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If a photon of wavelength 0.04250 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?), and (d) the energy gained by the electron.

Short Answer

Expert verified
The wavelength change is small due to the small angle; the photon loses energy, which the electron gains.

Step by step solution

01

Understanding the Compton Wavelength Shift Formula

The formula to calculate the change in wavelength of a photon after scattering is given by the Compton shift formula:\[\Delta \lambda = \lambda' - \lambda = \frac{h}{m_ec} (1 - \cos \theta)\]where \(h\) is Planck's constant \(6.626 \times 10^{-34}\) Js, \(m_e\) is the electron rest mass \(9.11 \times 10^{-31}\) kg, and \(c\) is the speed of light \(3 \times 10^8\) m/s.
02

Calculate the Compton Wavelength Shift

Substitute the given angle \(\theta = 35.0^{\circ}\) into the formula:\[\Delta \lambda = \frac{6.626 \times 10^{-34}}{9.11 \times 10^{-31} \times 3 \times 10^8} \times (1 - \cos 35^{\circ})\]Calculate \(\Delta \lambda\).
03

Calculate the Wavelength of the Scattered Photon

The wavelength of the scattered photon is given by:\[\lambda' = \lambda + \Delta \lambda\]Substitute \(\lambda = 0.04250\) nm and \(\Delta \lambda\) from Step 2 to find \(\lambda'\).
04

Calculate the Initial Energy of the Photon

The energy of a photon is related to its wavelength by:\[E = \frac{hc}{\lambda}\]Calculate the initial energy using \(\lambda = 0.04250\) nm.
05

Calculate the Energy of the Scattered Photon

Using the scattered wavelength \(\lambda'\) from Step 3, calculate the new energy using the same energy formula:\[E' = \frac{hc}{\lambda'}\]
06

Determine the Change in Energy of the Photon

Calculate the change in energy:\[\Delta E_{photon} = E - E'\]Determine if this indicates an energy loss or gain.
07

Calculate the Energy Gained by the Electron

The energy gained by the electron is equal to the loss in energy by the photon:\[\Delta E_{electron} = \Delta E_{photon}\]

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Key Concepts

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

Photon scattering
Photon scattering occurs when a photon, a particle of light, interacts with an electron or another particle, causing the photon to deviate from its original path. This deviation or deflection results in changes to the photon's properties. One significant aspect of photon scattering is the change in the photon's energy, which directly relates to its wavelength. In simpler terms, when a photon collides with a free electron, like in the Compton Effect, it can "bounce off" at an angle, and its energy and wavelength will shift slightly.

Scattering is crucial in understanding various phenomena in physics, including how light interacts with matter in different contexts, like in medical imaging or astronomical observations. By analyzing the scattering of photons, scientists can deduce information about the structures and properties of the scattering objects.
Wavelength shift
Wavelength shift refers to the change in the wavelength of a photon as it scatters from an electron. The Compton Effect is a classic example of a wavelength shift following photon scattering. When a photon collides with a free electron, part of its energy is transferred to the electron, altering the photon's wavelength.

The Compton wavelength shift can be mathematically described by the formula:
  • \[\Delta \lambda = \frac{h}{m_e c} (1 - \cos \theta)\]
Here,
  • \(h\) represents Planck's constant, \(6.626 \times 10^{-34}\) Js
  • \(m_e\) is the electron mass, \(9.11 \times 10^{-31}\) kg
  • \(c\) is the speed of light, \(3 \times 10^8\) m/s
  • \(\theta\) is the scattering angle.
This expression calculates the exact change in wavelength based on these constants and the angle of scattering, showing how energy gets transferred during the interaction.
Photon energy
Photon energy is fundamentally tied to a photon's wavelength. The shorter the wavelength, the higher its energy. This relationship can be expressed using the formula:
  • \[E = \frac{hc}{\lambda}\]
This equation states that the energy \(E\) of a photon is equal to Planck's constant \(h\) times the speed of light \(c\) divided by the wavelength \(\lambda\).

During the Compton Effect, as photons scatter off electrons, their energy and wavelength are both affected. If a photon loses energy to an electron, its corresponding wavelength increases (since energy is inverse to wavelength). Understanding this exchange is crucial for applications like X-ray production and the interpretation of astronomical data, where energy and wavelength shifts inform scientists about cosmic phenomena.
Compton scattering formula
The Compton scattering formula is crucial in calculating the change of wavelength of a photon due to scattering. This formula takes into account the principles of photon scattering outlined by Arthur Compton in 1923. The key to understanding this formula is recognizing that it models how energy and momentum transfer occur between photons and electrons during scattering.

The Compton scattering formula:
  • \[\Delta \lambda = \frac{h}{m_e c} (1 - \cos \theta)\]
is derived from conservation laws, predicting how much the photon's wavelength will change based on the angle \(\theta\) at which it scatters. By using this formula, one can precisely determine the shift in wavelength, which helps in identifying energy changes. This is central to many technologies and scientific domains, from medical imaging to material sciences, where knowledge of particle interactions plays a crucial role.

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Most popular questions from this chapter

\(\cdot\) What is the de Broglie wavelength of a red blood cell with a mass of \(1.00 \times 10^{-11} \mathrm{g}\) that is moving with a speed of 0.400 \(\mathrm{cm} / \mathrm{s} ?\) Do we need to be concerned with the wave nature of the blood cells when we describe the flow of blood in the body?

\(\bullet\) The neutral \(\pi^{\circ}\) meson is an unstable particle produced in high-energy particle collisions. Its mass is about 264 times that of the electron, and it exists for an average lifetime of \(8.4 \times 10^{-17}\) s before decaying into two gamma-ray photons. Assuming that the mass and energy of the particle are related by the Einstein relation \(E=m c^{2},\) find the uncertainty in the mass of the particle and express it as a fraction of the particle's mass.

Suppose that the uncertainty in position of an electron is equal to the radius of the \(n=1\) Bohr orbit, about \(0.5 \times 10^{-10} \mathrm{m} .\) Calculate the minimum uncertainty in the cor- responding momentum the minimum uncertainty in the cor- magnitude of the momentum of the electron in the \(n=1\) Bohr orbit.

\(\bullet\) (a) How much energy is needed to ionize a hydrogen atom that is in the \(n=4\) state? (b) What would be the wavelength of a photon emitted by a hydrogen atom in a transition from the \(n=4\) state to the \(n=2\) state?

\bullet A triply ionized beryllium ion, \(\mathrm{Be}^{3+}\) (a beryllium atom with three electrons removed), behaves very much like a hydrogen atom, except that the nuclear charge is four times as great. (a) What is the ground-level energy of \(\mathrm{Be}^{3+} ?\) How does this compare with the ground-level energy of the hydrogen atom? (b) What is the ionization energy of Be \(^{3+} ?\) How does this compare with the ionization energy of the hydrogen atom? (c) For the hydrogen atom, the wavelength of the photon emitted in the transition \(n=2\) to \(n=1\) is 122 nm. (See Example 28.6 . What is the wavelength of the photon emitted when a \(\mathrm{Be}^{3+}\) ion undergoes this transition? (d) For a given value of \(n,\) how does the radius of an orbit in \(\mathrm{Be}^{3+}\) compare with that for hydrogen?

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