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...
Related questions
Question
![**Expert Q&A**
---
**Question:** What is the energy of a photon of light with a wavelength of 469 μm?
---
This question involves calculating the energy of a photon based on its wavelength, which is 469 micrometers (μm). To find the energy, use the equation:
\[ E = \frac{h \cdot c}{\lambda} \]
Where:
- \( E \) is the energy,
- \( h \) is Planck's constant (\(6.626 \times 10^{-34} \, \text{J}\cdot\text{s}\)),
- \( c \) is the speed of light (\(3.00 \times 10^{8} \, \text{m/s}\)),
- \( \lambda \) is the wavelength in meters.
First, convert the wavelength from micrometers to meters:
\[ \lambda = 469 \, \mu\text{m} = 469 \times 10^{-6} \, \text{m} \]
Next, substitute the values into the equation:
\[ E = \frac{6.626 \times 10^{-34} \times 3.00 \times 10^{8}}{469 \times 10^{-6}} \]
Calculate to find the energy \( E \).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fce778a43-83cd-4b0c-a9c1-cb2530317a44%2F0894e875-b490-4c49-99d6-3f1c64369364%2Fwywkjt_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Expert Q&A**
---
**Question:** What is the energy of a photon of light with a wavelength of 469 μm?
---
This question involves calculating the energy of a photon based on its wavelength, which is 469 micrometers (μm). To find the energy, use the equation:
\[ E = \frac{h \cdot c}{\lambda} \]
Where:
- \( E \) is the energy,
- \( h \) is Planck's constant (\(6.626 \times 10^{-34} \, \text{J}\cdot\text{s}\)),
- \( c \) is the speed of light (\(3.00 \times 10^{8} \, \text{m/s}\)),
- \( \lambda \) is the wavelength in meters.
First, convert the wavelength from micrometers to meters:
\[ \lambda = 469 \, \mu\text{m} = 469 \times 10^{-6} \, \text{m} \]
Next, substitute the values into the equation:
\[ E = \frac{6.626 \times 10^{-34} \times 3.00 \times 10^{8}}{469 \times 10^{-6}} \]
Calculate to find the energy \( E \).
Expert Solution
Interpretation Introduction
Interpretation: The energy of a photon of light with a wavelength of is to be determined.
Concept Introduction: The energy of a photon is proportional to its frequency, as stated by the Planck's Einstein's equation,
Where,
E - the energy f the photon
h - Planck's constant =
v - The frequency of the photon
Step by step
Solved in 2 steps
Knowledge Booster
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.Recommended textbooks for you
Chemistry
Chemistry
ISBN:
9781305957404
Author:
Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:
Cengage Learning
Chemistry
Chemistry
ISBN:
9781259911156
Author:
Raymond Chang Dr., Jason Overby Professor
Publisher:
McGraw-Hill Education
Principles of Instrumental Analysis
Chemistry
ISBN:
9781305577213
Author:
Douglas A. Skoog, F. James Holler, Stanley R. Crouch
Publisher:
Cengage Learning
Chemistry
Chemistry
ISBN:
9781305957404
Author:
Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:
Cengage Learning
Chemistry
Chemistry
ISBN:
9781259911156
Author:
Raymond Chang Dr., Jason Overby Professor
Publisher:
McGraw-Hill Education
Principles of Instrumental Analysis
Chemistry
ISBN:
9781305577213
Author:
Douglas A. Skoog, F. James Holler, Stanley R. Crouch
Publisher:
Cengage Learning
Organic Chemistry
Chemistry
ISBN:
9780078021558
Author:
Janice Gorzynski Smith Dr.
Publisher:
McGraw-Hill Education
Chemistry: Principles and Reactions
Chemistry
ISBN:
9781305079373
Author:
William L. Masterton, Cecile N. Hurley
Publisher:
Cengage Learning
Elementary Principles of Chemical Processes, Bind…
Chemistry
ISBN:
9781118431221
Author:
Richard M. Felder, Ronald W. Rousseau, Lisa G. Bullard
Publisher:
WILEY