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Estimate the frequency of an electromagnetic wave whose wavelength is similar in size to an atom (about a nm). Referring back to figure o on p. 703 , in what part of the electromagnetic spectrum would such a wave lie (infrared, gamma-rays, ...)?

Short Answer

Expert verified
The wave has a frequency of 3 x 10^17 Hz, placing it in the X-ray spectrum region.

Step by step solution

01

Understand the Relationship between Frequency and Wavelength

Electromagnetic waves have a frequency (\( f \)) and a wavelength (\( \lambda \)). The speed of light (\( c \)) in a vacuum is given by the product of frequency and wavelength: \( c = f \times \lambda \). We need to solve for the frequency to estimate the correct part of the spectrum.
02

Substitute Known Values

The speed of light, \( c \),is approximately \( 3 \times 10^8 \, ext{m/s}\). The given wavelength \( \lambda \) is 1 nm, which is \( 1 \times 10^{-9} \) meters. Using the formula: \( f = \frac{c}{\lambda} \).
03

Calculate the Frequency

Substitute \( c \) and \( \lambda \) into the frequency equation to calculate \( f \): \( f = \frac{3 \times 10^8}{1 \times 10^{-9}} \). This calculates to \( 3 \times 10^{17} \) Hz.
04

Identify the Electromagnetic Spectrum Region

According to the electromagnetic spectrum, a frequency of \( 3 \times 10^{17} \) Hz corresponds to the X-ray region. Waves with this frequency are within the X-ray range, right between ultraviolet light and gamma rays.

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Key Concepts

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

Frequency and Wavelength Relationship
Electromagnetic waves consist of energy traveling through space. Each wave has key characteristics: **frequency** and **wavelength**. But what's vital is how they relate, particularly through the speed of light (symbolized as \( c \)).
Frequency, denoted by \( f \), is the number of wave cycles that pass a point per second, measured in Hertz (Hz). Wavelength, \( \lambda \), is the distance between two wave peaks, usually measured in meters (m).

The relationship is united in the equation:
\[ c = f \times \lambda \]

This formula highlights how frequency and wavelength are inversely proportional. This means:
  • As frequency increases, wavelength decreases.
  • As wavelength decreases, frequency increases.
When they multiply, they match the speed of light, a constant at roughly \( 3 \times 10^8 \) meters per second (m/s).
By understanding this relationship, we can solve for either frequency or wavelength, critical for identifying where on the electromagnetic spectrum a wave lies.
X-rays
In the vast ocean of the electromagnetic spectrum, X-rays hold a unique and important place. They have higher energy than visible light, making them especially useful.

**Understanding X-rays' Attributes:**
  • Wavelengths typically range from 0.01 nm to 10 nm.
  • Frequencies range from \( 3 \times 10^{16} \) Hz to \( 3 \times 10^{19} \) Hz.
Positioned between ultraviolet light and gamma rays, X-rays excel in practical applications due to their penetrating ability.

**Applications of X-rays:**
  • In medicine: Create images of the inside of the body, allowing for non-invasive diagnoses.
  • In industry and security: Inspect materials and structures without causing damage.
  • In astronomy: Help observe phenomena like black holes and distant celestial bodies.
By knowing that a frequency of \( 3 \times 10^{17} \) Hz as calculated falls in the X-ray category, we see how understanding electromagnetic waves aids both science and society.
Speed of Light
The **speed of light** is often considered one of nature's most fundamental constants, pivotal in comprehending electromagnetic waves. In a vacuum, it's approximately \( 3 \times 10^8 \) meters per second.

**Importance of Speed of Light:**
  • Acts as a cornerstone for everything that encompasses electromagnetic waves.
  • Serves in various theoretical frameworks, including Einstein's theory of relativity.
  • Plays a crucial role in calculations for phenomena like time dilation, GPS technology, and more.
When using the speed of light, it's instrumental in finding frequency or wavelength with the formula mentioned earlier: \[ c = f \times \lambda \]

**More Than Just Speed:**
Its constancy makes it critical for synchronization in modern technology and a universal standard for defining units of measurement.
With \( c = 3 \times 10^8 \), this can also be interpreted as the light traveling around Earth more than 7 times in just a second. Recognizing the speed of light as a universal constant not only allows us to navigate the electromagnetic spectrum but also expands our understanding of the universe.

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Most popular questions from this chapter

(a) A converging mirror is being used to create a virtual image. What is the range of possible magnifications? (b) Do the same for the other types of images that can be formed by curved mirrors (both converging and diverging).

There are two main types of telescopes, refracting (using a lens) and reflecting (using a mirror, as in figure i on p. 754 ). (Some telescopes use a mixture of the two types of elements: the light first encounters a large curved mirror, and then goes through an eyepiece that is a lens. To keep things simple, assume no eyepiece is used.) What implications would the colordependence of focal length have for the relative merits of the two types of telescopes? Describe the case where an image is formed of a white star. You may find it helpful to draw a ray diagram.

Here's a game my kids like to play. I sit next to a sunny window, and the sun reflects from the glass on my watch, making a disk of light on the wall or floor, which they pretend to chase as I move it around. Is the spot a disk because that's the shape of the sun, or because it's the shape of my watch? In other words, would a square watch make a square spot, or do we just have a circular image of the circular sun, which will be circular no matter what?

Fred's eyes are able to focus on things as close as \(5.0 \mathrm{~cm}\). Fred holds a magnifying glass with a focal length of \(3.0 \mathrm{~cm}\) at a height of \(2.0 \mathrm{~cm}\) above a flatworm. (a) Locate the image, and find the magnification. (b) Without the magnifying glass, from what distance would Fred want to view the flatworm to see its details as well as possible? With the magnifying glass? (c) Compute the angular magnification.

If the user of an astronomical telescope moves her head closer to or farther away from the image she is looking at, does the magnification change? Does the angular magnification change? Explain. (For simplicity, assume that no eyepiece is being used.)

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