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An x ray with a wavelength of 0.100 nm collides with an electron that is initially at rest. The x ray's final wavelength is 0.110 nm. What is the final kinetic energy of the electron?

Short Answer

Expert verified
The electron's final kinetic energy is the difference between the initial and final x-ray energies.

Step by step solution

01

Understanding the Problem

This problem involves Compton scattering, where an x-ray photon collides with a stationary electron, and the wavelength of the photon increases after the collision. We need to find the kinetic energy of the electron after the collision.
02

Identifying Given Values and Formulas

We have the initial wavelength of the x-ray, \(\lambda_i = 0.100\, \text{nm}\), and the final wavelength of the x-ray, \(\lambda_f = 0.110\, \text{nm}\). The change in wavelength \(\Delta \lambda\) is given by the formula: \[ \Delta \lambda = \lambda_f - \lambda_i = 0.110\, \text{nm} - 0.100\, \text{nm} = 0.010\, \text{nm} = 0.010 \times 10^{-9}\, \text{m}.\]
03

Using Compton's Wavelength Shift Formula

Compton's formula for the change in wavelength is: \[ \Delta \lambda = \frac{h}{m_e c} (1 - \cos \theta), \] where \(h\) is Planck's constant, \(m_e\) is the electron's rest mass, and \(c\) is the speed of light. Substituting the values, \(\Delta \lambda = 0.010 \times 10^{-9} \text{ m}\).
04

Solving for the Scattering Angle \(\theta\)

Using \( \Delta \lambda = 0.010 \times 10^{-9} \text{ m} = \frac{h}{m_e c} (1 - \cos \theta) \), we need to determine \(\theta\) first. However, solving for \(\theta\) usually isn't necessary if directly targeting kinetic energy. Let's use the energy perspective next.
05

Calculating the Energy Change of the Photon

The energy of a photon is given by \( E = \frac{hc}{\lambda} \). The initial energy of the x-ray is \( E_i = \frac{hc}{\lambda_i} \) and the final energy is \( E_f = \frac{hc}{\lambda_f} \). Substitute \(h = 6.626 \times 10^{-34} \text{ Js}\), \(c = 3 \times 10^8 \text{ m/s}\) to calculate these energies.
06

Calculate the Initial and Final Photon Energies

The initial energy of the photon: \( E_i = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{0.100 \times 10^{-9}} \text{ J}. \)The final energy of the photon: \( E_f = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{0.110 \times 10^{-9}} \text{ J}. \)
07

Calculate the Change in Photon's Energy

The change in energy \( \Delta E = E_i - E_f \). Compute this energy difference to determine the kinetic energy gained by the electron, since energy is transferred from photon to electron.
08

Converting Energy to Electron's Kinetic Energy

The kinetic energy of the electron is equivalent to the decrease in the x-ray photon energy: \( KE = \Delta E \). Calculate \(\Delta E\) using the results from Step 6.

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Key Concepts

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

Wavelength Shift
In the context of Compton scattering, the wavelength shift refers to the change in wavelength of a photon, such as an x-ray, after it scatters off an electron. When the photon collides with a stationary electron, part of its energy is transferred to the electron.
This causes the photon to lose some of its initial energy, leading to an increased wavelength. This phenomenon is quantified using the change in wavelength, \( \Delta \lambda \).
The formula for the wavelength shift is: \[ \Delta \lambda = \lambda_f - \lambda_i \] where \( \lambda_i \) is the initial wavelength, and \( \lambda_f \) is the final wavelength of the photon.
In Compton's formula, the shift is given by: \[ \Delta \lambda = \frac{h}{m_e c} (1 - \cos \theta) \] where \( h \) is Planck's constant, \( m_e \) is the electron rest mass, \( c \) is the speed of light, and \( \theta \) is the scattering angle.
This shift directly correlates to the energy transferred to the electron, which is key to determining the electron's kinetic energy after such a collision.
Kinetic Energy
Kinetic energy is the energy an object possesses due to its motion. In the case of Compton scattering, the electron initially at rest gains kinetic energy from the x-ray photon. This transfer of energy results from the reduction in the photon's energy, which is then utilized to move the electron.
To find the kinetic energy of the electron, note that it is equivalent to the energy lost by the photon. The energy of a photon is inversely proportional to its wavelength, so when the photon loses energy, its wavelength increases.
The change in the photon's energy, \( \Delta E \), corresponds directly to the kinetic energy gained by the electron. By calculating the initial \( E_i \) and final \( E_f \) energies of the photon using their respective wavelengths, you can determine the kinetic energy as follows: \[ KE = \Delta E = E_i - E_f \]
This relationship highlights the fundamental principles of energy conservation in particle physics, where the lost photon energy becomes the kinetic energy of the electron.
Photon Energy
The energy of a photon is a crucial aspect of understanding Compton scattering. Photons are packets of light energy, and their energy is directly related to their wavelength. The energy, \( E \), of a photon is expressed using Planck's principle: \[ E = \frac{hc}{\lambda} \]
Here, \( h \) is Planck's constant, and \( \lambda \) is the wavelength of the photon, while \( c \) stands for the speed of light. Since photon energy is inversely proportional to wavelength, a shorter wavelength means higher energy, and vice versa.
During Compton scattering, the x-ray photon initially has a high energy and short wavelength (\( \lambda_i \)). After colliding with an electron, some of this energy is lost, resulting in a longer wavelength (\( \lambda_f \)). By calculating the initial and final energies: \[ E_i = \frac{hc}{\lambda_i} \] \[ E_f = \frac{hc}{\lambda_f} \]
We can find the change in energy \( \Delta E = E_i - E_f \), which represents the amount of energy transferred to the electron.
Planck's Constant
Planck's constant is a fundamental physical constant that plays a central role in quantum mechanics. Denoted by \( h \), its value is approximately \( 6.626 \times 10^{-34} \) joule seconds. This constant is crucial in explaining how energy is quantized, particularly in the realm of particles such as photons.
In Compton scattering, Planck's constant relates the energy of a photon to its frequency or its wavelength. The energy relationship is expressed as: \[ E = \frac{hc}{\lambda} \]
This dependency indicates that small changes in wavelength result in significant changes in energy at the quantum level. Understanding this relationship helps illustrate why photons, which follow the principles outlined by Planck's constant, behave the way they do in scattering interactions.
As photons scatter, and their energy changes, Planck's constant ensures that these changes follow a predictable and quantifiable pattern. This keeps the behavior of light consistent with observed quantum phenomena.

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Most popular questions from this chapter

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. In your analysis, you use \(c\) = 2.998 \(\times\) 10\(^8\) m/s and \(e\) = 1.602 \(\times\) 10\(^{-19}\) 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 eV?

A photon of green light has a wavelength of 520 nm. Find the photon's frequency, magnitude of momentum, and energy. Express the energy in both joules and electron volts.

The human eye is most sensitive to green light of wavelength 505 nm. Experiments have found that when people are kept in a dark room until their eyes adapt to the darkness, a \(single\) photon of green light will trigger receptor cells in the rods of the retina. (a) What is the frequency of this photon? (b) How much energy (in joules and electron volts) does it deliver to the receptor cells? (c) To appreciate what a small amount of energy this is, calculate how fast a typical bacterium of mass 9.5 \(\times\) 10\(^{-12}\) g would move if it had that much energy.

Higher-energy photons might be desirable for the treatment of certain tumors. Which of these actions would generate higher-energy photons in this linear accelerator? (a) Increasing the number of electrons that hit the tungsten target; (b) accelerating the electrons through a higher potential difference; (c) both (a) and (b); (d) none of these.

When ultraviolet light with a wavelength of 400.0 nm falls on a certain metal surface, the maximum kinetic energy of the emitted photoelectrons is measured to be 1.10 eV. What is the maximum kinetic energy of the photoelectrons when light of wavelength 300.0 nm falls on the same surface?

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