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A photon has momentum of magnitude \(8.24 \times 10^{-28} \mathrm{kg} \cdot \mathrm{m} / \mathrm{s}\) . (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) Energy: \(2.472 \times 10^{-19} \text{ J}\) or \(1.543 \text{ eV}\). (b) Wavelength: \(8.04 \times 10^{-7} \text{ m}\), in the infrared region.

Step by step solution

01

Understand the Photon Energy-Momentum Relationship

The energy of a photon can be calculated using its momentum. The relationship between energy (E) and momentum (p) for a photon is given by the equation: \[ E = pc \]where \( c \) is the speed of light, approximately \( 3.00 \times 10^8 \text{ m/s} \). We will use this formula to find the energy in joules.
02

Calculate the Energy in Joules

Substitute the given momentum and the speed of light into the formula:\[E = p \cdot c = (8.24 \times 10^{-28} \, \text{kg} \cdot \text{m/s}) \times (3.00 \times 10^8 \, \text{m/s})\]Calculate the product to find the energy in joules:\[E = 2.472 \times 10^{-19} \, \text{J}\]
03

Convert Energy from Joules to Electron Volts

To convert the energy from joules to electron volts, use the conversion factor:\[1 \, \text{eV} = 1.602 \times 10^{-19} \, \text{J}\]Divide the energy in joules by the conversion factor:\[E = \frac{2.472 \times 10^{-19} \, \text{J}}{1.602 \times 10^{-19} \, \text{J/eV}} \]Calculate this to find:\[E \approx 1.543 \, \text{eV}\]
04

Use Photon Wavelength-Momentum Relationship

The wavelength (\( \lambda \)) of a photon can be found using its momentum with the formula:\[ \lambda = \frac{h}{p} \]where \( h \) is the Planck constant, approximately \( 6.626 \times 10^{-34} \, \text{J} \cdot \text{s} \). Substitute the values to find the wavelength.
05

Calculate the Wavelength

Substitute the Planck constant and the given momentum into the formula:\[\lambda = \frac{6.626 \times 10^{-34} \, \text{J} \cdot \text{s}}{8.24 \times 10^{-28} \, \text{kg} \cdot \text{m/s}}\]Calculate the division to find the wavelength:\[\lambda \approx 8.04 \times 10^{-7} \, \text{m}\]
06

Identify the Electromagnetic Spectrum Region

The calculated wavelength of \( 8.04 \times 10^{-7} \, \text{m} \) is equivalent to \( 804 \, \text{nm} \). This wavelength falls within the visible light spectrum, specifically in the near-infrared region. Therefore, the photon lies in the infrared part of the electromagnetic spectrum.

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Key Concepts

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

Photon Momentum
Photons, the fundamental particles of light, do not have mass in the traditional sense but they do possess momentum. Unlike classical objects, photon momentum is directly related to its energy and wavelength, which can be quite intriguing. The momentum (\( p \)) of a photon is given by:
\( p = \frac{E}{c} \)
where \( E \) is the energy of the photon and \( c \) denotes the speed of light. This equation reveals that a photon's momentum is a result of its energy and not mass, defying the usual relationship between mass and momentum seen in traditional physics.

The concept of photon momentum is crucial in understanding phenomena such as radiation pressure, where light exerts a force on an object upon striking it, and is foundational in technologies like solar sails used for space propulsion. Understanding photon momentum enriches our grasp of how light behaves both in physics and in practical applications.
Electromagnetic Spectrum
The electromagnetic spectrum encompasses all types of electromagnetic radiation, ranging from gamma rays with tiny wavelengths to radio waves with expansive ones. Each segment of the spectrum has its own unique properties and uses in science and technology.
  • Gamma Rays: Highest energy radiation, often originating from nuclear reactions.
  • X-rays: Utilized in medical imaging for internal body scans.
  • Ultraviolet (UV): Has higher energy than visible light, can cause skin tanning or burning.
  • Visible Light: The portion of the spectrum detectable to the human eye.
  • Infrared (IR): Employed in heat sensing and thermal imaging.
  • Microwaves: Used in cooking and certain communication technologies.
  • Radio Waves: Longest wavelengths, encompass AM, FM, and TV signals.
The electromagnetic spectrum is fundamental to many scientific fields, providing insights into the behavior of different kinds of light and their interactions with matter. Applications of these waves are varied, making them invaluable in everyday technology, from telecommunications to medical diagnostics.
Visible Light
Visible light is a small portion of the electromagnetic spectrum that is detectable by the human eye. It ranges from about 400 nm to 700 nm in wavelength, encompassing all the colors visible to the naked eye - violet, indigo, blue, green, yellow, orange, and red.
  • Violet: Shortest wavelength.
  • Red: Longest wavelength visible.
Colors we perceive are a result of light waves being absorbed or reflected by materials. For instance, a red apple reflects red wavelengths and absorbs others, causing us to see it as red.

Visible light plays a pivotal role in our daily lives, from helping us perceive our surroundings to enabling photosynthesis in plants. It's the basis of photography, vision technologies, and countless scientific studies. By understanding visible light, we appreciate the way our world appears and functions, deepening our connection with the universe around us.

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Most popular questions from this chapter

From the kinetic-molecular theory of an ideal gas (Chapter 15\()\) we know that the average kinetic energy of an atom is \(\frac{3}{2} k T\) . What is the wavelength of a photon that has this energy for a temperature of \(27^{\circ} \mathrm{C} ?\)

\(\cdot\) Removing birthmarks. Pulsed dye lasers emit light of wavelength 585 \(\mathrm{nm}\) in 0.45 \(\mathrm{ms}\) pulses to remove skin blemishes such as birthmarks. The beam is usually focused onto a circular spot 5.0 \(\mathrm{mm}\) in diameter. Suppose that the output of one such laser is 20.0 \(\mathrm{W}\) . (a) What is the energy of each photon, in eV? (b) How many photons per square millimeter are delivered to the blemish during each pulse?

\(\bullet\) Protons are accelerated from rest by a potential difference of 4.00 \(\mathrm{kV}\) and strike a metal target. If a proton produces one photon on impact, what is the minimum wavelength of the resulting \(\mathrm{x}\) rays? How does your answer compare to the minimum wavelength if 4.00 \(\mathrm{keV}\) electrons are used instead? Why do x-ray tubes use electrons rather than protons to produce \(x\) rays?

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. 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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