aj47 Posted February 20, 2006 Posted February 20, 2006 In the bohr model of an atom, orbiting electrons exist in orbits that have discrete quantized energy levels. Do these levels correspond to the s p and d orbitals of an atom?
swansont Posted February 21, 2006 Posted February 21, 2006 Yes. No. All of the orbits in the Bohr atom are circular; they correspond to no orbitals, and the closest you can get is an s orbital from the symmetry. But the angular momentum is all wrong.
YT2095 Posted February 21, 2006 Posted February 21, 2006 the Bohr model is only ever really used in Dot/Cross diagrams for simplicity and ease on the eye, S,D,P and F is really only explained in Chem as a way of making sense as to why the PTOE is organised the way it is.
aj47 Posted February 21, 2006 Author Posted February 21, 2006 So when a photon is absorbed by an electron in say an S orbital, it won't be exited into a p orbital, but just into a higher energy state?
Norman Albers Posted February 25, 2006 Posted February 25, 2006 Relevant quantum numbers: <n,l,m> . For n=1, l must be zero. For n=2, l may be zero or one; m may vary between +/- l. Photons carry spin one; this must be exchanged. States of l=n-1 are circular; the s,p,d,f...are l=0,1,2,3. What the heck were those selection rules? Energy depends on n and l.
swansont Posted February 25, 2006 Posted February 25, 2006 So when a photon is absorbed by an electron in say an S orbital, it won't be exited into a p orbital, but just into a higher energy state? Actually a photon must go to a p orbital in that scenario; the photon has one unit of angular momentum. You can't get an S-to-S transition with a single photon under normal conditions. The energy state will also be higher if you absorb the photon, though n doesn't have to change (i.e. the p state has higher energy than the s state) selection rules
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