GR8677 #30



Alternate Solutions 
a_coiled_atom 20110710 12:58:10  Another way of thinking about it that might be helpful is that , in general, no orbital angular momentum spherical symmetry.  

Comments 
joshuaprice153 20190809 02:49:54  Here has some effective and valuable information. Thanks for the information provided in this blog. This is the third time I visited in this blog. keep posting such stuff. Thanks a lot for the information. Randy Shine   a_coiled_atom 20110710 12:58:10  Another way of thinking about it that might be helpful is that , in general, no orbital angular momentum spherical symmetry.   suvas 20100909 16:23:08  I would choose E. The electronic charge distribution of atom is determined mainly by valence electron.   wittensdog 20090725 17:54:02  ETS is good at thinking of traps. I almost wanted to go for C. I saw the 4s1 electron being the least tightly bound, and in my mind a connection was made with the number 4. Clearly, l = 4 is supposed to trick you, when of course the electron actually has n = 4, not l = 4. It's value for orbital angular momentum is l = 0, of course.
jmason86 20090816 16:36:12 
I must have fallen for a trap too. I chose (A) immediately and without hesitation.
I was thinking that the n=3 shell only means the ssubshell. So when I saw that we were given , I assumed that (A) was correct. However, the n=3 is a total shell consisting of the s, p, and d subshells... If (A) were true then we would be talking about Zinc with [Ar] .

neon37 20101101 09:56:59 
same here @ wittendog. Ofcourse reading E, changed made me realize my mistake.

  kyros 20071030 21:42:24  Shouldn't that be Sigma_4,0,0, since m can only range from l to +l anyway?   simpsoxe 20061130 21:20:14  well the valence shell is , but my question is about the other electrons in the lower shells. don't we have to consider them as part of the electron distribution? or do we assume they all have spherically symmetric distributions? (and why would that be?)
Blake7 20070723 18:26:38 
The lower shells all appear to be completely filled
2 2 6 2 6
if I'm not mistaken that will make them all symmetric

kronotsky 20181023 03:51:23 
Filled subshells are the singlet component of the direct product of the component orbitals, so they have zero orbital angular momentum. So does 4s. The rotation operator(s) is an exponent of the orbital angular momentum operator, so it is equal to 1 (i.e. rotational symmetry).

 

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