Problem 6 - University Physics with Modern Physics

Question

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?

Step-by-Step Solution

Answer

The photon's energy is \(2.472 \times 10^{-19} \; \text{J}\) or \(1.54 \; \text{eV}\). Its wavelength is 804 nm, placing it in the infrared spectrum.

Step 1: Understanding Photon Momentum and Energy Relationship

Photons have no mass, but they have energy and momentum related by the equation \( E = pc \), where \( p \) is momentum and \( c \) is the speed of light (\(3 \times 10^8 \; \text{m/s}\)). We can use this to find the energy.

Step 2: Calculate Energy in Joules

Use the formula \( E = pc \) to calculate the energy in joules: \[ E = (8.24 \times 10^{-28} \; \text{kg} \cdot \text{m/s}) \times (3 \times 10^8 \; \text{m/s}) = 2.472 \times 10^{-19} \; \text{J} \].

Step 3: Convert Energy to Electron Volts

Use the conversion factor \( 1 \; \text{eV} = 1.602 \times 10^{-19} \; \text{J} \) to convert joules to electron volts: \[ E = \frac{2.472 \times 10^{-19} \; \text{J}}{1.602 \times 10^{-19} \; \text{J/eV}} \approx 1.54 \; \text{eV} \].

Step 4: Calculate Wavelength of the Photon

We use the relation \( \lambda = \frac{h}{p} \), where \( h \) is Planck's constant \( (6.626 \times 10^{-34} \; \text{m}^2 \cdot \text{kg/s}) \). Calculate the wavelength: \[ \lambda = \frac{6.626 \times 10^{-34} \; \text{m}^2 \cdot \text{kg/s}}{8.24 \times 10^{-28} \; \text{kg} \cdot \text{m/s}} = 8.04 \times 10^{-7} \; \text{m} \].

Step 5: Determine Region in the Electromagnetic Spectrum

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

Key Concepts

Photon Momentum

Photons are unique because despite having no mass, they still possess momentum. This might seem puzzling at first, but it's a fundamental concept in physics. The momentum of a photon is related to its energy and the speed of light, using the formula \( p = \frac{E}{c} \), where \( p \) is the momentum, \( E \) is energy, and \( c \) is the speed of light. In cases where the momentum is given, like in our exercise, you can easily calculate the energy using the formula \( E = pc \). Here, the challenge is to grasp that momentum for massless particles isn't tied to mass as it is for traditional particles like baseballs or cars. Instead, it's related to how the photon moves through space and the energy it carries.

Electromagnetic Spectrum

The electromagnetic spectrum is like a map that categorizes all forms of light by wavelength and energy. It ranges from very short wavelengths (like gamma rays) to very long wavelengths (like radio waves). In the middle, we find visible light, the tiny segment of the spectrum that human eyes can see. Different wavelengths have different energies and properties. For example:

• Gamma rays have the shortest wavelengths and the highest energies.
• Infrared light, like the one described in our example, is below visible red light and is often experienced as heat.
• Radio waves have the longest wavelengths.

Understanding where a photon lies on this spectrum helps in anticipating its behavior and potential applications, from medical imaging with X-rays to communication technologies using microwaves.

Wavelength Calculation

Calculating the wavelength of a photon requires using Planck's constant and the photon's momentum. This process uses the formula \( \lambda = \frac{h}{p} \), where \( \lambda \) is the wavelength, \( h \) is Planck's constant \( (6.626 \times 10^{-34} \; \text{m}^2 \cdot \text{kg/s}) \), and \( p \) is the momentum. When you plug in these values, you'll get the wavelength, which tells you more about how this photon will act and its place on the electromagnetic spectrum. Shorter wavelengths mean more energy and thus has different applications and effects than longer wavelengths. In our example, performing this calculation leads to the conclusion that the photon's wavelength is 804 nm, placing it firmly in the infrared range.

Photon Wavelength

Photon wavelength is a fundamental property that reveals a lot about a photon's energy and the type of electromagnetic radiation it is part of. Understanding photon wavelength is crucial because:

• It determines the photon's position in the electromagnetic spectrum.
• Shorter wavelengths (like ultraviolet) indicate more energetic photons.
• Longer wavelengths (like infrared) are less energetic.

The 804 nanometer wavelength calculated from our photon falls within the infrared region, which is not visible to the human eye but useful for various technologies, from remote controls to thermal imaging. Recognizing the relationship between wavelength and frequency is crucial because it helps in comprehending how photons interact with matter and can be manipulated or used in practical applications.