An atomic emission spectrum of hydrogen shows three wavelengths: 656.5 nm, 486.3 nm, and 434.2 nm. Assign these wavelengths to transitions in the hydrogen atom. Match the items in the left column to the appropriate blanks in the sentences on the right. AE-2.18 x 10-18 longer than equal to n=7 n=3 ΔΕ = 1.1 × 107 n=1 1 final n=2 n=8 AE-2.18 x 10-18 n=5 1 final n=4 n=6 1 initial shorter than 1 initial 1 nanitial 1 n² final The wavelengths of the three transitions are region. Therefore, it can be assumed that the electrons return to the Reset Help the wavelengths of the visual level. The difference in energy between the two levels initial and nfinal is given by the equation for 434.2nm. After calculating the energy difference using the given wavelengths, solving the equation for ninitial, and subtituting the proper values the following initial energy levels can be obtained: for 656.5nm, for 486.3nm, and
An atomic emission spectrum of hydrogen shows three wavelengths: 656.5 nm, 486.3 nm, and 434.2 nm. Assign these wavelengths to transitions in the hydrogen atom. Match the items in the left column to the appropriate blanks in the sentences on the right. AE-2.18 x 10-18 longer than equal to n=7 n=3 ΔΕ = 1.1 × 107 n=1 1 final n=2 n=8 AE-2.18 x 10-18 n=5 1 final n=4 n=6 1 initial shorter than 1 initial 1 nanitial 1 n² final The wavelengths of the three transitions are region. Therefore, it can be assumed that the electrons return to the Reset Help the wavelengths of the visual level. The difference in energy between the two levels initial and nfinal is given by the equation for 434.2nm. After calculating the energy difference using the given wavelengths, solving the equation for ninitial, and subtituting the proper values the following initial energy levels can be obtained: for 656.5nm, for 486.3nm, and
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
![## Hydrogen Emission Spectrum Problem
An atomic emission spectrum of hydrogen shows three wavelengths: 656.5 nm, 486.3 nm, and 434.2 nm. Assign these wavelengths to transitions in the hydrogen atom.
### Match the items in the left column to the appropriate blanks in the sentences on the right.
**Equation and Options:**
### Energy Difference Equation
\[
\Delta E = -2.18 \times 10^{-18} \left( \frac{1}{n_{\text{final}}^2} - \frac{1}{n_{\text{initial}}^2} \right)
\]
- longer than
- equal to
- \( n = 7 \)
- \( n = 3 \)
\[
\Delta E = 1.1 \times 10^7 \left( \frac{1}{n_{\text{final}}^2} - \frac{1}{n_{\text{initial}}^2} \right)
\]
- \( n = 1 \)
- \( n = 2 \)
- \( n = 8 \)
\[
\Delta E = -2.18 \times 10^{-18} \left( \frac{1}{n_{\text{initial}}^2} - \frac{1}{n_{\text{final}}^2} \right)
\]
- \( n = 5 \)
- \( n = 4 \)
- \( n = 6 \)
- shorter than
**Sentences:**
1. The wavelengths of the three transitions are ___ the wavelengths of the visual region. Therefore, it can be assumed that the electrons return to the ___ level.
2. The difference in energy between the two levels \( n_{\text{initial}} \) and \( n_{\text{final}} \) is given by the equation ___.
3. After calculating the energy difference using the given wavelengths, solving the equation for \( n_{\text{initial}} \), and substituting the proper values, the following initial energy levels can be obtained: ___ for 656.5 nm, ___ for 486.3 nm, and ___ for 434.2 nm.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F61643fcf-6c9c-4b30-9cd8-97d11e5075d4%2F3dbb76c7-d69b-46cf-a4c5-054835b537c8%2Fevgcjb_processed.jpeg&w=3840&q=75)
Transcribed Image Text:## Hydrogen Emission Spectrum Problem
An atomic emission spectrum of hydrogen shows three wavelengths: 656.5 nm, 486.3 nm, and 434.2 nm. Assign these wavelengths to transitions in the hydrogen atom.
### Match the items in the left column to the appropriate blanks in the sentences on the right.
**Equation and Options:**
### Energy Difference Equation
\[
\Delta E = -2.18 \times 10^{-18} \left( \frac{1}{n_{\text{final}}^2} - \frac{1}{n_{\text{initial}}^2} \right)
\]
- longer than
- equal to
- \( n = 7 \)
- \( n = 3 \)
\[
\Delta E = 1.1 \times 10^7 \left( \frac{1}{n_{\text{final}}^2} - \frac{1}{n_{\text{initial}}^2} \right)
\]
- \( n = 1 \)
- \( n = 2 \)
- \( n = 8 \)
\[
\Delta E = -2.18 \times 10^{-18} \left( \frac{1}{n_{\text{initial}}^2} - \frac{1}{n_{\text{final}}^2} \right)
\]
- \( n = 5 \)
- \( n = 4 \)
- \( n = 6 \)
- shorter than
**Sentences:**
1. The wavelengths of the three transitions are ___ the wavelengths of the visual region. Therefore, it can be assumed that the electrons return to the ___ level.
2. The difference in energy between the two levels \( n_{\text{initial}} \) and \( n_{\text{final}} \) is given by the equation ___.
3. After calculating the energy difference using the given wavelengths, solving the equation for \( n_{\text{initial}} \), and substituting the proper values, the following initial energy levels can be obtained: ___ for 656.5 nm, ___ for 486.3 nm, and ___ for 434.2 nm.
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
This is a popular solution!
Trending now
This is a popular solution!
Step by step
Solved in 4 steps with 12 images
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