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A photon has momentum of magnitude 8.24 \(\times\) 10\(^{-28}\) kg \(\bullet\) m/s. (a) What is the energy of this photon? Give your answer in joules and in electron volts. (b) What is the wavelength of this photon? In what region of the electromagnetic spectrum does it lie?

Short Answer

Expert verified
(a) 2.472 × 10⁻¹⁹ J, 1.543 eV. (b) Wavelength: 804 nm, Infrared region.

Step by step solution

01

Recall the Energy-Momentum Relationship

The energy of a photon can be determined using its momentum. We use the equation \( E = pc \), where \( E \) is the energy, \( p \) is the momentum, and \( c \) is the speed of light (approximately \( 3 \times 10^8 \) m/s).
02

Calculate the Energy in Joules

Substitute \( p = 8.24 \times 10^{-28} \) kg·m/s and \( c = 3 \times 10^8 \) m/s into the equation:\[E = (8.24 \times 10^{-28}) \times (3 \times 10^8)\]This gives:\[E = 2.472 \times 10^{-19} \text{ J}\]
03

Convert Energy from Joules to Electron Volts

Use the conversion factor where 1 eV = \( 1.602 \times 10^{-19} \) J. Thus, the energy in eV is:\[E = \frac{2.472 \times 10^{-19} \text{ J}}{1.602 \times 10^{-19} \text{ J/eV}}\]This gives:\[E = 1.543 \, eV\]
04

Use the De Broglie Wavelength Formula

The wavelength \( \lambda \) of a photon is determined by the relationship \( \lambda = \frac{h}{p} \), where \( h \) is Planck's constant \( 6.626 \times 10^{-34} \text{ J·s}\).Substitute the values into the equation:\[\lambda = \frac{6.626 \times 10^{-34}}{8.24 \times 10^{-28}}\]This gives:\[\lambda = 8.04 \times 10^{-7} \, \text{m}\]
05

Determine the Region of the Electromagnetic Spectrum

The calculated wavelength \(8.04 \times 10^{-7} \, \text{m}\) corresponds to \(804 \, \text{nm}\), which falls within the infrared region of the electromagnetic spectrum (700 nm to 1 mm).

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Key Concepts

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

Momentum-Energy Relationship
The connection between momentum and energy is pivotal when examining photons. Photons, even though they are massless, possess momentum. This stems from their intrinsic energy as described by the equation \( E = pc \). Here, \( E \) stands for energy, \( p \) is the momentum, and \( c \), the speed of light, is approximately \( 3 \times 10^8 \) meters per second.
  • This equation shows that a photon's energy is directly proportional to its momentum.
  • Understanding this relationship helps in calculating the energy when the momentum is known and vice versa.
In this exercise, the momentum of the photon was given as \( 8.24 \times 10^{-28} \) kg·m/s. By using the equation \( E = pc \), we can determine the energy of the photon in both joules and electron volts. This transformation helps in understanding the photon's interaction with other particles or fields.
Moreover, this relation underlines a key aspect of quantum mechanics, where particles like photons defy some classical intuitions about mass and momentum because they are energy-packets traveling through space.
Electromagnetic Spectrum
The electromagnetic spectrum encompasses all types of electromagnetic radiation, sorted by wavelength or frequency. Radiation ranges from gamma rays with the shortest wavelengths to radio waves, which have the longest.
  • Visible light is just one small part of this spectrum, typically from 380 nm to 750 nm.
  • Beyond visible light, the spectrum includes ultraviolet, infrared, microwaves, and radio waves.
Both wavelength and frequency are critical for identifying the type of radiation. The energy and properties of radiation depend on its position in the spectrum.
For instance, the calculated wavelength of this photon's electromagnetic radiation was \( 804 \text{ nm} \), placing it in the infrared region. The infrared region typically involves wavelengths from 700 nm to 1 mm and is mostly associated with heat radiation. These infrared waves are longer than visible light, meaning they have lower energy, which is paramount in applications such as thermal imaging and remote controls.
De Broglie Wavelength
The concept of the De Broglie wavelength bridges the gap between classical and quantum physics, proposing that particles exhibit wave-like properties. Louis de Broglie suggested that every particle, including photons, has an associated wavelength \( \lambda \), calculated by \( \lambda = \frac{h}{p} \). Here, \( h \) is Planck's constant \( 6.626 \times 10^{-34} \text{ J·s} \) and \( p \) is the momentum.
  • For photons, this concept emphasizes their dual nature as both particles and waves.
  • Calculating the wavelength provides insight into how these particles interact with matter.
In practical applications, the De Broglie wavelength helps explain patterns in phenomena like diffraction and interference, where even particles traditionally seen as solid show wave-like behavior.
For the given photon momentum of \( 8.24 \times 10^{-28} \) kg·m/s, the De Broglie wavelength calculation resulted in \( 8.04 \times 10^{-7} \text{ m} \). Knowing the wavelength allows us to discern which region of the electromagnetic spectrum the photon belongs to, thus giving insight into its energy and potential uses in technology or natural sciences.

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Most popular questions from this chapter

X rays with an initial wavelength of 0.900 \(\times\) 10\(^{-10}\) m undergo Compton scattering. For what scattering angle is the wavelength of the scattered x rays greater by 1.0\(\%\) than that of the incident x rays?

A laser used to weld detached retinas emits light with a wavelength of 652 nm in pulses that are 20.0 ms in duration. The average power during each pulse is 0.600 W. (a) How much energy is in each pulse in joules? In electron volts? (b) What is the energy of one photon in joules? In electron volts? (c) How many photons are in each pulse?

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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?

An x-ray photon is scattered from a free electron (mass \(m\)) at rest. The wavelength of the scattered photon is \(\lambda'\), and the final speed of the struck electron is \(v\). (a) What was the initial wavelength \(\lambda\) of the photon? Express your answer in terms of \(\lambda\), \(v\), and \(m\). (\(Hint\): Use the relativistic expression for the electron kinetic energy.) (b) Through what angle \(\phi\) is the photon scattered? Express your answer in terms of \(\lambda\), \(\lambda'\), and \(m\). (c) Evaluate your results in parts (a) and (b) for a wavelength of 5.10 \(\times\) 10\(^{-3}\) nm for the scattered photon and a final electron speed of 1.80 \(\times\) 10\(^8\) m/s. Give \(\phi\) in degrees.

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