Problem 19 - Chemistry: The Central Science in SI Units, Global Edition

Question

(a) What is the frequency of radiation that has a wavelength of \(10 \mu \mathrm{m}\), about the size of a bacterium? \((\mathbf{b})\) 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 \(50.0 \mu \mathrm{s} ?\)

Step-by-Step Solution

Answer

(a) The frequency of radiation with a wavelength of \(10 \mu\mathrm{m}\) is \(3.00 \times 10^{13} \mathrm{s}^{-1}\). (b) The wavelength of radiation with a frequency of \(5.50 \times 10^{14}\mathrm{s}^{-1}\) is \(5.45 \times 10^{-7} \mathrm{m}\). (c) Only the radiation in part (b) would be visible to the human eye. (d) Electromagnetic radiation travels \(15,000 \mathrm{m}\) in \(50.0 \mu\mathrm{s}\).

Step 1: (Step 1: Finding the frequency of radiation given the wavelength)

To find the frequency of radiation, we will use the equation \(c = \lambda\nu\), where \(c\) is the speed of light, \(\lambda\) is the wavelength, and \(\nu\) is the frequency. We can rearrange this equation to find the frequency: \(\nu = \frac{c}{\lambda}\) Using the given wavelength, \(10 \mu\mathrm{m}\), we can plug it into the equation: \(\nu = \frac{3.00\times10^8 \mathrm{m/s}}{10 \times 10^{-6}\mathrm{m}} = 3.00\times10^{13} \mathrm{s}^{-1}\)

Step 2: (Step 2: Finding the wavelength of radiation given the frequency)

Using the frequency given in the problem, \(5.50 \times 10^{14} \mathrm{s}^{-1}\), we will use the same equation \(\lambda = \frac{c}{\nu}\) to find the wavelength: \(\lambda = \frac{3.00\times10^8 \mathrm{m/s}}{5.50 \times 10^{14}\mathrm{s}^{-1}} = 5.45\times10^{-7} \mathrm{m}\)

Step 3: (Step 3: Determining if the radiation is visible to the human eye)

The visible light spectrum has a wavelength range of approximately \(400 \times 10^{-9} \mathrm{m}\) to \(700 \times 10^{-9} \mathrm{m}\). Comparing the values calculated in steps 1 and 2: For step 1: \(\lambda = 10 \times 10^{-6} \mathrm{m}\), which is not within the visible spectrum. For step 2: \(\lambda = 5.45\times10^{-7} \mathrm{m}\), which is within the visible spectrum. So, only the radiation in part (b) would be visible to the human eye.

Step 4: (Step 4: Finding the distance traveled by electromagnetic radiation in a given time)

To find the distance traveled by electromagnetic radiation in the given time, we'll use the equation \(d = ct\), where \(d\) is the distance, \(c\) is the speed of light, and \(t\) is the time. Given the time, \(50.0 \mu\mathrm{s}\), we can plug it into the equation: \(d = (3.00\times10^8 \mathrm{m/s})(50.0\times10^{-6} \mathrm{s}) = 15,000 \mathrm{m}\)

Key Concepts

Frequency

Frequency refers to how many waves pass a point in one second. It is measured in hertz (Hz).
Each wave in electromagnetic radiation is characterized by its frequency. Higher frequencies mean more waves pass by in a given time.
The relationship between frequency () and wavelength () is given by:

• The formula: \( c = \lambdau \)
• \( c \) is the speed of light (\(3.00 \times 10^8 \) m/s).
• \( \lambda \) is the wavelength.
• \( u \) (nu) represents frequency in Hertz.

For instance, if you know the wavelength of a radiation, you can find its frequency using \( u = \frac{c}{\lambda} \) as shown in the original solution.

Wavelength

Wavelength is the distance between two consecutive peaks of a wave. It is usually measured in meters (m).
Wavelengths can vary greatly across the electromagnetic spectrum, from very short gamma rays to long radio waves.

• Shorter wavelengths correspond to higher frequencies.
• Longer wavelengths correspond to lower frequencies.

Using the formula \( \lambda = \frac{c}{u} \), you can determine the wavelength if the frequency is known. This relationship allows us to understand the characteristics of different electromagnetic waves.

Visible Spectrum

The visible spectrum is a small portion of the electromagnetic spectrum that can be seen by the human eye. It includes a range of about 400 to 700 nanometers (nm).
The visible light spectrum contains the colors of the rainbow, red having the longest wavelength and violet the shortest.

• Wavelengths shorter than 400 nm fall into ultraviolet.
• Wavelengths longer than 700 nm fall into infrared.

For example, the original solution found that a wavelength of \(5.45 \times 10^{-7} \mathrm{m}\) (or 545 nm) falls within this visible range and could be perceived as colored light.

Speed of Light

The speed of light is a fundamental constant in physics, denoted by \( c \). It is the speed at which all electromagnetic waves, including light, travel in a vacuum.
The speed of light is approximately \(3.00 \times 10^8 \) meters per second (m/s).

• Light speed is used to calculate distances in space and time.
• It's crucial in formulas that relate frequency and wavelength, due to the equation \( c = \lambdau \).

For practical applications, like the original problem, you can use this constant to determine how far light travels in a given time, such as the distance calculated over 50 microseconds.