(p\) and energy \(E\) applies. (In this extreme situation, the kinetic energy of an electron is much greater than its rest energy.) (a) What is \(\lambda ?\) (b) If the target nucleus has radius \(R=5.0 \mathrm{fm},\) what is the ratio \(R / \lambda ?\)

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