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Chapter 38: Problem 14

A photon has momentum of magnitude \(8.24 \times 10^{-28} \mathrm{kg}\) . \(\mathrm{m} / \mathrm{s}\) (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
Energy is \(2.472 \times 10^{-19} \text{ J}\) or \(1.543 \text{ eV}\); wavelength is 804 nm (infrared spectrum).

Step by step solution

01

Calculate Energy in Joules

To find the energy of a photon in joules, you use the equation \( E = pc \), where \( p \) is the momentum and \( c \) is the speed of light (approximately \( 3 \times 10^8 \text{ m/s} \)). Plug in the momentum \( p = 8.24 \times 10^{-28} \text{ kg} \cdot \text{m/s} \) and the value of \( c \) to calculate:\[E = 8.24 \times 10^{-28} \times 3 \times 10^8 = 2.472 \times 10^{-19} \text{ J}\]
02

Convert Energy to Electron Volts

To convert joules to electron volts, use the conversion factor \( 1 ext{ eV} = 1.602 \times 10^{-19} ext{ J} \). Divide the energy in joules by this factor:\[E_{ev} = \frac{2.472 \times 10^{-19}}{1.602 \times 10^{-19}} \approx 1.543 \text{ eV}.\]
03

Calculate Wavelength

Use the equation \( \lambda = \frac{h}{p} \), where \( h = 6.626 \times 10^{-34} \text{ J} \cdot \text{s} \) is Planck's constant and \( p \) is the momentum. Substitute the known values:\[\lambda = \frac{6.626 \times 10^{-34}}{8.24 \times 10^{-28}} \approx 8.04 \times 10^{-7} \text{ m}.\]
04

Determine Electromagnetic Spectrum Region

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

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Key Concepts

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

Momentum and Energy Relationship
Photons, the fundamental particles of light, have a unique relationship between momentum and energy. In classical mechanics, particles with mass can have momentum and energy, but photons defy this norm because they are massless on their own. Yet, they possess both momentum and energy. This connection is represented by the equation \( E = pc \), where \( E \) is energy, \( p \) is momentum, and \( c \) is the speed of light.
  • Photons travel at the speed of light, \( c = 3 \times 10^8 \text{ m/s} \), which is crucial in this equation.
  • Since photons are massless, their momentum is not defined by mass and velocity as in classical mechanics, but rather by their energy content.
By calculating the photon energy using its momentum and the speed of light, we can express energy in joules. For convenience, energy can also be converted to electron volts (eV), a unit often used in atomic and quantum physics.
Electromagnetic Spectrum
The electromagnetic spectrum is a broad range of all types of electromagnetic radiation, from radio waves to gamma rays. Photons of different energies correspond to different types of radiation along this spectrum.
  • The spectrum covers wavelengths from kilometers (in the case of radio waves) to less than a nanometer (like gamma rays).
  • Energy increases across the spectrum from radio waves to gamma rays, while wavelength decreases.
When a photon's wavelength is calculated, it helps determine its position within this spectrum. For instance, a wavelength of 804 nm falls in the infrared region, known for its heat-related properties. Each region on the spectrum possesses unique characteristics and applications across various scientific and technological fields.
Wavelength Calculation
The wavelength of a photon is a critical parameter that relates to its energy and position within the electromagnetic spectrum. Wavelength is calculated using the equation \( \lambda = \frac{h}{p} \), where \( \lambda \) is the wavelength, \( h \) is Planck's constant, and \( p \) is the photon鈥檚 momentum.
  • Planck's constant, \( h = 6.626 \times 10^{-34} \text{ J} \cdot \text{s} \), is a crucial constant in quantum mechanics.
  • The calculated wavelength helps identify the type of electromagnetic wave formed by the photon.
Knowing the wavelength allows scientists and engineers to understand the photon's energy and predict its behavior across media and systems. Accurate wavelength calculations are pivotal in fields ranging from telecommunications to astronomy, as they affect everything from data transmission to the visibility of celestial objects.

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Most popular questions from this chapter

A hydrogen atom is in a state with energy \(-1.51\) eV. In the Bohr model, what is the angular momentum of the electron in the atom, with respect to an axis at the nucleus?

X rays with initial wavelength 0.0665 nm undergo Compton scattering. What is the longest wavelength found in the scattered x rays? At which scattering angle is this wavelength observed?

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 \(\mathbf{x}\) . rays, at what rate is energy being delivered to the target? (b) If the target has mass \(m\) and specific heat capacity \(c(\text { in } \mathrm{J} / \mathrm{kg} \cdot \mathrm{K}),\) 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 \(\mathrm{kV}\) and 60.0 \(\mathrm{mA}\) that converts 1.0\(\%\) of the electric power into \(\mathrm{x}\) rays. Assume that the \(0.250-\mathrm{kg}\) target is made of lead \((c=130 \mathrm{J} / \mathrm{kg} \cdot \mathrm{K}) .\) (d) What must the physical properties of a practical target material be? What would be some suitable target elements?

The photoelectric threshold wavelength of a tungsten surface is 272 \(\mathrm{nm}\) . 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.

An excited nucleus emits a gamma-ray photon with an energy of 2.45 \(\mathrm{MeV}\) . (a) What is the photon frequency?(b) What is the photon wavelength? (c) How does the wavelength compare with a typical nuclear diameter of \(10^{-14} \mathrm{m} ?\)
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