Problem 14 - University Physics with Modern Physics

Question

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?

Step-by-Step Solution

Answer

Energy is \(2.472 \times 10^{-19} \text{ J}\) or \(1.543 \text{ eV}\); wavelength is 804 nm (infrared spectrum).

Step 1: 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:\[E = 8.24 \times 10^{-28} \times 3 \times 10^8 = 2.472 \times 10^{-19} \text{ J}\]

Step 2: 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:\[E_{ev} = \frac{2.472 \times 10^{-19}}{1.602 \times 10^{-19}} \approx 1.543 \text{ eV}.\]

Step 3: 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:\[\lambda = \frac{6.626 \times 10^{-34}}{8.24 \times 10^{-28}} \approx 8.04 \times 10^{-7} \text{ m}.\]

Step 4: 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.

Key Concepts

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’s 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.