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Chapter 32: Problem 9

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 \(6.50 \times 10 ^ { 21 } \mathrm { Hz }\) and (ii) an AM station radio wave of frequency 590\(\mathrm { kHz }\) ?

Short Answer

Expert verified
(a) Frequencies: (i) \(6.00 \times 10^4\, \text{Hz}\), (ii) \(6.00 \times 10^{13}\, \text{Hz}\), (iii) \(6.00 \times 10^{16}\, \text{Hz}\). (b) Wavelengths: (i) \(4.62 \times 10^{-14}\, \text{m}\), \(4.62 \times 10^{-5}\, \text{nm}\); (ii) \(508.47 \, \text{m}\), \(508.47 \times 10^9 \, \text{nm}\).

Step by step solution

01

Understanding the Relationship Between Frequency and Wavelength

The speed of light in a vacuum (and approximately in air) is denoted as \( c \) and is approximately \( 3.00 \times 10^8 \text{ m/s} \). The relationship between the speed of light \( c \), wavelength \( \lambda \), and frequency \( f \) is given by the equation: \( c = \lambda f \). We'll use this equation to solve both parts of (a) and (b).
02

Part (a)(i): Frequency for Wavelength 5.0 km

Convert the wavelength from kilometers to meters: \( 5.0\, \text{km} = 5.0 \times 10^3 \, \text{m} \).Apply the formula: \[ f = \frac{c}{\lambda} = \frac{3.00 \times 10^8}{5.0 \times 10^3} \approx 6.00 \times 10^4 \, \text{Hz} \]
03

Part (a)(ii): Frequency for Wavelength 5.0 渭m

Convert the wavelength from micrometers to meters: \( 5.0\, \mu\text{m} = 5.0 \times 10^{-6} \, \text{m} \).Apply the formula: \[ f = \frac{c}{\lambda} = \frac{3.00 \times 10^8}{5.0 \times 10^{-6}} = 6.00 \times 10^{13} \, \text{Hz} \]
04

Part (a)(iii): Frequency for Wavelength 5.0 nm

Convert the wavelength from nanometers to meters: \( 5.0\, \text{nm} = 5.0 \times 10^{-9} \, \text{m} \).Apply the formula: \[ f = \frac{c}{\lambda} = \frac{3.00 \times 10^8}{5.0 \times 10^{-9}} = 6.00 \times 10^{16} \, \text{Hz} \]
05

Part (b)(i): Wavelength for Gamma Rays of Frequency 6.50 x 10^21 Hz in Meters and Nanometers

Given frequency \( f = 6.50 \times 10^{21} \text{ Hz} \).Apply the formula: \[ \lambda = \frac{c}{f} = \frac{3.00 \times 10^8}{6.50 \times 10^{21}} \approx 4.62 \times 10^{-14} \, \text{m} \]Convert to nanometers: \[ 4.62 \times 10^{-14} \, \text{m} = 4.62 \times 10^{-5} \, \text{nm} \]
06

Part (b)(ii): Wavelength for AM Radio of Frequency 590 kHz in Meters and Nanometers

Convert frequency from kilohertz to hertz: \( 590 \, \text{kHz} = 590 \times 10^3 \, \text{Hz} \).Apply the formula: \[ \lambda = \frac{c}{f} = \frac{3.00 \times 10^8}{590 \times 10^3} \approx 508.47 \, \text{m} \]Convert to nanometers: \[ 508.47 \, \text{m} = 508.47 \times 10^9 \, \text{nm} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Wavelength-Frequency Relationship
In the study of electromagnetic waves, one key relationship is between wavelength and frequency. These two quantities are intimately connected through the speed of light. The speed of light in a vacuum (and nearly the same in air) is approximately \( c = 3.00 \times 10^8 \text{ m/s} \). The fundamental equation that links wavelength \( \lambda \), frequency \( f \), and the speed of light \( c \) is:\[ c = \lambda f \].This means that the speed of light in meters per second equals the wavelength in meters multiplied by the frequency in hertz (cycles per second). In simple terms, higher frequencies correspond to shorter wavelengths, and vice versa. For example:
  • Visible light has a wavelength in the range of hundreds of nanometers and corresponding frequencies in the hundreds of terahertz.
  • Radio waves can have wavelengths from meters to kilometers, resulting in much lower frequencies.
The relationship allows us to calculate one of these parameters if the other and the speed of light are known. This principle is foundational to understanding light and other electromagnetic waves.
Speed of Light
Light is a form of electromagnetic radiation that travels incredibly fast. The speed of light in a vacuum is one of the fundamental constants of nature, defined as \( c = 3.00 \times 10^8 \text{ m/s} \). This value is crucial for calculating the behavior of electromagnetic waves, from radio waves that stretch kilometers long to gamma rays that are fractions of a nanometer.The constant speed of light facilitates the accurate prediction of how light and other forms of electromagnetic radiation behave during propagation through space. It underpins the ideas of relativity where nothing can travel faster than this speed. For calculations:
  • Use \( c = \lambda f \) to relate wavelength and frequency.
  • Convert frequencies from units like kilohertz or gigahertz to hertz.
Understanding the speed of light helps in appreciating how we perceive time and space around us, and also in technologies like GPS and communications systems.
Gamma Rays
Gamma rays are a type of electromagnetic radiation with the shortest wavelength and highest energy in the electromagnetic spectrum. These rays arise from nuclear reactions and other astronomical phenomena. Due to their extremely high frequency, gamma rays carry a large amount of energy, which allows them to penetrate many materials, making them useful in medical imaging and treatments like cancer radiotherapy.The formula \( \lambda = \frac{c}{f} \) lets us find the wavelength of gamma rays when we know their frequency. For instance:
  • A gamma-ray frequency of \( 6.50 \times 10^{21} \text{ Hz} \) corresponds to a wavelength of approximately \( 4.62 \times 10^{-14} \text{ meters} \), showing how minuscule these wavelengths are.
Gamma rays demonstrate the extreme end of the electromagnetic spectrum where physics meets chemistry and biology in applications such as sterilization processes and astrophysical observations.
AM Radio Waves
AM radio waves are a type of electromagnetic radiation used for transmitting audio signals over long distances. These waves operate in the kilohertz frequency range, which corresponds to wavelengths of hundreds of meters. Their longer wavelengths allow them to travel further than higher frequency radio waves, making them ideal for communication across vast areas.By using the relationship \( \lambda = \frac{c}{f} \), the wavelength for a typical AM station frequency of 590 kHz can be calculated. This comes out to about 508.47 meters. This wavelength allows AM radio waves to diffract around obstacles and travel beyond the horizon, increasing their broadcast range:
  • Converting from hertz: 590 kHz is \( 590 \times 10^3 \) Hz.
  • The calculated wavelength is 508.47 meters or \( 508.47 \times 10^9 \) nanometers in conversion terms.
The propagation characteristics of AM radio waves make them perfect for broadcasting news and music, especially in rural and remote areas.

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Most popular questions from this chapter

High-Energy Cancer Treatment. Scientists are working on a new technique to kill cancer cells by zapping them with ultrahigh-energy (in the range of \(10 ^ { 12 } \mathrm { W } )\) pulses of light that last for an extremely short time (a few nanoseconds). These short pulses scramble the interior of a cell without causing it to explode, as long pulses would do. We can model a typical such cell as a disk 5.0\(\mu \mathrm { m }\) in diameter, with the pulse lasting for 4.0\(\mathrm { ns }\) with an average power of \(2.0 \times 10 ^ { 12 } \mathrm { W }\) . We shall assume that the energy is spread uniformly over the faces of 100 cells for each pulse. (a) How much energy is given to the cell during this pulse? (b) What is the intensity (in \(\mathrm { W } / \mathrm { m } ^ { 2 }\) ) delivered to the cell? (c) What are the maximum values of the electric and magnetic fields in the pulse?

An electromagnetic standing wave in air of frequency 750\(\mathrm { MHz }\) is set up between two conducting planes 80.0\(\mathrm { cm }\) apart. At which positions between the planes could a point charge be placed at rest so that it would remain at rest? Explain.

An electromagnetic wave of wavelength 435\(\mathrm { nm }\) is traveling in vacuum in the \(- z\) -direction. The electric field has amplitude \(2.70 \times 10 ^ { - 3 } \mathrm { V } / \mathrm { m }\) and is parallel to the \(x\) -axis. What are (a) the frequency and (b) the magnetic-field amplitude? (c) Write the vector equations for \(\vec { E } ( z , t )\) and \(\vec { \boldsymbol { B } } ( z , t )\)

Electromagnetic waves propagate much differently in conductors than they do in dielectrics or in vacuum. If the resistivity of the conductor is sufficiently low (that is, it is a sufficiently good conductor), the oscillating electric field of the wave gives rise to an oscillating conduction current that is much larger than the displacement current. In this case, the wave equation for an electric field \(\vec { E } ( x , t ) = E _ { y } ( x , t ) \hat { J }\) propagating in the \(+ x\) -direction within a conductor is $$\frac { \partial ^ { 2 } E _ { y } ( x , t ) } { \partial x ^ { 2 } } = \frac { \mu } { \rho } \frac { \partial E _ { y } ( x , t ) } { \partial t }\( where \)\mu$$ is the permeability of the conductor and \(\rho\) is its resistivity. (a) A solution to this wave equation is $$E _ { y } ( x , t ) = E _ { \max } e ^ { - k _ { c } x } \cos \left( k _ { C } x - \omega t \right)$$ where \(k _ { \mathrm { C } } = \sqrt { \omega \mu / 2 \rho } .\) Verify this by substituting \(E _ { y } ( x , t )\) into the above wave equation. (b) The exponential term shows that the electric field decreases in amplitude as it propagates. Explain why this happens. (Hint: The field does work to move charges within the conductor. The current of these moving charges causes \(i ^ { 2 } R\) heating within the conductor, raising its temperature. Where does the energy to do this come from? \(( \mathrm { c } )\) Show that the electric-field amplitude decreases by a factor of 1\(/ e\) in a distance \(1 / k _ { \mathrm { C } } = \sqrt { 2 p / \omega \mu } ,\) and calculate this distance for a radio wave with frequency \(f = 1.0 \mathrm { MHz }\) in copper (resistivity \(1.72 \times 10 ^ { - 8 } \Omega \cdot\) m; permeability \(\mu = \mu _ { 0 } )\) . since this distance is so short, electromagnetic waves of this frequency can hardly propagate at all into copper. Instead, they are reflected at the surface of the metal. This is why radio waves can- not penetrate through copper or other metals, and why radio reception is poor inside a metal structure.

Consider each of the following electric- and magnetic-field orientations. In each case, what is the direction of propagation of the wave? (a) \(\vec { E } = E \hat { \imath } , \vec { B } = - B \hat { J } ;\) (b) \(\vec { E } = E \hat { J } , \vec { B } = B \hat { i } ;\) (c) \(\vec { E } =\) \(- E \hat { k } , \vec { B } = - B \hat { \imath } ; ( \mathrm { d } ) \vec { E } = E \hat { \imath } , \vec { B } = - B \hat { k }\)
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