Problem 2
Question
What is the energy (in electron volts) of a \(100-\mathrm{GHz}\) \(\left(1 \text { gigahertz }=10^{9} \mathrm{Hz}\right)\) microwave photon?
Step-by-Step Solution
Verified Answer
The energy of a 100 GHz microwave photon is approximately \(4.14 \times 10^{-4}\) eV.
1Step 1: Understand the Formula for Photon Energy
To calculate the energy of a photon, we use the formula: \( E = h \cdot f \), where \( E \) is the energy of the photon, \( h \) is Planck's constant \( (6.626 \times 10^{-34} \text{ J s}) \), and \( f \) is the frequency of the photon in hertz.
2Step 2: Convert Frequency to Hertz
Given that the frequency of the microwave photon is \( 100 \text{ GHz} \), we first convert this to hertz:\[ 100 \text{ GHz} = 100 \times 10^9 \text{ Hz} = 1.0 \times 10^{11} \text{ Hz}. \]
3Step 3: Calculate Energy in Joules
Now, substitute the frequency \( f = 1.0 \times 10^{11} \text{ Hz} \) into the formula:\[ E = (6.626 \times 10^{-34} \,\text{J s}) \times (1.0 \times 10^{11} \,\text{Hz}) = 6.626 \times 10^{-23} \,\text{J}. \]
4Step 4: Convert Joules to Electron Volts
To convert the energy from joules to electron volts, use the conversion: \( 1 \text{ eV} = 1.602 \times 10^{-19} \text{ J} \).Thus, \[ E = \frac{6.626 \times 10^{-23} \text{ J}}{1.602 \times 10^{-19} \text{ J/eV}} \approx 4.14 \times 10^{-4} \text{ eV}. \]
5Step 5: Verify Units and Result
Finally, ensure the calculated energy is correctly converted to electron volts and reasonable for a microwave photon, which usually falls in this range.
Key Concepts
Planck's ConstantFrequency to Hertz ConversionEnergy Conversion to Electron VoltsMicrowave Photon Frequency
Planck's Constant
To calculate the energy of a photon, the concept of Planck's constant is essential. Planck's constant, represented as \( h \), is a fundamental constant in physics, with a value of \( 6.626 \times 10^{-34} \text{ J s} \). It connects the amount of energy a photon carries with its frequency. This constant is crucial in quantum mechanics, serving as a bridge between the wave and particle nature of light.
When using Planck's constant, we calculate the photon energy using the formula: \[ E = h \cdot f \]. Here, \( E \) is the energy of the photon, \( h \) is Planck's constant, and \( f \) is the frequency of the photon. This simple formula helps us explore the quantum behavior of light and other electromagnetic radiation.
When using Planck's constant, we calculate the photon energy using the formula: \[ E = h \cdot f \]. Here, \( E \) is the energy of the photon, \( h \) is Planck's constant, and \( f \) is the frequency of the photon. This simple formula helps us explore the quantum behavior of light and other electromagnetic radiation.
Frequency to Hertz Conversion
Frequency is a key parameter when calculating photon energy. It tells us how fast a wave oscillates and is typically measured in hertz (Hz). In some cases, such as with microwave frequencies, we might get the frequency in gigahertz (GHz). Converting this to hertz is often a necessary first step.
Given that \(1 \text{ GHz} = 10^9 \text{ Hz}\), we convert the frequency of a microwave photon from GHz to Hz by multiplying by \(10^9\). For instance, a frequency of \(100 \text{ GHz}\) can be converted as follows:
Given that \(1 \text{ GHz} = 10^9 \text{ Hz}\), we convert the frequency of a microwave photon from GHz to Hz by multiplying by \(10^9\). For instance, a frequency of \(100 \text{ GHz}\) can be converted as follows:
- \(100 \text{ GHz} = 100 \times 10^9 \text{ Hz} = 1.0 \times 10^{11} \text{ Hz}\).
Energy Conversion to Electron Volts
After calculating the energy in joules, it is often helpful to convert the result into electron volts (eV), especially when dealing with microscopic processes like photon interactions. The electron volt is a more convenient unit for such small scales.
To convert energy from joules to electron volts, we use the relationship between these units. One electron volt equals \(1.602 \times 10^{-19} \text{ J}\). Therefore, converting energy from joules (in this case, \(6.626 \times 10^{-23} \text{ J}\) for a microwave photon) to electron volts involves dividing by this conversion factor:
To convert energy from joules to electron volts, we use the relationship between these units. One electron volt equals \(1.602 \times 10^{-19} \text{ J}\). Therefore, converting energy from joules (in this case, \(6.626 \times 10^{-23} \text{ J}\) for a microwave photon) to electron volts involves dividing by this conversion factor:
- \[ E = \frac{6.626 \times 10^{-23} \text{ J}}{1.602 \times 10^{-19} \text{ J/eV}} \approx 4.14 \times 10^{-4} \text{ eV}. \]
Microwave Photon Frequency
Microwave photons have specific characteristics due to their frequencies, which typically range from 300 MHz to 300 GHz. These frequencies are much lower than visible light, placing microwaves within the electromagnetic spectrum below infrared.
When discussing a specific microwave frequency, such as \(100 \text{ GHz}\), it is important to comprehend how this affects the energy of the photon. Using the photon energy equation \( E = h \cdot f \), the calculated energy is relatively low when compared to visible light photons. The lower energy levels associated with microwave frequencies are why microwaves are used effectively for heating in microwave ovens and in various communication applications.
Understanding microwave frequencies, including converting and applying them in energy calculations, helps in both technological applications and theoretical physics.
When discussing a specific microwave frequency, such as \(100 \text{ GHz}\), it is important to comprehend how this affects the energy of the photon. Using the photon energy equation \( E = h \cdot f \), the calculated energy is relatively low when compared to visible light photons. The lower energy levels associated with microwave frequencies are why microwaves are used effectively for heating in microwave ovens and in various communication applications.
Understanding microwave frequencies, including converting and applying them in energy calculations, helps in both technological applications and theoretical physics.
Other exercises in this chapter
Problem 1
What is the energy (in electron volts-see More Precisely 4-1) of a 450 -nm blue photon? A 200 -nm ultraviolet photon?
View solution Problem 3
What are the wavelengths of (a) a 2 -eV red photon, (b) an \(0.1-\mathrm{eV}\) infrared photon, and \((\mathrm{c})\) a 5000 -eV ( 5 -keV ) X-ray?
View solution Problem 4
How many times more energy has a 1 -nm gamma ray than a 10 -MHz radio photon?
View solution Problem 5
List all the spectral lines of hydrogen that lie in the visible range (taken to run from 400 to 700 nm in wavelength).
View solution