/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 23 An ultrashort pulse has a durati... [FREE SOLUTION] | 91影视

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Chapter 38: Problem 23

An ultrashort pulse has a duration of 9.00 fs and produces light at a wavelength of 556 nm. What are the momentum and momentum uncertainty of a single photon in the pulse?

Short Answer

Expert verified
Momentum: \( 1.193 \times 10^{-27} \text{ kg m/s} \), Uncertainty: \( 1.95 \times 10^{-29} \text{ kg m/s} \).

Step by step solution

01

Calculate the Frequency of the Light

To find the frequency, we use the formula: \( c = \lambda \cdot f \), where \( c \) is the speed of light \( (3 \times 10^8 \text{ m/s}) \), \( \lambda \) is the wavelength in meters, and \( f \) is the frequency. First, convert the wavelength from nanometers to meters: \( 556 \text{ nm} = 556 \times 10^{-9} \text{ m} \). Rearrange the formula to solve for frequency: \( f = \frac{c}{\lambda} = \frac{3 \times 10^8}{556 \times 10^{-9}} \). This calculation gives \( f \approx 5.4 \times 10^{14} \text{ Hz} \).
02

Calculate the Energy of the Photon

The energy of a photon is given by \( E = h \cdot f \), where \( h \) is Planck's constant \( (6.626 \times 10^{-34} \text{ J s}) \) and \( f \) is the frequency found in Step 1. Substitute the values: \( E = 6.626 \times 10^{-34} \cdot 5.4 \times 10^{14} \approx 3.579 \times 10^{-19} \text{ J} \).
03

Calculate the Momentum of the Photon

The momentum of a photon is given by \( p = \frac{E}{c} \). Using the energy from Step 2 and the constant speed of light, \( p = \frac{3.579 \times 10^{-19}}{3 \times 10^8} \approx 1.193 \times 10^{-27} \text{ kg m/s} \).
04

Calculate the Uncertainty in Energy Using Time-energy Uncertainty Principle

Use the time-energy uncertainty principle \( \Delta E \cdot \Delta t \geq \frac{h}{4\pi} \) to find \( \Delta E \). With \( \Delta t = 9 \times 10^{-15} \text{ s} \), rearrange to find \( \Delta E \geq \frac{6.626 \times 10^{-34}}{4\pi \cdot 9 \times 10^{-15}} \approx 5.86 \times 10^{-21} \text{ J} \).
05

Calculate the Momentum Uncertainty of the Photon

The uncertainty in momentum \( \Delta p \) can be calculated using \( \Delta p = \frac{\Delta E}{c} \). Substitute the \( \Delta E \) from Step 4: \( \Delta p \geq \frac{5.86 \times 10^{-21}}{3 \times 10^8} \approx 1.95 \times 10^{-29} \text{ kg m/s} \).

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Key Concepts

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

Energy of a Photon
The energy of a photon is a fundamental concept in physics and is central to understanding light and electromagnetic radiation. The energy of a photon is determined by its frequency, which can be calculated using Planck's equation: \( E = h \cdot f \),where:
  • \( E \) is the energy of the photon.
  • \( h \) is Planck鈥檚 constant \((6.626 \times 10^{-34} \text{ J s})\).
  • \( f \) is the frequency of the photon.
More energetic photons have higher frequencies. Frequency and energy are directly proportional, meaning as one increases, so does the other. This is crucial in fields like quantum mechanics and optics, where understanding photon behavior is essential.
Wavelength Conversion
Wavelength is often measured in nanometers, especially in optics and laser physics. To work with it in scientific equations, you must convert it to meters because the speed of light, \( c \), uses meters per second. Conversion is straightforward:\( 1 \text{ nm} = 10^{-9} \text{ meters} \).For our photon of wavelength 556 nm:\( 556 \text{ nm} = 556 \times 10^{-9} \text{ m} \).This conversion helps to accurately calculate frequency and energy. Converting wavelengths is a fundamental skill necessary for understanding relationships between different physical quantities of light.
Time-Energy Uncertainty Principle
In quantum mechanics, the time-energy uncertainty principle is a fundamental relation with profound implications. It is expressed as:\( \Delta E \cdot \Delta t \geq \frac{h}{4\pi} \),where:
  • \( \Delta E \) is the uncertainty in energy.
  • \( \Delta t \) is the uncertainty in time 鈥� here, the pulse duration.
  • \( h \) is Planck鈥檚 constant.
This principle means that the longer a pulse lasts, the less certain we are about its energy, and vice versa. For an ultrashort pulse, like in our problem, we find a significant energy uncertainty, leading to the analysis short, high-energy interactions in quantum phenomena.
Photon Momentum Uncertainty
Photon momentum is linked to its energy by the relationship:\( p = \frac{E}{c} \). Uncertainties in photon momentum can be derived using the time-energy uncertainty principle through:\( \Delta p = \frac{\Delta E}{c} \),where:
  • \( \Delta p \) is the momentum uncertainty.
  • \( \Delta E \) is the energy uncertainty found using the time-energy uncertainty principle.
This gives insights into the precision of photon-based measurements. Knowing uncertainty aids in better designs for experiments and helps understand the limitations of measurements at quantum scales.
Frequency Calculation
Calculating the frequency of light involves the relation between frequency, speed, and wavelength, expressed as:\( c = \lambda \cdot f \),where:
  • \( c \) is the speed of light \((3 \times 10^8 \text{ m/s})\).
  • \( \lambda \) is the wavelength.
  • \( f \) is the frequency.
To find the frequency (\( f \)), rearrange the formula:\( f = \frac{c}{\lambda} \).For a light with a wavelength of 556 nm:\( f = \frac{3 \times 10^8}{556 \times 10^{-9}} \approx 5.4 \times 10^{14} \text{ Hz} \).Deriving this frequency is crucial in applications like telecommunications and spectroscopy, where precise control and understanding of light properties are essential.

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Most popular questions from this chapter

An x-ray tube is operating at voltage \(V\) and current \(I\). (a) If only a fraction \(p\) of the electric power supplied is converted into x rays, at what rate is energy being delivered to the target? (b) If the target has mass \(m\) and specific heat \(c\) (in J/kg \(\bullet\) K), at what average rate would its temperature rise if there were no thermal losses? (c) Evaluate your results from parts (a) and (b) for an x-ray tube operating at 18.0 kV and 60.0 mA that converts 1.0\(\%\) of the electric power into x rays. Assume that the 0.250-kg target is made of lead (\(c\) = 130 J/kg \(\bullet\) K). (d) What must the physical properties of a practical target material be? What would be some suitable target elements?

A photon of wavelength 4.50 pm scatters from a free electron that is initially at rest. (a) For \(\phi = 90.0^\circ\), what is the kinetic energy of the electron immediately after the collision with the photon? What is the ratio of this kinetic energy to the rest energy of the electron? (b) What is the speed of the electron immediately after the collision? (c) What is the magnitude of the momentum of the electron immediately after the collision? What is the ratio of this momentum value to the nonrelativistic expression \(mv\)?

The photoelectric threshold wavelength of a tungsten surface is 272 nm. Calculate the maximum kinetic energy of the electrons ejected from this tungsten surface by ultraviolet radiation of frequency 1.45 \(\times\) 10\(^{15}\) Hz. Express the answer in electron volts.

A 2.50-W beam of light of wavelength 124 nm falls on a metal surface. You observe that the maximum kinetic energy of the ejected electrons is 4.16 eV. Assume that each photon in the beam ejects a photoelectron. (a) What is the work function (in electron volts) of this metal? (b) How many photoelectrons are ejected each second from this metal? (c) If the power of the light beam, but not its wavelength, were reduced by half, what would be the answer to part (b)? (d) If the wavelength of the beam, but not its power, were reduced by half, what would be the answer to part (b)?

(a) What is the minimum potential difference between the filament and the target of an x-ray tube if the tube is to produce x rays with a wavelength of 0.150 nm? (b) What is the shortest wavelength produced in an x-ray tube operated at 30.0 kV?
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