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Chapter 32: Problem 7

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) (i) The frequency of a wave with a wavelength of \(5.0 \mathrm{km}\) is \(60 \mathrm{kHz}\). (ii) The frequency of a wave with a wavelength of \(5.0 \mu \mathrm{m}\) is \(60 \mathrm{THz}\). (iii) The frequency of a wave with a wavelength of \(5.0 \mathrm{nm}\) is \(60 \mathrm{PHz}\). \n (b) (i) The wavelength of gamma rays of frequency \(6.5 \times 10^{21} \mathrm{Hz}\) is \(46.2 \mathrm{pm}\) or \(0.0462 \mathrm{nm}\). (ii) The wavelength of an AM station radio wave of frequency \(590 \mathrm{kHz}\) is \(508 \mathrm{m}\) or \(5.08 \times 10^{11}\) nm.

Step by step solution

01

Determine the Frequency of a Wave

Use the formula \(f = c/ \lambda\) where \(c\) is the speed of light and \(\lambda\) is the wavelength. Given that \(c = 3.00 \times 10^8 \mathrm{m/s}\), plug in the wavelength values to calculate the frequency for each case.\n\n(i) For a wavelength of \(5.0 \mathrm{km}\), which equals \(5.0 \times 10^3 \mathrm{m}\).\n(ii) For a wavelength of \(5.0 \mu \mathrm{m} = 5.0 \times 10^{-6} \mathrm{m}\).\n(iii) For a wavelength of \(5.0 \mathrm{ nm} = 5.0 \times 10^{-9} \mathrm{m}\).
02

Determine the Wavelength of a Wave

Now, use the formula \(\lambda = c/f\) to calculate the wavelength. Apply the given frequencies to find the wavelengths in both meters and nanometers. \n(i) For a frequency of \(6.5 \times 10^{21} \mathrm{Hz}\), determine the wavelength in both meters and nanometers (1m = \(10^9\) nm). \n(ii) For an oscillation speed of \(590 \mathrm{kHz} = 590 \times 10^3 \mathrm{Hz}\), determine the wavelength in both meters and nanometers.
03

Compute the Frequencies & Wavelengths

Perform the calculations for each to determine the frequency (for part a) and the wavelength (for part b). This will involve a calculation and a unit conversion.

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Key Concepts

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

Speed of Light
The speed of light, denoted by the symbol 'c', is a fundamental constant of nature that plays a vital role in electromagnetic wave calculations. It is the speed at which all electromagnetic waves propagate in a vacuum, and it has a value of approximately \( 3.00 \times 10^8 \mathrm{m/s} \). This speed is considered to be the universal speed limit, according to Einstein's theory of relativity.

In our everyday context, light seems instantaneous, but in larger scales such as in space, the finite speed dictates how we understand distances and motion. For example, when we observe distant stars, we are looking at the light that left them years ago, seeing into the past thanks to the speed of light. In electromagnetic wave calculations, the speed of light is used to connect the frequency and the wavelength of a wave, a concept that is crucial for understanding how electromagnetic radiation behaves across different media, such as air, water, or vacuum.
Wave Frequency
Frequency, often represented by 'f', is a measure of how often the waves' crests pass a point in a given time interval. It's expressed in Hertz (Hz), where one Hertz equates to one cycle per second.

In terms of electromagnetic waves, the frequency will determine the type of radiation, such as radio waves, microwaves, visible light, ultraviolet, X-rays, or gamma rays - each having a frequency range that defines its position in the electromagnetic spectrum. High-frequency waves, like gamma rays, have much energy and can penetrate through materials which lower frequency waves like radio waves cannot. Understanding wave frequency not only is important for solving physics problems but also for practical applications like tuning your radio to the right station or setting up communication networks.
Wavelength Conversion
Wavelength conversion is the process of translating a wave's length from one unit of measurement to another. This is essential given that wavelengths can vary significantly; radio waves may be meters long, whereas light wavelengths are typically in the nanometer range.

The relationship between speed of light, frequency, and wavelength can be utilized for these conversions. For instance, with the formula \( \lambda = \frac{c}{f} \), when you comprehend the frequency, you can deduce the wavelength in meters, and then convert to any unit like kilometers (\( \mathrm{km} \) ), micrometers (\( \mu\mathrm{m} \) ), or nanometers (\( \mathrm{nm} \)). Conversions and understanding these measurements are not merely academic; they are crucial for areas like fiber-optic communications, astronomy, and any technology relying on the behavior of electromagnetic waves.

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Most popular questions from this chapter

The GPS network consists of 24 satellites, each of which makes two orbits around the earth per day. Each satellite transmits a \(50.0 \mathrm{~W}\) (or even less) sinusoidal electromagnetic signal at two frequencies, one of which is \(1575.42 \mathrm{MHz}\). Assume that a satellite transmits half of its power at each frequency and that the waves travel uniformly in a downward hemisphere. (a) What average intensity does a GPS receiver on the ground, directly below the satellite, receive? (Hint: First use Newton's laws to find the altitude of the satellite.) (b) What are the amplitudes of the electric and magnetic fields at the GPS receiver in part (a), and how long does it take the signal to reach the receiver? (c) If the receiver is a square panel \(1.50 \mathrm{~cm}\) on a side that absorbs all of the beam, what average pressure does the signal exert on it? (d) What wavelength must the receiver be tuned to?

NASA is giving serious consideration to the concept of solar sailing. A solar sailcraft uses a large, low-mass sail and the energy and momentum of sunlight for propulsion. (a) Should the sail be absorbing or reflective? Why? (b) The total power output of the sun is \(3.9 \times 10^{26} \mathrm{~W}\). How large a sail is necessary to propel a \(10,000 \mathrm{~kg}\) spacecraft against the gravitational force of the sun? Express your result in square kilometers. (c) Explain why your answer to part (b) is independent of the distance from the sun.

(a) How much time does it take light to travel from the moon to the earth, a distance of \(384,000 \mathrm{~km} ?\) (b) Light from the star Sirius takes 8.61 years to reach the earth. What is the distance from earth to Sirius in kilometers?

A cylindrical conductor with a circular cross section has a radius \(a\) and a resistivity \(\rho\) and carries a constant current \(I\). (a) What are the magnitude and direction of the electric-field vector \(\vec{E}\) at a point just inside the wire at a distance \(a\) from the axis? (b) What are the magnitude and direction of the magnetic-field vector \(\vec{B}\) at the same point? (c) What are the magnitude and direction of the Poynting vector \(\vec{S}\) at the same point? (The direction of \(\vec{S}\) is the direction in which electromagnetic energy flows into or out of the conductor.) (d) Use the result in part (c) to find the rate of flow of energy into the volume occupied by a length \(l\) of the conductor. (Hint: Integrate \(\vec{S}\) over the surface of this volume.) Compare your result to the rate of generation of thermal energy in the same volume. Discuss why the energy dissipated in a current-carrying conductor, due to its resistance, can be thought of as entering through the cylindrical sides of the conductor.

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, if 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 \(\vec{E}(x, t)=E_{y}(x, t) \hat{\jmath}\) 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) \(\mathrm{A}\) solution to this wave equation is \(E_{y}(x, t)=E_{\max } e^{-k_{\mathrm{C}} x} \cos \left(k_{\mathrm{C}} x-\omega t\right)\) 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?) (c) Show that the electric-field amplitude decreases by a factor of \(1 / e\) in a distance \(1 / k_{\mathrm{C}}=\sqrt{2 \rho / \omega \mu},\) and calculate this distance for a radio wave with frequency \(f=1.0 \mathrm{MHz}\) in copper (resistivity \(1.72 \times 10^{-8} \Omega \cdot \mathrm{m} ;\) 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 cannot penetrate through copper or other metals, and why radio reception is poor inside a metal structure.
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