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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

He-Ne lasers are often used in physics demonstrations. They produce light of wavelength 633 nm and a power of 0.500 mW spread over a cylindrical beam 1.00 mm in diameter (although these quantities can vary). (a) What is the intensity of this laser beam? (b) What are the maximum values of the electric and magnetic fields? (c) What is the average energy density in the laser beam?

An electromagnetic wave with frequency 5.70 \(\times\) 10\(^{14}\) Hz propagates with a speed of 2.17 \(\times\) 10\(^8\) m/s in a certain piece of glass. Find (a) the wavelength of the wave in the glass; (b) the wavelength of a wave of the same frequency propagating in air; (c) the index of refraction \(n\) of the glass for an electromagnetic wave with this frequency; (d) the dielectric constant for glass at this frequency, assuming that the relative permeability is unity.

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 \(\vec{E} (x, t) = E_y(x, t)\hat{\jmath}\) en propagating in the +\(x\)-direction within a conductor is $${\partial^2E_y(x, t)\over \partial x^2} = {\mu \over \rho} {\partial Ey(x, t)\over \partial t}$$ where \(\mu\) is the permeability of the conductor and \(\rho\) is its resistivity. (a) A solution to this wave equation is \(E_y(x, t) = E_{max} e^{-k_C x} cos(k_Cx - \omega t)\), where \(k_C = \sqrt{(\omega \mu/2\rho}\). 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 \(i^2R\) 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 \(1/k_C = \sqrt{2\rho/\omega\mu}\), and calculate this distance for a radio wave with frequency \(f\) = 1.0 MHz in copper (resistivity 1.72 \(\times\) 10\(^{-8 } \Omega \bullet m\); 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.

For a sinusoidal electromagnetic wave in vacuum, such as that described by Eq. (32.16), show that the \(average\) energy density in the electric field is the same as that in the magnetic field.

A circular wire loop has a radius of \(7.50 \mathrm{~cm}\). A sinusoidal electromagnetic plane wave traveling in air passes through the loop, with the direction of the magnetic field of the wave perpendicular to the plane of the loop. The intensity of the wave at the location of the loop is \(0.0275 \mathrm{~W} / \mathrm{m}^{2}\), and the wavelength of the wave is \(6.90 \mathrm{~m}\). What is the maximum emf induced in the loop?

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