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Chapter 20: Problem 34

For electromagnetic waves with each of these frequencies, find the wavelength in vacuum. In what region of the electromagnetic spectrum does each lie? (a) \(9.4 \times 10^{12} \mathrm{~Hz}\) (b) \(6.2 \times 10^{14} \mathrm{~Hz}\) (c) \(2.9 \times 10^{15} \mathrm{~Hz}\) (d) \(5.0 \times 10^{17} \mathrm{~Hz}\)

Short Answer

Expert verified
(a) Infrared (\(3.19 \times 10^{-5} \text{ m}\)), (b) Visible light (\(4.84 \times 10^{-7} \text{ m}\)), (c) Ultraviolet (\(1.03 \times 10^{-7} \text{ m}\)), (d) X-ray (\(6.00 \times 10^{-10} \text{ m}\)).

Step by step solution

01

Understand the Relationship

The wavelength of electromagnetic waves in a vacuum can be calculated using the formula:\[\lambda = \frac{c}{f}\]where \(\lambda\) is the wavelength, \(c\) is the speed of light in a vacuum (\(c = 3.00 \times 10^8 \) m/s), and \(f\) is the frequency. This formula arises because the speed of light in a vacuum is a constant.
02

Solve for (a)

Use the formula to find the wavelength for \(9.4 \times 10^{12} \text{ Hz}\):\[\lambda = \frac{3.00 \times 10^8 \text{ m/s}}{9.4 \times 10^{12} \text{ Hz}} = 3.19 \times 10^{-5} \text{ m}\]This frequency lies in the infrared region of the electromagnetic spectrum.
03

Solve for (b)

Use the formula to find the wavelength for \(6.2 \times 10^{14} \text{ Hz}\):\[\lambda = \frac{3.00 \times 10^8 \text{ m/s}}{6.2 \times 10^{14} \text{ Hz}} = 4.84 \times 10^{-7} \text{ m} \]This frequency lies in the visible light region of the electromagnetic spectrum.
04

Solve for (c)

Use the formula to find the wavelength for \(2.9 \times 10^{15} \text{ Hz}\):\[\lambda = \frac{3.00 \times 10^8 \text{ m/s}}{2.9 \times 10^{15} \text{ Hz}} = 1.03 \times 10^{-7} \text{ m} \]This frequency lies in the ultraviolet region of the electromagnetic spectrum.
05

Solve for (d)

Use the formula to find the wavelength for \(5.0 \times 10^{17} \text{ Hz}\):\[\lambda = \frac{3.00 \times 10^8 \text{ m/s}}{5.0 \times 10^{17} \text{ Hz}} = 6.00 \times 10^{-10} \text{ m} \]This frequency lies in the X-ray region of the electromagnetic spectrum.

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Key Concepts

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

Electromagnetic Waves
Electromagnetic waves are a fascinating phenomenon that travel through space without needing a material medium. These waves carry energy and information across vast distances, making communication and observations of the universe possible. Examples include radio waves, microwaves, infrared light, visible light, ultraviolet light, X-rays, and gamma rays. They differ mainly in their wavelengths and frequencies, which determine how they interact with matter and the effects they produce. Understanding electromagnetic waves is crucial for technologies like radio, television, and medical imaging.
Wavelength Calculation
To calculate the wavelength () of an electromagnetic wave, the basic formula used is: \[\lambda = \frac{c}{f}\]where \(c\) is the speed of light in vacuum, approximately \(3.00 \times 10^8\,\text{m/s}\), and \(f\) is the frequency of the electromagnetic wave.
  • Wavelength is inversely proportional to frequency. Higher frequencies mean shorter wavelengths.
  • This relationship is fundamental for determining how waves interact with different materials and regions of space.
Calculating wavelength helps us classify the wave into the appropriate region of the electromagnetic spectrum.
Frequency and Wavelength Relationship
The relationship between frequency and wavelength is pivotal in understanding how electromagnetic waves behave. Frequency refers to the number of wave cycles that pass a given point per second. Measured in Hertz (Hz), it directly affects the wavelength of the wave.
  • As frequency increases, wavelength decreases, making the waves shorter.
  • Conversely, a lower frequency results in longer wavelengths.
  • Waves with high frequency and short wavelengths (like X-rays) can penetrate materials more deeply than those with low frequency and long wavelengths (like radio waves).
This concept is essential in identifying the wave's position in the electromagnetic spectrum.
Infrared Region
The infrared region of the electromagnetic spectrum features waves with longer wavelengths than visible light but shorter than microwaves. These are typically in the range of \(700\,\text{nm}\) to \(1\,\text{mm}\) in wavelength. Infrared waves are often associated with heat, as many objects emit infrared radiation as they warm up.
  • Common applications include remote controls, night-vision equipment, and thermal imaging cameras.
  • Infrared light is not visible to the human eye, but sensors can detect it.
Understanding this region helps in developing technologies that detect heat and produce images based on thermal radiation.
Visible Light
Visible light is a small part of the electromagnetic spectrum that is detectable by the human eye. It ranges from wavelengths of approximately \(400\,\text{nm}\) (violet) to \(700\,\text{nm}\) (red).
  • This range of wavelengths is responsible for the colors we perceive in everyday life.
  • Different wavelengths within this range correspond to different colors, creating the vibrant spectrum from red through violet.
  • Technologies like cameras and projectors make use of visible light to capture and display images.
Visible light plays a crucial role in enabling us to observe our environment and artistic creations.
Ultraviolet Region
The ultraviolet (UV) region features electromagnetic waves with wavelengths shorter than visible light but longer than X-rays, typically ranging from \(10\,\text{nm}\) to \(400\,\text{nm}\). This region is known for its ability to cause certain materials to fluoresce and its role in producing vitamin D in the skin.
  • Overexposure to UV radiation can cause skin damage and increase the risk of skin cancer.
  • UV light is used in sterilization and disinfection processes due to its ability to destroy bacteria and viruses.
It's important to recognize the utility as well as the dangers associated with UV radiation.
X-ray Region
The X-ray region consists of electromagnetic waves with very short wavelengths, typically ranging from \(0.01\,\text{nm}\) to \(10\,\text{nm}\). X-rays possess high energy and can penetrate most substances, including human tissue.
  • This makes X-rays invaluable in medical imaging, allowing us to see inside the body without surgery.
  • X-ray astronomy allows scientists to observe celestial phenomena in greater detail.
  • Safety measures are crucial during X-ray usage to prevent unnecessary exposure to their high energy.
X-rays have revolutionized diagnostics and our understanding of the universe.

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Most popular questions from this chapter

A proton is accelerated from rest through a potential difference of \(100 \mathrm{MV}\). Find the proton's kinetic energy and rest energy What's the ratio of kinetic to rest energy?

If the speed of a spacecraft if its clock, relative to clocks on Earth, runs at a rate (a) \(90 \%\) as fast; (b) \(60 \%\) as fast; (c) \(10 \%\) as fast.

A spaceship approaching Earth at \(0.5 c\) fires a particle beam toward Earth. The particles move at \(0.6 c\) relative to the spaceship. At what speed does Earth receive the particles? (a) \(0.75 c\) (b) \(0.8 c^{*},\) (c) \(0.85 c\); (d) 0.90 c.

A relativistic train with proper length \(80 \mathrm{~m}\) passes through a tunnel with a proper length \(75 \mathrm{~m}\) (a) If the train's speed is \(0.45 c\) is the train ever completely inside the tunnel? (b) At this speed, how much time elapses for a cross-country trip of \(5000 \mathrm{~km},\) as measured on the train?

A spacecraft heading to the outer solar system at \(0.20 c\) sends radio signals back to Earth. The spacecraft transmits at a wavelength of \(2.75 \mathrm{~cm}\). What's the wavelength received on Earth? (a) \(3.37 \mathrm{~cm}\) (b) \(3.75 \mathrm{~cm}\) (c) \(4.13 \mathrm{~cm}\) (d) \(5.50 \mathrm{~cm}\)
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