elength, 410 nm, corresponds to the highest-energy the least energy. 3. Determine the Rydberg constant from the data. Since 1 1\| n- =R Equation 1 %3D 2. . 4. R, Н --R 22 H. 4. RH . = -RH Equation 2 4. 4. which is of the form y = mx + h 4.78x10-7 25 6. 3e 4.10x10-7 Calculate the change in energy for each transitic and give the results in units of joules. 4. hc AE = = hcR,-) Equation %3D H. n² -19 3 2: AE = -3.02Sx10 J-2.178x1 %3D -19 <-4.0838XID 1-2.17axIp- -2.173*ID

Chemistry: Principles and Reactions
8th Edition
ISBN:9781305079373
Author:William L. Masterton, Cecile N. Hurley
Publisher:William L. Masterton, Cecile N. Hurley
Chapter6: Electronic Structure And The Periodic Table
Section: Chapter Questions
Problem 66QAP: Consider the following transitions 1. n=3 to n=1 2. n=2 to n=33. n=4 to n=34. n=3 to n=5(a) For...
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Derive equation 3 from equation 1 

elength, 410 nm, corresponds to the highest-energy
the least energy.
3. Determine the Rydberg constant from the data. Since
1 1\|
n-
=R
Equation 1
%3D
2.
.
4.
R,
Н
--R
22
H.
4.
RH
.
= -RH
Equation 2
4.
4.
which is of the form y = mx + h
Transcribed Image Text:elength, 410 nm, corresponds to the highest-energy the least energy. 3. Determine the Rydberg constant from the data. Since 1 1\| n- =R Equation 1 %3D 2. . 4. R, Н --R 22 H. 4. RH . = -RH Equation 2 4. 4. which is of the form y = mx + h
4.78x10-7
25
6.
3e
4.10x10-7
Calculate the change in energy for each transitic
and give the results in units of joules.
4.
hc
AE =
= hcR,-) Equation
%3D
H.
n²
-19
3 2: AE = -3.02Sx10
J-2.178x1
%3D
-19
<-4.0838XID 1-2.17axIp-
-2.173*ID
Transcribed Image Text:4.78x10-7 25 6. 3e 4.10x10-7 Calculate the change in energy for each transitic and give the results in units of joules. 4. hc AE = = hcR,-) Equation %3D H. n² -19 3 2: AE = -3.02Sx10 J-2.178x1 %3D -19 <-4.0838XID 1-2.17axIp- -2.173*ID
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