Problem 55 - Chemistry Matter and Change

Question

Radio Waves If your favorite FM radio station broadcasts at a frequency of 104.5 \(\mathrm{MHz}\) , what is the wavelength of the station's signal in meters? What is the energy of a photon of the station's electromagnetic signal?

Step-by-Step Solution

Answer

The wavelength is 2.87 meters, and the photon energy is approximately \(6.92 \times 10^{-26} \) Joules.

Step 1: Understand the Relationship between Frequency and Wavelength

The speed of light (c) in a vacuum is a constant approximately equal to \(3.00 \times 10^8\) meters per second (m/s). The relationship between the speed of light, frequency \( (f) \), and wavelength \( (\lambda) \) is given by the equation \(c = f\lambda\). Given the frequency of 104.5 MHz, first convert it to Hz: \(104.5 \, \text{MHz} = 104.5 \times 10^6 \, \text{Hz}\).

Step 2: Calculate the Wavelength

Using the formula \(\lambda = \frac{c}{f}\), substitute the values: \(c = 3.00 \times 10^8 \,\text{m/s}\) and \(f = 104.5 \times 10^6 \, \text{Hz}\). Perform the calculation: \[\lambda = \frac{3.00 \times 10^8}{104.5 \times 10^6} = 2.87 \, \text{meters}\]. The wavelength of the station's signal is 2.87 meters.

Step 3: Understand the Relationship between Frequency and Photon Energy

The energy of a photon is directly related to its frequency by the formula \(E = hf\), where \(h\) is Planck's constant \(6.626 \times 10^{-34} \, \text{Joule seconds}\) and \(f\) is the frequency.

Step 4: Calculate the Photon Energy

Using the frequency in Hz from Step 1 and Planck's constant, substitute into the energy formula: \(E = (6.626 \times 10^{-34} \, \text{J s}) \times (104.5 \times 10^6 \, \text{Hz})\). This results in \(E = 6.92 \times 10^{-26} \, \text{Joules}\). The energy of a photon of the station's signal is approximately \(6.92 \times 10^{-26} \) Joules.

Key Concepts

Speed of Light

In the realm of electromagnetic waves, the speed of light stands as a fundamental constant. Its value is approximately \(3.00 \times 10^8\) meters per second (m/s) when traveling through a vacuum. This constancy enables waves such as radio, microwaves, and visible light to traverse space at incredible speeds.

Understanding that light speed is a constant helps in relating other wave properties, like frequency and wavelength. By knowing any two of these properties, the third can always be determined. For example, when we know the frequency of a radio wave, we can determine its wavelength using the formula \(c = f\lambda\). In this equation, \(c\) is the speed of light, \(f\) is the frequency, and \(\lambda\) is the wavelength.

Electromagnetic waves encompass much more than just visible light. Radio waves also travel at this speed, making it essential in tasks like broadcasting where signals need to travel across long distances quickly and consistently. Remembering that this constant applies is a key first step in solving many wave equations.

Wavelength Calculation

Wavelength calculation is a vital aspect of understanding electromagnetic phenomena, especially when dealing with radio waves. The primary formula connecting wavelength \((\lambda)\) to speed of light \((c)\) and frequency \((f)\) is \(\lambda = \frac{c}{f}\).

To start a calculation, you'll need to know the frequency of the wave. Radio stations often broadcast in megahertz, like 104.5 MHz. To use this in calculations, convert it to hertz, giving \(104.5 \times 10^6\) Hz.

Then, plug this value into the formula to find the wavelength:

• \(c = 3.00 \times 10^8\) m/s
• \(f = 104.5 \times 10^6\) Hz
• Using \(\lambda = \frac{c}{f}\), we calculate:

The wavelength is thus \(2.87\) meters.

This calculation is integral not only to understanding the size of the waves but also to their applications, such as tuning your radio to the correct frequency to hear your chosen station.

Frequency to Energy Conversion

Energy conversion from frequency is a fascinating concept that ties into the quantum nature of electromagnetic radiation. Every photon, or particle of light, carries energy that can be calculated using its frequency.

This relationship is expressed with the formula \(E = hf\), where:

• \(E\) is the energy in joules
• \(h\) is Planck's constant, \(6.626 \times 10^{-34}\) joule seconds
• \(f\) is the frequency in hertz

Given the frequency of a radio station, the energy of their signal can then be determined. Continuing with our example of a 104.5 MHz wave, first convert the frequency to hertz, giving \(104.5 \times 10^6\) Hz.

Incorporate that into the formula:

• \(E = (6.626 \times 10^{-34} \text{ J s}) \times (104.5 \times 10^6 \text{ Hz})\)

This results in an energy value of \(6.92 \times 10^{-26}\) joules.

Understanding this conversion is crucial for fields like quantum mechanics and electronics, where the energies being discussed, while very small, have substantial implications in how materials and signals are used and understood.