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idoubt 20151019 06:55:38  I think the answer has to do with the fact that photons are a spin 1 particle and it cannot have the m = 0 state. So to emit a photon the two wave functions must have an angular momentum difference . This is not possible if both are S shell.  
jgramm 20140923 21:58:30  If they're both spherical, they're both in the s shell (think back to those silly pictures of orbitals in high school chem).
If transitioned to , both electrons (fermions) would occupy the same state, which is not allowed by the Pauli exclusion principle.
Thus, D is the obviously the answer without having to worry about any of that selection rule nonsense that nobody ever remembers (or maybe it's just me).
Baharmajorana 20141015 06:04:04 
I agree with u, I can't remember too

James6M 20141015 20:36:49 
Um, what?
There is only one electron in a hydrogen atom, so I don't know why you are talking about "both." And just because both wavefunctions are sshells, that doesn't mean they are the same wavefunction. They can have different n.
So, no, the Pauli exclusion principle has nothing to do with this question.

 
a_coiled_atom 20110716 12:35:34  I thought of it in terms of the selection rules , . If the initial state has , which corresponds to spherical symmetry in the wavefunction, then according to these the final state must have , which is not spherically symmetric.  
a_coiled_atom 20110716 12:34:28  I thought of it in terms of the selection rules , . If the initial state has , which corresponds to spherical symmetry in the wavefunction, then according to these the final state must have , which is not spherically symmetric.
pam d 20110924 11:31:55 
This is how it's done.

 
thebigshow500 20081013 22:12:48  Can anyone explain if other choices are something to do with wave function? (e.g. parity, orthogonality, etc.) Or are these choices just simply nonsense? :(
kroner 20091016 12:06:41 
(A): I'm not really sure what this is talking about.
(B): Any two different eigenstates are mutually orthogonal. So in any transition this is required.
(C): The eigenstates that are zero at the center are those with l>0. This is allowed (for example l could change from 3 to 2).
(E): The selection rules require l to change.

alemsalem 20100921 04:32:38 
the states with even l have even parity(1) and those with odd l have odd parity(1) since in transitions l changes by plus or minus 1 the parity must change in any transition, so (A) is out

 
lattes 20081013 09:18:54  it might be useful to make an analogy with electromagnetism. It's known that a pulsating sphere does not emit radiation because of its spherical symmetry. An analogy can be made with wavefunction. (D). For further reading check out Beiser's book on modern physics pag. 165.  
curie 20081008 05:55:58  Why does sphericallysymmetric imply same angular momentum?
rkraman 20081015 14:35:02 
This has to do with the symmetry properties in mechanics. If the Lagrangian (and hence the Hamiltonian) describing a system is symmetrical with respect to any coordinate, then the canonical momentum corresponding to that coordinate is conserved. So if there is translation symmetry in , the the canonical momentum which is the linear momentum is conserved. Similarly if the Hamiltonian is rotationally invariant, it is symmetrical with respect to , then the corresponding angular momentum , which is nothing but the angular momentum , is conserved. This is a specific instance of a fundamental theorem in Physics called the Noether's theorem.

spacemanERAU 20091015 18:04:53 
VERY VERY good explanation rkraman! kudos!

flyboy621 20101114 20:52:31 
Also, if the wavefunction is spherically symmetric, that means .

 
evanb 20080620 11:56:30  "completely unique relative..."
I'm not sure exactly what that means, but if it means that m must also change that's not quite right.
The selection rules are that NO TRANSITION OCCURS UNLESS:
m = {1, 0, +1} and
l = {1 , +1}.
has ore options.  