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An atom which has a valance electron in an extremely high principal quantum
number state is in a Rydberg state. These states are called circular
states for the following reason. (We use the simpler hydrogen case due
to the fact that it is not a terrible approximation to hydrogenic atoms.)
The radial equation (see the compendium on the hydrogen atom) was found
to be simplifiable to,
![$\displaystyle -\frac{\hbar^2}{2m}u'' + \left(-\frac{e^2}{4\pi \epsilon_0}\frac{1}{r} +
\frac{\hbar^2}{2m}\frac{l(l+1)}{r^2}\right)u = Eu$](img55.png) |
(2.1) |
With,
![$\displaystyle E=-\frac{m}{2\hbar^2}\left(\frac{e^2}{4\pi \epsilon_0}\right)^2\frac{1}{n^2}$](img56.png) |
(2.2) |
and
, we can rewrite this equation as,
![$\displaystyle u''=\left(\frac{l(l+1)}{r^2} - \frac{2}{ar} + \frac{1}{n^2 a^2}\right)u$](img58.png) |
(2.3) |
We then consider that (using integration by parts and knowing that the infinite limit of the radial component must be zero)
Algebraic exercise yields Kramer's relation,
![$\displaystyle \frac{s+1}{n^2}\langle r^s \rangle - (2s+1)a\langle r^{s-1}\rangle + \frac{s}{4}\left((2l+1)^2 - s^2\right)a^2\langle r^{s-2}\rangle =0$](img68.png) |
(2.5) |
When we set
we find,
![$\displaystyle \frac{1}{n^2}\langle 1 \rangle - a \langle \frac{1}{r}\rangle = 0 \ \ni \ \langle \frac{1}{r}\rangle = \frac{1}{n^2 a}$](img70.png) |
(2.6) |
Thus setting
yields,
![$\displaystyle \frac{2}{n^2}\langle r \rangle - 3 a\langle 1 \rangle + \frac{1}{...
...r}\rangle = 0 \ \ni \ \langle r \rangle = \frac{a}{2}\left(3n^2 - l(l+1)\right)$](img72.png) |
(2.7) |
When we use the states such that
and
, i.e. the maximum allowed, we have
![$\displaystyle \langle r \rangle = an^2 + \frac{an}{2}$](img75.png) |
(2.8) |
For very large n, this approximates the classical result for circular orbits,
,
thus the name circular Rydberg states.
Next: Selection Rules and Available
Up: The Aufbau Principal, Kramers
Previous: From the Hydrogenic: The
tim jones
2007-04-09