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?

Yes.

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.

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.

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?

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.

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

aha great help thanks