Question

Without doing detailed calculations, arrange the following electromagnetic radiation sources in order of increasing frequency: (a) a red traffic light, (b) a \(91.9 \mathrm{MHz}\) radio transmitter, (c) light with a frequency of \(3.0 \times 10^{14} \mathrm{s}^{-1}\) (d) light with a wavelength of \(49 \mathrm{nm}\).

Short answer

The arranged order of increasing frequency is: a 91.9 MHz radio transmitter (b), light with a frequency of \(3.0 \times 10^{14} s^{-1}\) (c), a red traffic light (a), and light with a wavelength of \(49 nm\) (d).

Step-by-step solution

Frequency of a radio transmitter

Given that (b) a radio transmitter operates at \(91.9 \mathrm{MHz}\), which is equal to \(91.9 \times 10^{6} \mathrm{s^{-1}}\).

Frequency of red light

Light from a red traffic light (a) is at the lower frequency end of the visible spectrum, with a frequency approximately around \(4.3 \times 10^{14} \mathrm{s^{-1}}\).

Frequency of light

For (c), we are already given light with a frequency of \(3.0 \times 10^{14} \mathrm{s}^{-1}\).

Frequency with a wavelength of 49 nm

Using the wave equation and the value of the speed of light \( v = 3.0 \times 10^8 m/s \), the light of wavelength \(49 \mathrm{nm}\) (d) has a frequency \(f = v/\lambda = (3.0 \times 10^{8} m/s) / (49 \times 10^{-9} m) = 6.12 \times 10^{15} \mathrm{s^{-1}}\).

Arranging in order of increasing frequency

Arranging these in order of increasing frequency: a radio transmitter (b), light of frequency \(3.0 \times 10^{14} s^{-1}\) (c), red traffic light (a), and light of wavelength \(49 nm\) (d).

Key concepts

Frequency Calculation

The concept of frequency calculation is crucial when dealing with electromagnetic waves. Frequency, denoted as \( f \), tells you how many wave cycles pass through a point in one second. The unit for frequency is Hertz (\( Hz \)), which is equivalent to \( \text{s}^{-1} \).

In many cases, electromagnetic wave frequencies are given in megahertz (\( MHz \)), where \( 1 \text{ MHz} = 10^6 \text{ Hz} \). For instance, if a radio transmitter operates at \( 91.9 \text{ MHz} \), you can easily convert this to \( 91.9 \times 10^6 \text{ s}^{-1} \) or Hertz by multiplying by \( 10^6 \).

Understanding frequency allows you to make comparisons between different sources of electromagnetic radiation by determining how many cycles occur in any given time period. This way, you can assess the properties of the wave, such as energy, based on their frequency.

Wavelength and Frequency Relationship

Wavelength and frequency are interconnected through a fundamental relationship in physics. They are linked by the speed of light \( c \), given by \( c = 3.0 \times 10^8 \text{ m/s} \). This relationship is expressed in the wave equation:

\[ f = \frac{c}{\lambda} \]

Here, \( f \) represents frequency, and \( \lambda \) stands for wavelength. The equation shows that frequency is inversely related to wavelength; as the wavelength decreases, the frequency increases and vice versa.

For example, light with a wavelength of \( 49 \text{ nm} \) can have its frequency calculated using the wave equation. Converting \( 49 \text{ nm} \) to meters (\( 1 \text{ nm} = 10^{-9} \text{ m} \)), you find that the frequency equals:

\[ f = \frac{3.0 \times 10^8 \text{ m/s}}{49 \times 10^{-9} \text{ m}} = 6.12 \times 10^{15} \text{ s}^{-1} \]

This fundamental relationship helps to understand how electromagnetic radiation is characterized and how different wavelengths fit into the electromagnetic spectrum.

Visible Spectrum

The visible spectrum is a tiny portion of the electromagnetic spectrum that the human eye can detect. It ranges in wavelengths from approximately 400 nm (violet) to 700 nm (red). This range corresponds to frequency values of roughly \( 4.3 \times 10^{14} \text{ Hz} \) (red) to \( 7.5 \times 10^{14} \text{ Hz} \) (violet).

The visible spectrum contains all the colors that we perceive, from red to violet. Within this range, red light has the longest wavelength and the lowest frequency, while violet light has the shortest wavelength and the highest frequency.

Understanding the visible spectrum helps in identifying how light behaves and interacts with materials. It helps in knowing why we see different colors and how these colors are related to their respective wavelengths and frequencies. For tasks such as identifying occurrences like the light of a red traffic light, this knowledge assists in contextualizing where different sources fall within the spectrum.