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Chapter 7: Problem 27

The wavelength used by citizen's band radio is \(21 \mathrm{~m}\). Calculate the frequency (in \(\mathrm{s}^{-1}\) ) of this electromagnetic radiation.

Short Answer

Expert verified
The frequency is approximately \(1.43 \times 10^7\) Hz.

Step by step solution

01

Understand the Relationship

To find the frequency, we need to know the relationship between wavelength and frequency of electromagnetic waves. The formula connecting them is \( c = \lambda u \), where \( c \) is the speed of light (approximately \(3 \times 10^8\) meters per second), \( \lambda \) is the wavelength, and \( u \) is the frequency.
02

Rearrange the Formula

Rearrange the formula to solve for frequency \( u \). The relationship becomes \( u = \frac{c}{\lambda} \), meaning the frequency is equal to the speed of light divided by the wavelength.
03

Substitute the Wavelength and Speed of Light

Substitute the given wavelength \(\lambda = 21\) meters and the speed of light \(c = 3 \times 10^8\) meters per second into the equation: \( u = \frac{3 \times 10^8}{21} \).
04

Calculate the Frequency

Perform the division to find the frequency: \( u = \frac{3 \times 10^8}{21} = 1.43 \times 10^7 \) Hz (hertz or \(s^{-1}\)).

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Key Concepts

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

Frequency Calculation
To calculate the frequency of an electromagnetic wave, we begin by understanding the direct relationship between frequency and wavelength. The frequency (\( u \)) can be calculated by dividing the speed of light (\( c \)) by the wavelength (\( \lambda \)) of the wave. The formula for this calculation is:
  • \( u = \frac{c}{\lambda} \)
This formula highlights an inverse relationship: as the wavelength increases, the frequency decreases, and vice versa. Such is the nature of electromagnetic waves. By performing the calculation, you get the total number of wave cycles passing a point per second. In our exercise, with a wavelength of 21 meters and using the speed of light, we find the frequency to be approximately \(1.43 \times 10^7\) Hz.
Wavelength-Frequency Relationship
Electromagnetic waves reveal an important relationship between wavelength and frequency. This relationship is governed by the equation \(c = \lambda u\). In this formula:
  • \(c\) is the speed of light, a fundamental constant.
  • \(\lambda\) is the wavelength, the distance between two consecutive peaks of a wave.
  • \(u\) is the frequency, the number of cycles the wave completes in a second.
These quantities are tightly intertwined, meaning a change in one affects the others. Waves with larger wavelengths have lower frequencies, and those with shorter wavelengths are much higher in frequency. This principle helps in understanding how waves behave, such as radio waves, microwaves, and more.
Speed of Light
The speed of light (\(c\)) is a crucial constant in the world of physics and forms the backbone of many calculations involving electromagnetic waves. It is approximately \(3 \times 10^8\) meters per second. This means that light can travel roughly 300 million meters, or about 186,000 miles, in just one second. This incredible speed is part of why light and other electromagnetic waves can travel vast distances in what seems like an instant. When calculating frequency or wavelength, the speed of light is used in formulas because electromagnetic waves travel at this speed in a vacuum. Keep in mind this speed is consistent across all electromagnetic waves in a vacuum, linking wavelength and frequency precisely. Understanding this speed helps unravel many mysteries surrounding the behavior and essence of light and other waves.

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Most popular questions from this chapter

Microwave ovens operate using radiation that has a frequency of \(2.45 \times 10^{9} \mathrm{~Hz}\). What is the wavelength of this radiation? What name is used for the spectral region of this radiation?

How did de Broglie justify Bohr's assumption that angular momentum was quantized?

The absorption spectra of ions have been used to identify the presence of the elements in the atmospheres of the Sun and other stars. (In fact, the element helium was discovered in the spectrum of the Sun before it was identified on Earth, hence its name.) What is the wavelength of light (in \(\mathrm{nm}\) ) that is absorbed by \(\mathrm{He}^{+}\) ions when they are excited from the Bohr orbit with \(n=3\) to the \(n=4\) state?

A The charge of an electron is \(-1.602 \times 10^{-19} \mathrm{C}\). How many microamperes \((1 \mathrm{~A}=1 \mathrm{C} / \mathrm{s})\) of electrical current are produced by a photoelectric cell that ejects \(2.50 \times 10^{13}\) electrons each second? How many photons must be absorbed each second to produce this number of photoelectrons, assuming that each photon causes an electron to be ejected?

In extremely energetic systems such as the Sun, hydrogen emission lines can be seen from shells as high as \(n=40\). The spectrum emitted is quite striking because the energy levels become spaced quite closely. (a) Calculate the difference in energy (in J) between the \(n=2\) and \(n=3\) levels, and compare it with the difference in energy between the \(n=32\) and \(n=33\) levels. (b) Calculate the largest energy difference (in J) that can be observed in the hydrogen atom. (c) Explain why there is a limit to the energy difference. (d) Is there a physical phenomenon that corresponds to this difference? Explain your answer.
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