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Consider electromagnetic waves propagating in air. (a) Determine the frequency of a wave with a wavelength of (i) 5.0 km, (ii) 5.0 mm, (iii) 5.0 nm. (b) What is the wavelength (in meters and nanometers) of (i) gamma rays of frequency 6.50 \(\times\) 10\(^{21}\) Hz and (ii) an AM station radio wave of frequency 590 kHz?

Short Answer

Expert verified
(a) (i) 60 kHz, (ii) 60 GHz, (iii) 60 PHz; (b) (i) 4.62 × 10^{-14} m or 0.0462 nm, (ii) 508.47 m or 5.0847 × 10^{11} nm.

Step by step solution

01

Understanding the formula

The relationship between the speed of light \(c\), wavelength \(\lambda\), and frequency \(f\) is given by the formula \(c = \lambda f\), where \(c = 3 \times 10^8\) m/s in air.
02

Solving Problem (a) - Finding Frequency

To find the frequency, rearrange the formula to \(f = \frac{c}{\lambda}\). We will substitute the given wavelengths one by one.
03

Calculating Frequency for 5.0 km Wavelength

Convert 5.0 km to meters: 5.0 km = 5000 m. Calculate frequency: \[f = \frac{3 \times 10^8}{5000} = 6 \times 10^4 \text{ Hz}\].
04

Calculating Frequency for 5.0 mm Wavelength

Convert 5.0 mm to meters: 5.0 mm = 0.005 m. Calculate frequency: \[f = \frac{3 \times 10^8}{0.005} = 6 \times 10^{10} \text{ Hz}\].
05

Calculating Frequency for 5.0 nm Wavelength

Convert 5.0 nm to meters: 5.0 nm = 5 \times 10^{-9} m. Calculate frequency: \[f = \frac{3 \times 10^8}{5 \times 10^{-9}} = 6 \times 10^{16} \text{ Hz}\].
06

Solving Problem (b) - Finding Wavelength

Rearrange the formula to find wavelength \(\lambda = \frac{c}{f}\) given the frequency.
07

Calculating Wavelength for Gamma Rays

Frequency is given as 6.50 \(\times\) 10\(^{21}\) Hz. Calculate wavelength: \[\lambda = \frac{3 \times 10^8}{6.50 \times 10^{21}} \approx 4.62 \times 10^{-14} \text{ m}\]. Convert to nanometers: 4.62 \(\times\) 10\(^{-14}\) m = 0.0462 nm.
08

Calculating Wavelength for AM Radio Wave

Frequency is given as 590 kHz. First convert to Hz: 590 kHz = 590,000 Hz. Calculate wavelength: \[\lambda = \frac{3 \times 10^8}{590,000} \approx 508.47 \text{ m}\]. Convert to nanometers: 508.47 m = 5.0847 \times 10^{11}\text{ nm}.

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Key Concepts

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

Wavelength
Wavelength is a key concept in understanding electromagnetic waves. It is the distance between two consecutive peaks or troughs in a wave. Wavelength is typically denoted by the Greek letter \(\lambda\).

In an electromagnetic wave, the wavelength can vary widely depending on the type of wave. For example:
  • Radio waves have long wavelengths, often measured in kilometers.
  • Visible light has wavelengths in the nanometer range.
  • Gamma rays have extremely short wavelengths, even smaller than visible light.
Wavelength helps determine the wave's energy and its frequency. The shorter the wavelength, the higher the frequency and energy of the wave.

To calculate the wavelength when you know the frequency and speed of light, you can use: \(\lambda = \frac{c}{f}\), where \(c\) is the speed of light.
Frequency
Frequency is another fundamental attribute of electromagnetic waves. It tells us how many wave crests pass a certain point in one second. Frequency is denoted by the letter \(f\) and is measured in Hertz (Hz).

A wave’s frequency is inversely related to its wavelength: as frequency increases, wavelength decreases, and vice versa. This relationship is captured in the equation: \(f = \frac{c}{\lambda}\).

Frequency can be very different across the electromagnetic spectrum:
  • Radio waves have frequencies ranging from about 3 kHz to 300 GHz.
  • Visible light frequencies are in the range of hundreds of THz.
  • Gamma rays exhibit extremely high frequencies, in the range of billions of GHz.
Understanding frequency helps in applications such as broadcasting, where radio and TV signals are transmitted at specific frequencies.
Speed of Light
The speed of light is a constant that is fundamental to physics, particularly in the study of electromagnetic waves. Denoted as \(c\), the speed of light is approximately \(3 \times 10^8\) meters per second.

In a vacuum, light travels at this speed, and it remains remarkably constant except in different media where it may slow down. However, for most calculations in air, especially in introductory physics, the vacuum speed of light is used.

This speed is crucial for calculations involving electromagnetic waves, as demonstrated in the core equation: \(c = \lambda f\). This relationship allows us to determine either the frequency or wavelength when the other is known. The constancy of the speed of light also provides a basis for measuring astronomical distances and exploring the vast universe.
Gamma Rays
Gamma rays are a form of electromagnetic radiation with the shortest wavelength and highest frequency. This gives them extremely high energy, making them more powerful than X-rays.

Gamma rays are emitted in nuclear reactions, such as those inside stars, and during radioactive decay processes. Their high energy makes them capable of penetrating most materials, which is why they are used in medical treatment to target cancer cells in radiotherapy.

Despite their utility, gamma rays can be harmful to living organisms as they can damage cellular structures and DNA.

Dealing with gamma rays often involves understanding their wavelength to assess their energy level: the shorter the wavelength, the more energy the gamma ray possesses. This information is crucial in both protecting against and utilizing gamma rays safely.
Radio Waves
Radio waves are a type of electromagnetic wave with long wavelengths and low frequencies. They are usually in the range from about 1 millimeter to 100 kilometers in wavelength, corresponding to frequencies from 300 GHz to as low as 3 kHz.

These waves are primarily used in communication systems. They form the backbone of broadcasting technologies such as radio, television, and mobile phone signals.

One significant feature of radio waves is their ability to travel long distances and penetrate through the atmosphere, making them ideal for wireless communication.

Understanding the frequency of radio waves is essential for tuning broadcast devices to the correct channel or signal. Moreover, radio waves are crucial in many modern technologies like radar, satellite communications, and even astronomy observations.

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Most popular questions from this chapter

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 kg/m\(^3\). Find the particle radius R such that the gravitational and radiation forces acting on the particle are equal in magnitude. The luminosity of the sun is 3.9 \(\times\) 10\(^{26}\) W. Does your answer depend on the distance of the particle from the sun? Why or why not? (d) Explain why dust particles with a radius less than that found in part (c) are unlikely to be found in the solar system. [\(Hint\): Construct the ratio of the two force expressions found in parts (a) and (b).]

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A sinusoidal electromagnetic wave emitted by a cellular phone has a wavelength of 35.4 cm and an electric-field amplitude of 5.40 \(\times\) 10\(^{-2}\) V/m at a distance of 250 m from the phone. Calculate (a) the frequency of the wave; (b) the magnetic-field amplitude; (c) the intensity of the wave.

Medical x rays are taken with electromagnetic waves having a wavelength of around 0.10 nm in air. What are the frequency, period, and wave number of such waves?

The intensity of a cylindrical laser beam is 0.800 W/m\(^2\). The cross- sectional area of the beam is 3.0 \(\times\) 10\(^{-4}\) m\(^2\) and the intensity is uniform across the cross section of the beam. (a) What is the average power output of the laser? (b) What is the rms value of the electric field in the beam?

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