Problem 5
Question
\(\bullet\) Consider electromagnetic waves propagating in air. (a) Determine the frequency of a wave with a wavelength of (i) \(5.0 \mathrm{km},\) (ii) \(5.0 \mu \mathrm{m},\) (iii) 5.0 \(\mathrm{nm}\) . (b) What is the wavelength (in meters and nanometers) of (i) gamma rays of frequency \(6.50 \times 10^{21} \mathrm{Hz}\) , (ii) an AM station radio wave of frequency 590 \(\mathrm{kHz} ?\)
Step-by-Step Solution
Verified Answer
(a) Frequencies: (i) \(6.0 \times 10^4\) Hz, (ii) \(6.0 \times 10^{13}\) Hz, (iii) \(6.0 \times 10^{16}\) Hz. (b) Wavelengths: (i) 4.62 x 10^-14 m (4.62 x 10^-5 nm), (ii) 508.47 m (5.0847 x 10^11 nm).
1Step 1: Understanding the Relationship Between Frequency and Wavelength
The relationship between the frequency \(f\) and wavelength \(\lambda\) of electromagnetic waves is given by the formula:\[ c = \lambda \cdot f \]where \(c\) is the speed of light in a vacuum, approximately \(3 \times 10^8\) m/s.
2Step 2: Calculate Frequency for Wavelength of 5.0 km
Given the wavelength \(\lambda = 5.0\) km, convert it to meters first, \(5.0 \text{ km} = 5000 \text{ m}\). Now, use the relation:\[ f = \frac{c}{\lambda} = \frac{3 \times 10^8 \text{ m/s}}{5000 \text{ m}} = 6.0 \times 10^4 \text{ Hz} \]
3Step 3: Calculate Frequency for Wavelength of 5.0 micrometers
Convert the wavelength \(\lambda = 5.0 \mu m\) to meters, \(5.0 \mu \text{m} = 5.0 \times 10^{-6} \text{ m}\). Using the formula:\[ f = \frac{3 \times 10^8 \text{ m/s}}{5.0 \times 10^{-6} \text{ m}} = 6.0 \times 10^{13} \text{ Hz} \]
4Step 4: Calculate Frequency for Wavelength of 5.0 nm
Convert the wavelength \(\lambda = 5.0 \text{ nm}\) to meters, \(5.0 \text{ nm} = 5.0 \times 10^{-9} \text{ m}\). Using the formula:\[ f = \frac{3 \times 10^8 \text{ m/s}}{5.0 \times 10^{-9} \text{ m}} = 6.0 \times 10^{16} \text{ Hz} \]
5Step 5: Calculate Wavelength for Gamma Rays Frequency 6.50 x 10^21 Hz
Using the frequency \(f = 6.50 \times 10^{21} \text{ Hz}\), the wavelength \(\lambda\) is:\[ \lambda = \frac{c}{f} = \frac{3 \times 10^8 \text{ m/s}}{6.50 \times 10^{21} \text{ Hz}} = 4.62 \times 10^{-14} \text{ m} \]To convert to nanometers, \(\lambda = 4.62 \times 10^{-5} \text{ nm}\).
6Step 6: Calculate Wavelength for AM Station Frequency 590 kHz
Convert the frequency from kHz to Hz, \(590 \text{ kHz} = 590 \times 10^3 \text{ Hz}\). The wavelength \(\lambda\) is:\[ \lambda = \frac{c}{f} = \frac{3 \times 10^8 \text{ m/s}}{590 \times 10^3 \text{ Hz}} = 508.47 \text{ m} \]Converting to nanometers, \(\lambda = 5.0847 \times 10^{11} \text{ nm}\).
Key Concepts
Wavelength and Frequency RelationshipSpeed of Light ConstantGamma RaysRadio WavesUnit Conversion in Physics
Wavelength and Frequency Relationship
Electromagnetic waves exhibit a fascinating reciprocal relationship between wavelength and frequency. This relationship is governed by the equation \(c = \lambda \cdot f\), where \(c\) is the speed of light, \(\lambda\) is the wavelength, and \(f\) is the frequency. Simply put, if you know any two of these variables, you can easily determine the third one.
The speed of light (denoted by \(c\)) is constant, meaning that if the wavelength increases, the frequency must decrease, and vice versa.
The speed of light (denoted by \(c\)) is constant, meaning that if the wavelength increases, the frequency must decrease, and vice versa.
- A longer wavelength corresponds to a lower frequency.
- A shorter wavelength corresponds to a higher frequency.
Speed of Light Constant
The speed of light in a vacuum is an unfaltering constant: \(3 \times 10^8\) meters per second (m/s). This constant is crucial because it connects wavelength and frequency in the equation \(c = \lambda \cdot f\).
Wherever electromagnetic waves travel, this speed remains unchangeable when in a vacuum. However, it can be slightly reduced when passing through different media, like air or water. Despite these alterations in speed in various media, the relationship between wavelength and frequency stays intact because both are affected proportionally to maintain the product equal to the speed of light.
Knowing this constant allows for simplified calculations across the electromagnetic spectrum, from gamma rays to radio waves, making it an essential component of physics and engineering.
Wherever electromagnetic waves travel, this speed remains unchangeable when in a vacuum. However, it can be slightly reduced when passing through different media, like air or water. Despite these alterations in speed in various media, the relationship between wavelength and frequency stays intact because both are affected proportionally to maintain the product equal to the speed of light.
Knowing this constant allows for simplified calculations across the electromagnetic spectrum, from gamma rays to radio waves, making it an essential component of physics and engineering.
Gamma Rays
Gamma rays are a type of electromagnetic radiation with incredibly high frequencies and correspondingly short wavelengths. They fall at the extreme end of the electromagnetic spectrum. For instance, waves with a frequency as high as \(6.50 \times 10^{21}\) Hz are classified as gamma rays.
- Gamma rays can travel through most materials, making them useful in medical imaging and cancer treatment radiation therapy.
- They originate from radioactive decay and celestial phenomena.
Radio Waves
Radio waves are part of the electromagnetic spectrum that have the longest wavelength and the lowest frequency. Their frequencies typically range from thousands (kilohertz) to millions (megahertz) of cycles per second. These characteristics make them suitable for wireless communication such as AM and FM radio broadcasting, television, and mobile phones.
For example, an AM radio wave might have a frequency of \(590\) kHz, which can be converted to a wavelength by rearranging the light-speed equation to \(\lambda = \frac{c}{f}\). Understanding how to perform this conversion is integral to designing and analyzing communication systems that use radio waves.
For example, an AM radio wave might have a frequency of \(590\) kHz, which can be converted to a wavelength by rearranging the light-speed equation to \(\lambda = \frac{c}{f}\). Understanding how to perform this conversion is integral to designing and analyzing communication systems that use radio waves.
Unit Conversion in Physics
Physics often requires converting between different units to solve problems accurately, particularly in calculating electromagnetic wave properties. One common conversion is from kilometers or various sub-units to meters, as the standard unit for wavelength in physics problems. For example:
- Convert kilometers (km) to meters (m): multiply by \(1000\).
- Convert micrometers (\(\mu m\)) to meters: multiply by \(10^{-6}\).
- Convert nanometers (nm) to meters: multiply by \(10^{-9}\).
Other exercises in this chapter
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