quantum-mechanics electrons atomic-slrfc.org atoms orbitals
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edited Aug 8 "16 at 23:22
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asked Feb 23 "16 at 9:52
Anindya MahajanAnindya Mahajan
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The Rydberg formula just works for hydrogenic atoms, and also in hydrogenic atoms all the orbitals with the very same primary quantum number have (approximately) the very same power. The $2s$ and $2p$ have actually the same power as perform the $3s$, $3p$ and also $3d$, and so on.
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The Rydberg formula only works where the potential energy of the electron varies as $r^-1$. If we have actually more than one electron current then the electrons repel each other and they screen each other from the nucleus. As a result the potential is strictly speaking no longer also central, though to a good approximation we can treat the electron potential as main yet no much longer varying as $r^-1$ (even more on this below if you"re interested).
In hydrogenic atoms the various angular momentum says are only about of equal energy bereason relativistic effects cause a separating. For example in hydrogen the $2s$ is slightly higher in power than the $2p$, and also this is known as the Lamb shift. However this is a tiny result.
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edited Apr 13 "17 at 12:39
answered Feb 23 "16 at 15:23
John RennieJohn Rennie
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Rydberg"s formula have the right to be obtained from scattering concept between an electron and also a proton (using the known asymptotic actions of the consistent and ircontinual Coulomb functions. Using comparable arguments, you deserve to describe the scattering in between an electron and an singly charged ion, e.g. He$^+$. As stated by John Rennie, the potential is not strictly a Coulomb potential anymore because of the other electron(s), but the principle is that this just matters at brief ranges between the electron and also ion core. By corresponding the asymptotic solutions of the hydrogen atom to the brief range solutions of the more-electron trouble (which consists of superpositions of regular and irconsistent Coulomb functions), you successfully acquire a collisional phase transition. When proceeding and trying to derive the Rydberg formula, you"ll check out that the expression alters slightly:
$$ ilde u=fracE_ extIPhc-fracmathcalR_M(n-mu_ell)^2$$
Here I slightly rewrite the Rydberg formula so that $ ilde u$ is the spectral position (energy) of the Rydberg state through principal quantum number $n$ and orbital angular momentum $ell$. As you have the right to view, the principle quantum number $n$ is reinserted by $n-mu_ell$, the reliable quantum number, wright here $mu_ell$ is the quantum defect which have the right to be pertained to the collision phase shift. The magnitude of the quantum defect depends on the orbital angular momentum of the Rydberg electron. For small $ell$, the electron "sees" a lot of the short-selection potential and $mu_ell$ is reasonably large, while for $ell gt 3$ the electron does not pass through the core anymore and $mu_ellapprox 0$. Each value of $ell$ specifies a channel. The nice point is that $mu_ell$ is a really smooth and also almost constant function in the power of the electron, even when the system is ionized.
Another thing that need to be adjusted in the Rydberg formula is the Rydberg continuous $mathcalR_infty$, as the Rydberg consistent assumes an unlimited mass of the nucleus, therefore you should use the mass-corrected Rydberg constant for the mechanism via mass $M$ this is defined as
$$mathcalR_M = mathcalR_infty fracM-m_ exteM $$
wbelow $m_ exte$ is the electron mass.
Things deserve to obtain complicated as various Rydberg networks might interact, resulting in shifts of the power levels. These effects can be accounted for making use of multichannel quantum-defect concept (MQDT), that basically expands the single channel collision trouble to more networks.
Even for straightforward molecules such as H$_2$ and also ammonia, civilization have actually efficiently used this technique. See Ross, AIP Conf. Proc. 225, 73 (1991) for a nice advent.