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

Short Answer

Expert verified
(a) Frequencies: (i) \(6.0 \times 10^4\) Hz, (ii) \(6.0 \times 10^{13}\) Hz, (iii) \(6.0 \times 10^{16}\) Hz. (b) Wavelengths: (i) 4.62 x 10^-14 m (4.62 x 10^-5 nm), (ii) 508.47 m (5.0847 x 10^11 nm).

Step by step solution

01

Understanding the Relationship Between Frequency and Wavelength

The relationship between the frequency \(f\) and wavelength \(\lambda\) of electromagnetic waves is given by the formula:\[ c = \lambda \cdot f \]where \(c\) is the speed of light in a vacuum, approximately \(3 \times 10^8\) m/s.
02

Calculate Frequency for Wavelength of 5.0 km

Given the wavelength \(\lambda = 5.0\) km, convert it to meters first, \(5.0 \text{ km} = 5000 \text{ m}\). Now, use the relation:\[ f = \frac{c}{\lambda} = \frac{3 \times 10^8 \text{ m/s}}{5000 \text{ m}} = 6.0 \times 10^4 \text{ Hz} \]
03

Calculate Frequency for Wavelength of 5.0 micrometers

Convert the wavelength \(\lambda = 5.0 \mu m\) to meters, \(5.0 \mu \text{m} = 5.0 \times 10^{-6} \text{ m}\). Using the formula:\[ f = \frac{3 \times 10^8 \text{ m/s}}{5.0 \times 10^{-6} \text{ m}} = 6.0 \times 10^{13} \text{ Hz} \]
04

Calculate Frequency for Wavelength of 5.0 nm

Convert the wavelength \(\lambda = 5.0 \text{ nm}\) to meters, \(5.0 \text{ nm} = 5.0 \times 10^{-9} \text{ m}\). Using the formula:\[ f = \frac{3 \times 10^8 \text{ m/s}}{5.0 \times 10^{-9} \text{ m}} = 6.0 \times 10^{16} \text{ Hz} \]
05

Calculate Wavelength for Gamma Rays Frequency 6.50 x 10^21 Hz

Using the frequency \(f = 6.50 \times 10^{21} \text{ Hz}\), the wavelength \(\lambda\) is:\[ \lambda = \frac{c}{f} = \frac{3 \times 10^8 \text{ m/s}}{6.50 \times 10^{21} \text{ Hz}} = 4.62 \times 10^{-14} \text{ m} \]To convert to nanometers, \(\lambda = 4.62 \times 10^{-5} \text{ nm}\).
06

Calculate Wavelength for AM Station Frequency 590 kHz

Convert the frequency from kHz to Hz, \(590 \text{ kHz} = 590 \times 10^3 \text{ Hz}\). The wavelength \(\lambda\) is:\[ \lambda = \frac{c}{f} = \frac{3 \times 10^8 \text{ m/s}}{590 \times 10^3 \text{ Hz}} = 508.47 \text{ m} \]Converting to nanometers, \(\lambda = 5.0847 \times 10^{11} \text{ nm}\).

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Key Concepts

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

Wavelength and Frequency Relationship
Electromagnetic waves exhibit a fascinating reciprocal relationship between wavelength and frequency. This relationship is governed by the equation \(c = \lambda \cdot f\), where \(c\) is the speed of light, \(\lambda\) is the wavelength, and \(f\) is the frequency. Simply put, if you know any two of these variables, you can easily determine the third one.
The speed of light (denoted by \(c\)) is constant, meaning that if the wavelength increases, the frequency must decrease, and vice versa.
  • A longer wavelength corresponds to a lower frequency.
  • A shorter wavelength corresponds to a higher frequency.
This inverse relationship helps explain why different types of electromagnetic waves, like radio waves and gamma rays, have such varying properties. Understanding this principle is foundational when studying the behavior of electromagnetic waves.
Speed of Light Constant
The speed of light in a vacuum is an unfaltering constant: \(3 \times 10^8\) meters per second (m/s). This constant is crucial because it connects wavelength and frequency in the equation \(c = \lambda \cdot f\).
Wherever electromagnetic waves travel, this speed remains unchangeable when in a vacuum. However, it can be slightly reduced when passing through different media, like air or water. Despite these alterations in speed in various media, the relationship between wavelength and frequency stays intact because both are affected proportionally to maintain the product equal to the speed of light.
Knowing this constant allows for simplified calculations across the electromagnetic spectrum, from gamma rays to radio waves, making it an essential component of physics and engineering.
Gamma Rays
Gamma rays are a type of electromagnetic radiation with incredibly high frequencies and correspondingly short wavelengths. They fall at the extreme end of the electromagnetic spectrum. For instance, waves with a frequency as high as \(6.50 \times 10^{21}\) Hz are classified as gamma rays.
  • Gamma rays can travel through most materials, making them useful in medical imaging and cancer treatment radiation therapy.
  • They originate from radioactive decay and celestial phenomena.
Conversion between their high frequency and their incredibly small wavelength requires precise calculation. In most practical applications, we use the speed of light constant in \(\lambda = \frac{c}{f}\) to determine their minute wavelength measurements, as demonstrated in scientific studies and technologies that employ gamma rays.
Radio Waves
Radio waves are part of the electromagnetic spectrum that have the longest wavelength and the lowest frequency. Their frequencies typically range from thousands (kilohertz) to millions (megahertz) of cycles per second. These characteristics make them suitable for wireless communication such as AM and FM radio broadcasting, television, and mobile phones.
For example, an AM radio wave might have a frequency of \(590\) kHz, which can be converted to a wavelength by rearranging the light-speed equation to \(\lambda = \frac{c}{f}\). Understanding how to perform this conversion is integral to designing and analyzing communication systems that use radio waves.
Unit Conversion in Physics
Physics often requires converting between different units to solve problems accurately, particularly in calculating electromagnetic wave properties. One common conversion is from kilometers or various sub-units to meters, as the standard unit for wavelength in physics problems. For example:
  • Convert kilometers (km) to meters (m): multiply by \(1000\).
  • Convert micrometers (\(\mu m\)) to meters: multiply by \(10^{-6}\).
  • Convert nanometers (nm) to meters: multiply by \(10^{-9}\).
Frequency units might also change from kilohertz (kHz) to hertz (Hz) by multiplying by \(10^3\). These conversions ensure uniformity and prevent calculation errors. Familiarity with these unit conversions is an essential skill in physics academia and industry.

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Most popular questions from this chapter

\(\bullet\) Ultraviolet radiation. There are two categories of ultraviolet light. Ultraviolet A (UVA) has a wavelength ranging from 320 \(\mathrm{nm}\) to 400 nm. It is not so harmful to the skin and is necessary for the production of vitamin D. UVB, with a wavelength between 280 \(\mathrm{nm}\) and \(320 \mathrm{nm},\) is much more dangerous, because it causes skin cancer. (a) Find the frequency ranges of UVA and UVB. (b) What are the ranges of the wave numbers for UVA and UVB?

\(\bullet\) Visible light. The wavelength of visible light ranges from 400 nm to 700 nm. Find the corresponding ranges of this light's (a) frequency, (b) angular frequency, (c) wave number.

\(\bullet\) The indices of refraction for violet light \((\lambda=400 \mathrm{nm})\) and red light \((\lambda=700 \mathrm{nm})\) in diamond are 2.46 and \(2.41,\) respectively. A ray of light traveling through air strikes the diamond surface at an angle of \(53.5^{\circ}\) to the normal. Calculate the angular separation between these two colors of light in the refracted ray.

\(\bullet$$\bullet\) Heart sonogram. Physicians use high-frequency \((f=\) 1 MHz to 5 MHz) sound waves, called ultrasound, to image internal organs. The speed of these ultrasound waves is 1480 \(\mathrm{m} / \mathrm{s}\) in muscle and 344 \(\mathrm{m} / \mathrm{s}\) in air. We define the index of refraction of a material for sound waves to be the ratio of the speed of sound in air to the speed of sound in the material. Snell's law then applies to the refraction of sound waves. (a) At what angle from the normal does an ultrasound beam enter the heart if it leaves the lungs at an angle of \(9.73^{\circ}\) from the normal to the heart wall? (Assume that the speed of sound in the lungs is 344 \(\mathrm{m} / \mathrm{s} .\) ) (b) What is the critical angle for sound waves in air incident on muscle?

\(\bullet\) A beam of light strikes a sheet of glass at an angle of \(57.0^{\circ}\) with the normal in air. You observe that red light makes an angle of \(38.1^{\circ}\) with the normal in the glass, while violet light makes a \(36.7^{\circ}\) angle. (a) What are the indexes of refraction of this glass for these colors of light? (b) What are the speeds of red and violet light in the glass?

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