Problem 19 - Chemistry The Central Science In Si Units

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?

Step-by-Step Solution

Answer

(a) Frequency is \(3.00 \times 10^{13}\) Hz; (b) Wavelength is 545 nm; (c) Part (b) is visible; (d) Distance is 15.0 km.

Step 1: Calculate Frequency from Wavelength

To find the frequency (ufactured) of radiation from its wavelength (ufactured), use the equation \( c = \lambda \cdot u \), where \( c \) is the speed of light \( (3.00 \times 10^8 \text{ m/s}) \). For part (a) with \( \lambda = 10 \mu\text{m} = 10^{-5} \text{ m}\),\[ u = \frac{c}{\lambda} = \frac{3.00 \times 10^8 \text{ m/s}}{10^{-5} \text{ m}} = 3.00 \times 10^{13} \text{ Hz} \]

Step 2: Calculate Wavelength from Frequency

To find the wavelength (ufactured) of radiation from its frequency (ufactured), use the same equation \( c = \lambda \cdot u \). For part (b) with \( u = 5.50 \times 10^{14} \text{ Hz} \),\[ \lambda = \frac{c}{u} = \frac{3.00 \times 10^8 \text{ m/s}}{5.50 \times 10^{14} \text{ Hz}} = 5.45 \times 10^{-7} \text{ m} = 545 \text{ nm} \]

Step 3: Determine Visibility of Radiation

Visible light has a wavelength range of approximately 380 nm to 750 nm. From step 1, the radiation has a frequency corresponding to a wavelength of 10 µm, which is far longer than this range and thus, not visible. From step 2, the radiation has a wavelength of 545 nm, which falls within this range and is visible.

Step 4: Calculate Distance Traveled by Radiation

The distance (ufactured) traveled by electromagnetic radiation is given by the formula \( d = c \cdot t \), where \( t \) is the time. For part (d) with \( t = 50.0 \mu\text{s} = 50.0 \times 10^{-6} \text{s} \),\[ d = 3.00 \times 10^8 \text{ m/s} \times 50.0 \times 10^{-6} \text{ s} = 1.50 \times 10^4 \text{ m} = 15.0 \text{ km} \]

Key Concepts

Wavelength and Frequency Relationship

In the world of electromagnetic radiation, wavelength and frequency share a special connection. Imagine them like a seesaw; when one goes up, the other goes down. This relationship can be expressed through the equation:

• \( c = \lambda \cdot u \)

where \( c \) is the constant speed of light (\(3.00 \times 10^8 \text{ m/s}\)), \( \lambda \) is the wavelength, and \( u \) is the frequency.

If you've ever played with a slinky, you've seen how waves can change. When you stretch the slinky, the waves become longer and fewer. Electromagnetic radiation works similarly! The longer the wavelength (\( \lambda \)), the lower the frequency (\( u \)), and vice versa.

The answer to part (a) in our original exercise used this equation to find the frequency of radiation with a wavelength of \(10 \mu m\), which is critical in understanding how different forms of radiation behave and interact with the world around us.

Visible Spectrum

The visible spectrum is a slice of the electromagnetic spectrum that our eyes can see. This spectrum contains all the colors of the rainbow, ranging from violet to red. Light in this spectrum has wavelengths from about 380 nanometers to 750 nanometers.

In part (b) of our exercise, we explored a wavelength of 545 nanometers, associating it with green light. Since 545 nm falls within the visible spectrum, it is visible to our eyes.

Contrast this with the result from part (a), where the wavelength, at 10 micrometers, was far too long and falls into the infrared region, invisible to the human eye.

• Wavelength < 380 nm - Ultraviolet (invisible)
• Wavelength 380 nm to 750 nm - Visible light
• Wavelength > 750 nm - Infrared (invisible)

These distinctions make the visible spectrum a crucial concept in understanding both natural phenomena and technological applications like cameras and light sensors.

Speed of Light

The speed of light, an astonishing \(3.00 \times 10^8 \text{ m/s}\), is a universal constant that defines how fast electromagnetic waves, like light, travel in a vacuum. This speed is crucial for calculations involving wavelength and frequency and is part of the reason light is such a fascinating area of study.

In the problem in part (d), using this constant helps us understand how far light can travel in a given amount of time. For example, in 50 microseconds, light travels a distance of 15 kilometers.

Understanding the speed of light aids in many areas:

• Determining time delays in communication (like satellite signals)
• Exploring the vastness of space and distances between stars
• Enhancing technologies like fiber optics, which rely on the quick transmission of data

Knowing this, we appreciate how important the speed of light is, not just in calculations, but in comprehending the universe itself.