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Chapter 3: Problem 26

X-ray photons of wavelength \(0.02218 \mathrm{nm}\) are incident in a target and the Compton-scattered photons are observed at \(90.0^{\circ} .(a)\) What is the wavelength of the scattered photons? (b) What is the momentum of the incident photons and the scattered photons? (c) What is the kinetic energy of the scattered electrons? \((d)\) What is the momentum (magnitude and direction) of the scattered electrons?

Short Answer

Expert verified
Wavelength of scattered photon: 0.09966 nm. Momentum of incident photon: 3.00 x 10^-23 kg路m/s. Kinetic Energy of electron: 8.87 x 10^-15 J. Electron momentum: 9.65 x 10^-23 kg路m/s, at 90 degrees.

Step by step solution

01

- Calculate the Wavelength of Scattered Photons

Use the Compton wavelength shift formula: \[ \frac{\text{Compton Shift}}{\theta (angle)} = \frac{h}{m_ec} (1 - \text {cos}(\theta))\] In this problem, \( \theta = 90掳 \), so: \(\Delta \lambda = \frac{h}{m_ec} (1 - \text{cos}(90掳)) \) Convert the incoming wavelength (0.02218 nm) to meters and compute \( \Delta \lambda \). Then add \( \Delta \lambda \) to the original wavelength to find the scattered photon wavelength.
02

- Calculate the Momentum of the Incident Photons

Use the formula for the momentum of a photon: \(p = \frac{h}{\lambda}\). Plug in the values for Planck鈥檚 constant (\(h\)) and the original wavelength (\(\lambda\)).
03

- Calculate the Momentum of the Scattered Photons

The equation remains the same as Step 2, but use the scattered wavelength found in Step 1: \(p' = \frac{h}{\lambda'}\). Compute this momentum.
04

- Calculate the Kinetic Energy of the Scattered Electrons

Use the energy conservation principle for the Compton effect: \( K.E_e = E_i - E_s \) where \( E_i = \frac{hc}{\lambda} \) and \( E_s = \frac{hc}{\lambda'} \). Calculate the initial and scattered photon's energies and then find the kinetic energy of the scattered electrons.
05

- Calculate the Momentum of the Scattered Electrons

Use the conservation of momentum principle to find the magnitude and direction: \( p_e = \sqrt{p_i^2 + p_s^2 - 2p_i p_s \text{cos} (\theta)} \). Compute it with the initial and scattered momenta found in Steps 2 and 3. The direction can be found using vector addition.

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Key Concepts

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

Compton wavelength shift
In Compton scattering, the wavelength of an incoming X-ray photon increases after it collides with a stationary electron. This phenomenon is known as the Compton wavelength shift. The shift (\( \Delta \lambda \)) depends on the angle (\theta\) at which the photon is scattered. The relationship is given by the formula: \[ \Delta \lambda = \frac{h}{m_ec}(1 - cos\( \theta \)) \] where:
  • \( h \) is Planck's constant.
  • \( m_e \) is the electron rest mass.
  • \( c \) is the speed of light.
  • \( \theta \) is the scattering angle.
For this problem, the incident photon scatter at \(90掳\). Thus, \( cos(90掳) = 0 \). So, the shift solely depends on the constants \( h \), \( m_e \), and \( c \. You add \( \Delta \lambda \) to the initial wavelength to find the new wavelength of the scattered photon.
Photon momentum
Photons, though they have no mass, carry momentum. The momentum (\( p \)) of a photon is inversely proportional to its wavelength (\( \lambda \)). This can be expressed as:
\[ p = \frac{h}{\lambda} \]
  • Here, \( h \) is Planck's constant.
  • \( \lambda \) is the photon's wavelength.
By inserting the wavelength of the incident photon into this formula, we get its momentum. The same formula applies to the scattered photon; you just use the new wavelength obtained from the Compton wavelength shift.
Energy conservation
Energy conservation is a vital concept in physics. In Compton scattering, the total energy before and after interaction remains the same. Specifically, the initial energy (\( E_i \)) of the photon is partially transferred to the scattered photon (\( E_s \)) and partially to the electron as kinetic energy (\( KE_e \)). The relation can be written as: \[ KE_e = E_i - E_s \] where:
  • \( E_i = \frac{hc}{\lambda} \) is the energy of the incoming photon.
  • \( E_s = \frac{hc}{\boldsymbol{\text{new}}\boldsymbol{\text{wavelength}} \}} \) is the energy of the scattered photon.
We plug in the respective wavelengths to compute \( E_i \) and \( E_s \), then subtract to find the kinetic energy imparted to the electron.
Scattered electron kinetic energy
The electron ejected during Compton scattering gains kinetic energy from the photon. As we discussed, the kinetic energy (\( KE_e \)) is calculated using the energy conservation principle: \[ KE_e = E_i - E_s \] Remember,
  • \( E_i \) and \( E_s \) are computed as \( \frac{hc}{\boldsymbol{\text{wavelength}}} \), where \( h \) is Planck's constant and \( c \) is the speed of light.
This kinetic energy is the energy that the photon loses, which instead is gained by the electron, causing it to recoil.
Momentum conservation
Momentum conservation is a core principle in Compton scattering. Initially, the system's momentum consists only of the photon's. After scattering, both the photon and electron carry momentum, and the total system momentum remains conserved. For momentum magnitude:
\[ p_e = \sqrt{ p_i^2 + p_s^2 - 2 p_i p_s \text{cos} ( \theta ) } \] where:
  • \( p_i \) is the initial photon momentum.
  • \( p_s \) is the scattered photon momentum.
  • \( \theta \) is the scattering angle.
At \( 90掳 \), this equation simplifies because \( cos(90掳) = 0 \). Determining the direction involves breaking down the momenta into components and using vector addition principles.

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Most popular questions from this chapter

Before a positron and an electron annihilate, they form a sort of "atom" in which each orbits about their common center of mass with identical speeds. As a result of this motion, the photons emitted in the annihilation show a small Doppler shift. In one experiment, the Doppler shift in energy of the photons was observed to be \(2.41 \mathrm{keV}\). (a) What would be the speed of the electron or positron before the annihilation to produce this Doppler shift? (b) The positrons form these atom-like structures with the nearly "free" electrons in a solid. Assuming the positron and electron must have about the same speed to form this structure, find the kinetic energy of the electron. This technique, called "Doppler broadening," is an important method for learning about the energies of electrons in materials.

You have been hired by NASA to analyze a solar sail that uses the momentum of sunlight for interplanetary travel. A prototype sail of area \(1 \mathrm{km}^{2}\) has been developed and is made from a thin lightweight polymer that has a highly reflective aluminum coating on one side. This material has thickness \(2 \mu \mathrm{m}\) and density \(0.29 \mathrm{g} / \mathrm{cm}^{2}\). You are limited by the design which requires that the supporting frame and the cargo weigh no more than the film itself. Make any necessary assumptions about the parameters of the calculation, and estimate the travel time of this spacecraft from Earth to Mars.

(a) Assuming the human body (skin temperature \(34^{\circ} \mathrm{C}\) ) to behave like an ideal thermal radiator, find the wavelength where the intensity from the body is a maximum. In what region of the electromagnetic spectrum is radiation with this wavelength? (b) Making whatever (reasonable) assumptions you may need, estimate the power radiated by a typical person isolated from the surroundings. (c) Estimate the radiation power absorbed by a person in a room in which the temperature is \(20^{\circ} \mathrm{C}\).

A hydrogen atom is moving at a speed of \(125.0 \mathrm{m} / \mathrm{s}\). It absorbs a photon of wavelength \(97 \mathrm{nm}\) that is moving in the opposite direction. By how much does the speed of the atom change as a result of absorbing the photon?

A photon of energy \(E\) interacts with an electron at rest and undergoes pair production, producing a positive electron (positron) and an electron (in addition to the original electron): $$ \text { photon }+\mathrm{e}^{-} \rightarrow \mathrm{e}^{+}+\mathrm{e}^{-}+\mathrm{e}^{-} $$ The two electrons and the positron move off with identical momenta in the direction of the initial photon. Find the kinetic energy of the three final particles and find theenergy \(E\) of the photon. (Hint: Conserve momentum and total relativistic energy.)
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