Rydberg Equation
Johan Rydberg(1854-1919)
As a Swedish mathematian, 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.When white light passes through a prism, it is separated into components (visible 400-700 nm).


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
m = shell returning to
When m = 1 Lyman Series
m = 2 Balmer Series
m = 3 Paschen Series
Rydberg Equation written as:
1/l = [1.0974x107m-1][1/m2 - 1/n2]
1m/[l( 1.0974x107)] = [1/m2 - 1/n2]
1x10-9nm/[(l)( 1.0974x107) = [1/m2 - 1/n2]
91.1nm/l = [1/m2 - 1/n2]
Here's a handy table for n versus 1/n2 | n | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
| 1/n2 | 1 | 0.25 | 0.111 | 0.0625 | 0.0400 | 0.0278 | 0.0204 | 0.0156 | 0.0123 | 0.0100 |
As a Swedish mathematian, 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.
When white light passes through a prism, it is separated into components (visible 400-700 nm).


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
m = shell returning to
When m = 1 Lyman Series
Rydberg Equation written as:m = 2 Balmer Series
m = 3 Paschen Series
1/l = [1.0974x107m-1][1/m2 - 1/n2]
1m/[l( 1.0974x107)] = [1/m2 - 1/n2]
1x10-9nm/[(l)( 1.0974x107) = [1/m2 - 1/n2]
91.1nm/l = [1/m2 - 1/n2]
Here's a handy table for n versus 1/n2