Problem 19 - Chemistry: The Central Science

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 \(50.0 \mu \mathrm{s}\) ?

Step-by-Step Solution

Answer

(a) The frequency of radiation with a wavelength of \(10 \mu m\) is \(f = 3.00 \times 10^{14} Hz\). (b) The wavelength of radiation with a frequency of \(5.50 \times 10^{14} Hz\) is \(\lambda = 5.45 \times 10^{-7} m = 545 nm\). (c) The radiation in part (b) would be visible to the human eye, while the radiation in part (a) would not be visible. (d) The distance electromagnetic radiation travels in \(50.0 \mu s\) is \(d = 15,000 m\).

Step 1: Calculate the frequency of radiation

Recall the formula relating the speed of light (c), frequency (f), and wavelength (λ): \(c = fλ\) We are given the wavelength, λ, and we need to find the frequency, f. First, rewrite the equation to solve for frequency: \(f = \frac{c}{\lambda}\) Now, plug in the given values of the speed of light \(c = 3.00 \times 10^8 m/s\) and the wavelength \(λ = 10\mu m = 10 \times 10^{-6}m\): \(f = \frac{3.00 \times 10^8 m/s}{10 \times 10^{-6}m}\) Calculate the result: \(f = 3.00 \times 10^{14} Hz\)

Step 2: Calculate the wavelength of radiation

For this step, we're given the frequency and need to find the wavelength. We can still use the formula from step 1: \(c = f\lambda\) Now, we need to rewrite the equation to solve for the wavelength: \(\lambda = \frac{c}{f}\) Now, plug in the given values of the speed of light \(c = 3.00 \times 10^8 m/s\) and the frequency \(f = 5.50 \times 10^{14} Hz\): \(\lambda = \frac{3.00 \times 10^8 m/s}{5.50 \times 10^{14} Hz}\) Calculate the result: \(\lambda = 5.45 \times 10^{-7} m\)

Step 3: Discuss the visibility of the radiations for the human eye

For radiation to be visible to the human eye, it needs to be in the visible light spectrum, with wavelengths ranging from approximately \(380 nm\) to \(750 nm\). For part (a), we found a wavelength of \(10 \mu m = 10,000 nm\), which is outside the range of visible light. For part (b), we found a wavelength of \(5.45 \times 10^{-7}m = 545 nm\), which falls within the range of visible light. Therefore, the radiation in part (a) would not be visible to the human eye, while the radiation in part (b) would be visible.

Step 4: Calculate distance traveled by electromagnetic radiation

For this step, we are given the time and need to find the distance electromagnetic radiation travels. We can use the formula for distance: \(d = ct\) Plug in the given values of the speed of light \(c = 3.00 \times 10^8 m/s\) and the time \(t = 50.0 \mu s = 50.0 \times 10^{-6} s\): \(d = 3.00 \times 10^8 m/s \times 50.0 \times 10^{-6} s\) Calculate the result: \(d = 15,000 m\)

Key Concepts

Frequency of Radiation

In electromagnetic radiation calculations, the frequency of radiation is a fundamental concept that refers to the number of waves that pass a given point per second. It is measured in hertz (Hz). Higher frequencies correspond to shorter wavelengths and more energy per photon, a concept that's crucial when examining different kinds of electromagnetic waves, from radio waves to gamma rays.

To find the frequency given the wavelength, one uses the formula:
\( f = \frac{c}{\lambda} \)
Here, \( c \) is the speed of light in a vacuum, and \( \lambda \) represents the wavelength. For a wavelength of about the size of a bacterium (\(10 \mu\mathrm{m}\)), we calculated the frequency using the speed of light (\(3.00 \times 10^8 m/s\)), which gives us \(3.00 \times 10^{14} Hz\). This high number is indicative of the rapid oscillations of such radiation.

Wavelength of Radiation

Conversely, the wavelength of radiation is the distance between successive peaks (or troughs) of a wave. It is usually denoted by \( \lambda \) and is measured in meters (m). The wavelength is inversely related to the frequency, which means that as the wavelengths get longer, the frequencies get lower.

To determine the wavelength from the frequency, invert the formula used for frequency: \( \lambda = \frac{c}{f} \)
When we took the frequency of \(5.50 \times 10^{14} \mathrm{~s}^{-1}\), and with \(c\) being a constant, we found the wavelength to be around \(5.45 \times 10^{-7} m\), which is within the visible light spectrum. This is a good demonstration of how, by knowing one attribute (frequency or wavelength), you can readily calculate the other.

Visible Light Spectrum

The visible light spectrum is the portion of the electromagnetic spectrum that is detectable by the human eye. This part of the spectrum ranges from wavelengths of approximately \(380 nm\) to \(750 nm\). Colors ranging from violet to red can be found within this range, with different wavelengths corresponding to different colors.

When comparing the wavelengths attained in our problem, \(10 \mu\mathrm{m}\) far exceeds this range and hence is not visible, whereas the \(545 nm\) fits right into the spectrum and would be perceived as green light. Understanding the visible light spectrum is essential not just for scientific studies, but also for applications in art and communications, among others.

Speed of Light

As a universal physical constant, the speed of light (denoted as \(c\)) in a vacuum is about \(3.00 \times 10^8 m/s\). It's the fastest speed at which all energy, matter, and information in the universe can travel. It’s critical in many areas of physics and is the backbone of the theory of relativity.

In electromagnetic equations, the speed of light connects frequency and wavelength, allowing us to solve for one when the other is known. This constancy provides a stable foundation for calculations across astrophysics, optics, and beyond. Whether calculating cosmic distances or designing fiber-optic cables, the speed of light is a key parameter.

Calculating Distance Electromagnetic Radiation Travels

When it comes to figuring out how far electromagnetic radiation travels within a certain time frame, we look to the simple formula for distance: \( d = ct \)
Here, \( d \) is the distance, \( c \) the speed of light, and \( t \) the time in seconds. This calculation is straightforward yet immensely useful, such as when determining how far light travels from stars or communicating with satellites. For our problem, the radiation traveled \(15,000 m\) in just \(50.0 \mu \mathrm{s}\), illustrating the incredible speed at which light moves through space. This fundamental aspect of electromagnetic waves ties in with everything from time measurements in GPS systems to analyzing light from the early universe.