Question

(a) What is the frequency of radiation that has a wavelength of \(10 \mu \mathrm{m}\), about the size of a bacterium? (b) What is the wavelength of radiation that has a frequency of \(5.50 \times 10^{14} \mathrm{~s}^{-1} ?\) (c) Would the radiations in part (a) or part (b) be visible to the human eye? (d) What distance does electromagnetic radiation travel in \(50.0 \mu\) s?

Short answer

(a) The frequency of radiation with a wavelength of \(10 \mu m\) is \(3 \times 10^{13} s^{-1}\). (b) The wavelength of radiation with a frequency of \(5.50 \times 10^{14} s^{-1}\) is \(5.45\times 10^{-7} m\). (c) Radiation in part (a) is not visible, while radiation in part (b) is visible to the human eye. (d) Electromagnetic radiation travels a distance of \(1.5 \times 10^4 m\) in \(50.0 \mu s\).

Step-by-step solution

Find the frequency in part (a)

We are given the wavelength of radiation as \(10\mu m\), and we need to find its frequency. The equation relating frequency, wavelength and the speed of light (c) is: \[ c = \text{frequency} \times\text{wavelength}\] where c is the speed of light and is approximately equal to \(3 \times 10^8 m/s\). Rearrange this equation to solve for the frequency: \[ \text{frequency} = \frac{c}{\text{wavelength}}\] Now we can plug in the values and calculate the frequency: \[ \text{frequency} = \frac{3 \times 10^8 m/s}{10\mu m} = \frac{3 \times 10^8 m/s}{10\times 10^{-6}m}\]

Calculate the frequency in part (a)

Perform the calculation from the last step: \[ \text{frequency} = \frac{3 \times 10^8 m/s}{10\times 10^{-6}m} = 3 \times 10^{13} s^{-1}\] So, the frequency of radiation with a wavelength of \(10\mu m\) is \(3 \times 10^{13} s^{-1}\). #a# What is the frequency of radiation that has a wavelength of $10 \mu \mathrm{m}$, about the size of a bacterium? Frequency = \(3 \times 10^{13} \;s^{-1}\)

Find the wavelength in part (b)

Now we are given the frequency of radiation as \(5.50 \times 10^{14} s^{-1}\), and we need to find its wavelength. The equation relating frequency, wavelength and the speed of light (c) is: \[ c = \text{frequency} \times \text{wavelength}\] Rearrange this equation to solve for the wavelength: \[ \text{wavelength} = \frac{c}{\text{frequency}}\] Now we can plug in the values and calculate the wavelength: \[ \text{wavelength} = \frac{3 \times 10^8 m/s}{5.50 \times 10^{14} s^{-1}}\]

Calculate the wavelength in part (b)

Perform the calculation from the last step: \[ \text{wavelength} = \frac{3 \times 10^8 m/s}{5.50 \times 10^{14} s^{-1}} = 5.45\times 10^{-7} m\] So, the wavelength of radiation with a frequency of \(5.50 \times 10^{14} s^{-1}\) is \(5.45\times 10^{-7} m\). #b# What is the wavelength of radiation that has a frequency of \(5.50 \times 10^{14} \mathrm{~s}^{-1} ?\) Wavelength = \(5.45\times 10^{-7} m\)

Check whether the radiations in part (a) or part (b) are visible to the human eye

The range of wavelengths that are visible to the human eye is from approximately \(4\times10^{-7} m\) to \(7\times10^{-7} m\). We can now check if the wavelengths in part (a) and (b) fall within this range. For part (a), the wavelength is \(10\mu m\) which is equal to \(10\times10^{-6}m\). This is outside the range of visible light. For part (b), the wavelength is \(5.45\times 10^{-7} m\). This is within the range of visible light. #c#Would the radiations in part (a) or part (b) be visible to the human eye? Radiation in part (a) is not visible, while radiation in part (b) is visible to the human eye.

Calculate the distance traveled by electromagnetic radiation in the given time

We are given the time \(50.0\mu s\), and we need to find the distance traveled by the electromagnetic radiation. Since the speed of light (c) is constant, we can use the following equation: \[ \text{distance} = c \times \text{time} \] Now we can plug in the values and calculate the distance: \[ \text{distance} = (3 \times 10^8 m/s) \times (50.0\mu s) = (3 \times 10^8 m/s) \times (50\times10^{-6}s)\]

Calculate the distance in part (d)

Perform the calculation from the last step: \[ \text{distance} = (3 \times 10^8 m/s) \times (50\times10^{-6}s) = 1.5 \times 10^4 m\] So, the electromagnetic radiation travels a distance of \(1.5 \times 10^4 m\) in \(50.0\mu s\). #d#What distance does electromagnetic radiation travel in \(50.0 \mu\) s? Distance = \(1.5 \times 10^4 m\)

Key concepts

Wavelength and Frequency

Understanding the relationship between wavelength and frequency is essential in the study of electromagnetic radiation. These two properties are inversely related, meaning that as one increases, the other decreases.
Frequency, denoted as \( f \) and measured in hertz (Hz), refers to the number of wave cycles that pass a given point per second. Wavelength, often represented by \( \lambda \), is the distance between successive crests or troughs of a wave, measured in meters (m).
The speed of light (\( c \)) connects these two variables through the equation:

• \( c = f \times \lambda \)

This equation signifies that the product of frequency and wavelength equals the speed of light.
By rearranging this formula, we can determine either frequency or wavelength if the other variable and the speed of light are known. For example, to find frequency:

• \( f = \frac{c}{\lambda} \)

Similarly, to find wavelength, rearrange it to:

• \( \lambda = \frac{c}{f} \)

These equations are fundamental in identifying the properties of different electromagnetic waves.

Speed of Light

The speed of light, denoted by \( c \), is a constant that is central to the study of electromagnetic waves. In a vacuum, light travels at approximately \( 3 \times 10^8 \mathrm{m/s} \). This value is crucial when calculating the frequency or wavelength of electromagnetic radiation.
Since electromagnetic radiation consists of waves, the speed of light provides a necessary bridge between wavelength and frequency. Whenever calculations involve electromagnetic waves, the speed of light allows us to switch between these two properties using the simple equation:

• \( c = f \times \lambda \)

The constancy of the speed of light ensures that no matter where it travels, electromagnetic radiation maintains this speed in a vacuum, although it may vary slightly in different media (like air or water).
Apart from connecting frequency and wavelength, the speed of light is a cornerstone in physical theories, like Einstein's theory of relativity, which emphasizes its role as a universal constant essential for understanding the universe.

Visible Light Spectrum

The visible light spectrum is a portion of the electromagnetic spectrum that is visible to the human eye. It ranges approximately between wavelengths of \( 4\times10^{-7} \mathrm{m} \) (400 nm) and \( 7\times10^{-7} \mathrm{m} \) (700 nm).
These wavelengths correspond to frequencies between roughly \( 4.29 \times 10^{14} \mathrm{Hz} \) and \( 7.50 \times 10^{14} \mathrm{Hz} \). Within this spectrum, different wavelengths correspond to different colors. For example:

• Violet: around 400 nm
• Blue: around 450 nm
• Green: around 520 nm
• Yellow: around 570 nm
• Orange: around 590 nm
• Red: around 620 nm

These colors form the rainbow we can visualize, with violet having the shortest wavelength and red the longest. The visible spectrum is just a small part of the entire electromagnetic spectrum.
Understanding this range is essential, especially when determining whether given radiation is visible, as seen in problems like those in the original exercise. Different wavelengths or frequencies outside this range aren't perceivable as visible light by humans.

Electromagnetic Spectrum

The electromagnetic spectrum encompasses all types of electromagnetic radiation, varying by frequency and wavelength. This spectrum is broadly divided into several categories based on the energy, frequency, and wavelength of the waves.
Some primary categories include:

• Radio Waves: Longest wavelength and lowest frequency. Used in communication like radio and television broadcasts.
• Microwaves: Used in cooking and certain communication technologies.
• Infrared: Felt as heat and used in remote controls and some imaging technologies.
• Visible Light: Only part detectable by the human eye, spanning wavelengths from 400 to 700 nm.
• Ultraviolet: Beyond visible light, causes sunburn, and used in sterilization.
• X-Rays: High energy, used in medical imaging.
• Gamma Rays: Highest energy, produced in nuclear reactions, used in cancer treatments.

Understanding the electromagnetic spectrum is crucial as it illustrates how different types of radiation are connected through their frequency and wavelength. It gives insight into their various applications and effects. For instance, the concept of frequency and wavelength helps in determining which part of the spectrum a particular radiation belongs to, as seen in solving textbook problems.