Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Question
![**Understanding Light and Wavelengths: A Problem Example**
**Question:**
What is the wavelength of blue light that has a frequency of \(6.27 \times 10^{14} \, \text{s}^{-1}\)? (Given: \(c = 3.00 \times 10^8 \, \text{m/s}\))
**Solution:**
To find the wavelength (\(\lambda\)) of light, we use the formula:
\[
\lambda = \frac{c}{f}
\]
where:
- \(c\) is the speed of light, \(3.00 \times 10^8 \, \text{m/s}\)
- \(f\) is the frequency of light, \(6.27 \times 10^{14} \, \text{s}^{-1}\)
The wavelength is calculated as follows:
\[
\lambda = \frac{3.00 \times 10^8}{6.27 \times 10^{14}}
\]
This will give the wavelength in meters, which can be converted to nanometers (1 m = 1 billion nm).
**Interactive Component:**
- Below the question, there's a numerical keypad to input the answer. The display shows a placeholder for entering a number in scientific notation (e.g., \(5.9 \times 10^{14}\) nm).
- Buttons include numbers (0-9), +/- for sign adjustment, a decimal point, a backspace/delete button (⟵), and a button to input scientific notation (\(x10^{\Box}\)).
**Practical Application:**
Using this method allows students to understand the relationship between frequency and wavelength, essential concepts in physics, particularly in optics and electromagnetic theory.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F163dc8ba-851f-4764-b6aa-51728fdb757c%2F27417f93-b83b-4650-b898-1244432d3858%2Ftmy0an_processed.png&w=3840&q=75)
Transcribed Image Text:**Understanding Light and Wavelengths: A Problem Example**
**Question:**
What is the wavelength of blue light that has a frequency of \(6.27 \times 10^{14} \, \text{s}^{-1}\)? (Given: \(c = 3.00 \times 10^8 \, \text{m/s}\))
**Solution:**
To find the wavelength (\(\lambda\)) of light, we use the formula:
\[
\lambda = \frac{c}{f}
\]
where:
- \(c\) is the speed of light, \(3.00 \times 10^8 \, \text{m/s}\)
- \(f\) is the frequency of light, \(6.27 \times 10^{14} \, \text{s}^{-1}\)
The wavelength is calculated as follows:
\[
\lambda = \frac{3.00 \times 10^8}{6.27 \times 10^{14}}
\]
This will give the wavelength in meters, which can be converted to nanometers (1 m = 1 billion nm).
**Interactive Component:**
- Below the question, there's a numerical keypad to input the answer. The display shows a placeholder for entering a number in scientific notation (e.g., \(5.9 \times 10^{14}\) nm).
- Buttons include numbers (0-9), +/- for sign adjustment, a decimal point, a backspace/delete button (⟵), and a button to input scientific notation (\(x10^{\Box}\)).
**Practical Application:**
Using this method allows students to understand the relationship between frequency and wavelength, essential concepts in physics, particularly in optics and electromagnetic theory.
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