Rydberg Equation

Johan Rydberg(1854-1919)
As a Swedish mathematician, Rydberg analyzed many atomic spectra in an attempt to understand the periodic system of the elements. He eventually developed what has become known as the Rydberg Equation. Although nominated, the Nobel prize was never awarded to Rydberg.  (Please note that these sample calculations are done in Angstroms and the Rydberg constant is given in units of 1/meters.  The  Brady text gives the constant in 1/centimeters.)  
 
When white light passes through a prism, it is separated into components (visible 4000-7000Angstroms).
 

Heating hydrogen and passing the light through a prism gives the following spectrum:

Consider the excitation of hydrogen's electron from the ground state (n=1) to the 5th shell (n=5).
The blue line results when the electron returns directly from n=5 to n=2.
 

Rydberg developed the following equation to fit hydrogen's atomic spectrum:
R = Rydberg constant = 1.0974x107 m-1
 n = shell returning from also called n2
m = shell returning to also called n1
 

To express Rydberg equation in angstroms:
1/l = [1.0974x107m-1][1/n12 - 1/n22]
1m/[l( 1.0974x107)] = [1/n12 - 1/n22]
1x10-10A/[(l)( 1.0974x107) = [1/n12 - 1/n22]
911A/l = [1/n12 - 1/n22]


 

Practice Problems
 
 
 
Problem #1
Use Rydberg equation to calculate wavelength when electron returns from n=5 to n=2.
911A/l = [1/n12 - 1/n22]
911A/l = [1/22 - 1/52] = 0.25 - 0.0400 = 0.21
l = 911A/0.21 = 4340A  Blue
 
Problem #2
For l = 10,900A, use Rydberg equation to determine initial and final energy levels.
911A/l = [1/n12 - 1/n22]
911A/10900A = 0.0836 = DIF
DIF = [1/n12 - 1/n22] = 0.0836   check table for the best fit!
The best fit involves n=6 to n=3
[1/32 - 1/62] = 0.111- 0.0278 = 0.0832
 
Problem #3
One electron de-excites from n = 8 to n =2. Another electron de-excites from n = 2 to n = 1.
Which emits the greater energy?  Explain!
For n = 8 to n = 2:                            For n = 2 to n = 1:                        
911A/l = [1/22 - 1/82]                      911A/l = [1/12 - 1/22]
            = 0.25 - 0.0156 = 0.234                     = 1.00 - 0.25 = 0.75   
            = 911A/0.234 = 3890A                  = 911A/0.75 = 1215A                                    
                                                         (shorter l higher energy)