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Chapter 12: 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) \(6.00 \times 10^4 \mathrm{~Hz}\), (ii) \(6.00 \times 10^{13} \mathrm{~Hz}\), (iii) \(6.00 \times 10^{16} \mathrm{~Hz}\). (b) Wavelengths: (i) \(4.62 \times 10^{-14} \mathrm{~m}, 4.62 \times 10^{-5} \mathrm{~nm}\), (ii) \(508.47 \mathrm{~m}, 5.08 \times 10^{11} \mathrm{~nm}\).

Step by step solution

01

Understand the relationship between frequency and wavelength

The frequency \( f \) of an electromagnetic wave is related to its wavelength \( \lambda \) and the speed of light \( c \) by the equation:\[c = f \times \lambda\]where \( c = 3.00 \times 10^8 \mathrm{~m/s} \) is the speed of light in a vacuum. We can rearrange this equation to solve for frequency:\[f = \frac{c}{\lambda}\]
02

Calculate the frequency for different wavelengths

For each wavelength given, use the equation from Step 1:(i) For \( \lambda = 5.0 \mathrm{~km} = 5.0 \times 10^3 \mathrm{~m} \),\[f = \frac{3.00 \times 10^8 \mathrm{~m/s}}{5.0 \times 10^3 \mathrm{~m}} = 6.00 \times 10^4 \mathrm{~Hz}\](ii) For \( \lambda = 5.0 \mu \mathrm{m} = 5.0 \times 10^{-6} \mathrm{~m} \),\[f = \frac{3.00 \times 10^8 \mathrm{~m/s}}{5.0 \times 10^{-6} \mathrm{~m}} = 6.00 \times 10^{13} \mathrm{~Hz}\](iii) For \( \lambda = 5.0 \mathrm{~nm} = 5.0 \times 10^{-9} \mathrm{~m} \),\[f = \frac{3.00 \times 10^8 \mathrm{~m/s}}{5.0 \times 10^{-9} \mathrm{~m}} = 6.00 \times 10^{16} \mathrm{~Hz}\]
03

Calculate wavelength from frequency for gamma rays

Using the relationship from Step 1 in reverse to find wavelength:(i) For gamma rays with \( f = 6.50 \times 10^{21} \mathrm{~Hz} \),\[\lambda = \frac{3.00 \times 10^8 \mathrm{~m/s}}{6.50 \times 10^{21} \mathrm{~Hz}} = 4.62 \times 10^{-14} \mathrm{~m}\]In nanometers,\[\lambda = 4.62 \times 10^{-14} \mathrm{~m} \times 10^9 \frac{\mathrm{nm}}{\mathrm{m}} = 4.62 \times 10^{-5} \mathrm{~nm}\]
04

Calculate wavelength from frequency for AM radio waves

Similarly, for AM radio waves:(ii) For \( f = 590 \mathrm{kHz} = 590 \times 10^3 \mathrm{~Hz} \),\[\lambda = \frac{3.00 \times 10^8 \mathrm{~m/s}}{590 \times 10^3 \mathrm{~Hz}} = 508.47 \mathrm{~m}\]In nanometers,\[\lambda = 508.47 \mathrm{~m} \times 10^9 \frac{\mathrm{nm}}{\mathrm{m}} = 5.08 \times 10^{11} \mathrm{~nm}\]

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Key Concepts

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

Frequency Calculation
Electromagnetic waves have a relationship between frequency, wavelength, and the speed of light that allows us to calculate one if we know the others. The frequency of an electromagnetic wave (\( f \)) can be found using the formula: \[ f = \frac{c}{\lambda} \] where \( c \) is the speed of light, \( 3.00 \times 10^8 \) meters per second, and \( \lambda \) is the wavelength of the wave in meters.
To find the frequency:
  • Convert the wavelength from its given unit to meters, if necessary. For example, \( 5.0 \) kilometers equals \( 5.0 \times 10^3 \) meters.
  • Insert the values into the formula to get the frequency.
Understanding this relationship is crucial since it helps in converting between quantities, such as when dealing with radio stations or any other form of electromagnetic spectrum analysis.
Wavelength Calculation
Just as frequency can be calculated with wavelength, the reverse is true. You can also calculate the wavelength if the frequency of the wave is known. Use the formula:\[ \lambda = \frac{c}{f} \] where \( f \) is the frequency of the wave. The key steps for doing this are:
  • Ensure the frequency is in Hertz (Hz).
    Sometimes, you will need to convert from other units such as kilohertz (kHz) or megahertz (MHz). For instance, convert \( 590 \) kHz to \( 590 \times 10^3 \) Hz.
  • Calculate the wavelength in meters using the formula.
  • Optionally convert meters to another unit, such as nanometers by multiplying by \( 10^9 \).
This technique is beneficial for understanding the scale of wavelengths in different parts of the electromagnetic spectrum, from short gamma rays to long AM radio waves.
Speed of Light
The speed of light (\( c \)) is a constant that plays a vital role in the understanding of electromagnetic waves. It is approximately \( 3.00 \times 10^8 \) meters per second in a vacuum or air.
This constant allows us to simplify the relation between frequency and wavelength into an easily manageable form, a pillar of many calculations in physics.
It's important to remember:
  • It acts as a bridge between frequency and wavelength calculations.
  • Since light speed is the same for all electromagnetic waves in a vacuum, this constant simplifies many complex phenomena.
Therefore, leveraging the speed of light in formulas becomes a handy tool for determining the properties of electromagnetic waves, which are essential in fields like telecommunications, astronomy, and even medical imaging.

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Most popular questions from this chapter

Testing a Space Radio Transmitter. You are a NASA mission specialist on your first flight aboard the space shuttle. Thanks to your extensive training in physics, you have been assigned to evaluate the performance of a new radio transmitter on board the International Space Station (ISS). Perched on the shuttle's movable arm, you aim a sensitive detector at the ISS, which is \(2.5 \mathrm{~km}\) away. You find that the electric-field amplitude of the radio waves coming from the ISS transmitter is \(0.090 \mathrm{~V} / \mathrm{m}\) and that the frequency of the waves is \(244 \mathrm{MHz}\). Find the following: (a) the intensity of the radio wave at your location; (b) the magnetic-field amplitude of the wave at your location; (c) the total power output of the ISS radio transmitter. (d) What assumptions, if any, did you make in your calculations?

Television Broadcasting. Public television station KQED in San Francisco broadcasts a sinusoidal radio signal at a power of \(316 \mathrm{~kW}\). Assume that the wave spreads out uniformly into a hemisphere above the ground. At a home \(5.00 \mathrm{~km}\) away from the antenna, (a) what average pressure does this wave exert on a totally reflecting surface, (b) what are the amplitudes of the electric and magnetic fields of the wave, and (c) what is the average density of the energy this wave carries? (d) For the energy density in part (c), what percentage is due to the electric field and what percentage is due to the magnetic field?

A plane sinusoidal electromagnetic wave in air has a wavelength of \(3.84 \mathrm{~cm}\) and an \(\vec{E}\)-field amplitude of \(1.35 \mathrm{~V} / \mathrm{m}\). (a) What is the frequency? (b) What is the \(\overrightarrow{\boldsymbol{B}}\)-field amplitude? (c) What is the intensity? (d) What average force does this radiation exert on a totally absorbing surface with area \(0.240 \mathrm{~m}^{2}\) perpendicular to the direction of propagation?

In the 25 -ft Space Simulator facility at NASA's Jet Propulsion Laboratory, a bank of overhead arc lamps can produce light of intensity \(2500 \mathrm{~W} / \mathrm{m}^{2}\) at the floor of the facility. (This simulates the intensity of sunlight near the planet Venus.) Find the average radiation pressure (in pascals and in atmospheres) on (a) a totally absorbing section of the floor and (b) a totally reflecting section of the floor. (c) Find the average momentum density (momentum per unit volume) in the light at the floor.

Interplanetary space contains many small particles referred to as interplanetary dust. Radiation pressure from the sun sets a lower limit on the size of such dust particles. To see the origin of this limit, consider a spherical dust particle of radius \(R\) and mass density \(\rho\). (a) Write an expression for the gravitational force exerted on this particle by the sun (mass \(M\) ) when the particle is a distance \(r\) from the sun. (b) Let \(L\) represent the luminosity of the sun, equal to the rate at which it emits energy in electromagnetic radiation. Find the force exerted on the (totally absorbing) particle due to solar radiation pressure, remembering that the intensity of the sun's radiation also depends on the distance \(r\). The relevant area is the cross-sectional area of the particle, not the total surface area of the particle. As part of your answer, explain why this is so. (c) The mass density of a typical interplanetary dust particle is about \(3000 \mathrm{~kg} / \mathrm{m}^{3} .\) Find the particle radius \(R\) such that the gravitational and radiation forces acting on the particle are equal in mag nitude. The luminosity of the sun is \(3.9 \times 10^{26} \mathrm{~W}\). Does your answer depend on the distance of the particle from the sun? Why (d) Explain why dust particles with a radius less than or why not? that found in part (c) are unlikely to be found in the solar system.
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