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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For the part circled, why do I keep getting 4860 nm? I cannot figure out the calculation for nm to m and vise versa
![Class Example: Using the Bohr Model
Calculate the Wavelength of Light Emitted when an
electron in the H atom falls from the n = 4 to n = 2 level
[12
AE = -2.178 × 10-18]
22
121
= -4.084 x 10-191
%3D
%3D
42
hc
hc
AE = Ephoton= hv = →
%3D
%3D
JAE|
(6.626 × 10-34y s)(2.998 × 108 m/s) 1× 10° nm
23=
= 486 nm
%3D
|- 4.084 × 10-19/|
1m
Energy emitted falls in the visible light region
4.86mx/0-7
2014 Pearson Education, Inc.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F61b75115-d70f-4fe2-af93-2076876ad69a%2F96eff4e9-d0e1-433c-8d5b-8f96051eaa0c%2Fdl5zcrm_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Class Example: Using the Bohr Model
Calculate the Wavelength of Light Emitted when an
electron in the H atom falls from the n = 4 to n = 2 level
[12
AE = -2.178 × 10-18]
22
121
= -4.084 x 10-191
%3D
%3D
42
hc
hc
AE = Ephoton= hv = →
%3D
%3D
JAE|
(6.626 × 10-34y s)(2.998 × 108 m/s) 1× 10° nm
23=
= 486 nm
%3D
|- 4.084 × 10-19/|
1m
Energy emitted falls in the visible light region
4.86mx/0-7
2014 Pearson Education, Inc.
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