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GR9277 #76
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Alternate Solutions |
Mall-Saint
2016-10-23 00:32:38 | We have l=0+1+1=2. Each electron has s=1/2. If we have two electrons, the possible values of s are 0 and 1. Therefore for three electrons, the possible values of s are 1/2 and 3/2. Hence if we have (l=2, s=1/2) the possible j values are 3/2 and 5/2. But if we have (l=2, s=3/2) the possible j values are 1/2, 3/2, 5/2, 7/2. Hence the largest possible j is 7/2.\r\n\r\nAlso note l=2 because m_l = 0 for the 1s state, but m_l = {-1,0,1} for the 2p and 3p states, and so the total m_l can take any value from {-2,-1,0,1,2} -> l=2. |  |
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Comments |
Mall-Saint
2016-10-23 00:32:38 | We have l=0+1+1=2. Each electron has s=1/2. If we have two electrons, the possible values of s are 0 and 1. Therefore for three electrons, the possible values of s are 1/2 and 3/2. Hence if we have (l=2, s=1/2) the possible j values are 3/2 and 5/2. But if we have (l=2, s=3/2) the possible j values are 1/2, 3/2, 5/2, 7/2. Hence the largest possible j is 7/2.\r\n\r\nAlso note l=2 because m_l = 0 for the 1s state, but m_l = {-1,0,1} for the 2p and 3p states, and so the total m_l can take any value from {-2,-1,0,1,2} -> l=2. | | jeffray
2011-11-07 11:19:56 | If the state were instead 1s 2s3s, would the calculation change to 0 + 1 + 1 + 1/2 - 1/2 + 1/2 + 1/2? Do the two spins cancel out or do we just add another 1/2? |  | ashowmega
2010-04-02 13:01:05 | So, here the electrons are 1s 2p 3p. We know that for s, l=0, for 2p l=-1,0,1, for 3p l=-1,0,1. Hence we have seven values of quantum numbers. And each of them have 1/2 spin angular momentum. Hence, for seven azimutal quantum numbers, the total angular momentum = 7x(1/2) = 7/2 .
I guess what I am thinking is right.
flyboy621
2010-10-22 17:23:43 |
That's not right.
For s states, l=0. For p states, l=1. For spin 1/2 particles (i.e. electrons), s=1/2,m_s=1/2, -1/2.
For this problem we are not concerned with any of the m values, only l and s. We have 6 angular momentum quantities to add together:
0, 1, 1, 1/2, 1/2, 1/2
The outcome of adding these together can be any non-negative number you can get by adding or subtracting them. The maximum possible outcome is when they all add together (instead of subtracting). That gives 7/2, which is the answer. Other possible outcomes are 3, 5/2, 2, 3/2, 1, and 1/2. Zero is not possible.
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| | coke_man
2008-08-01 12:56:40 | I know this is a stupid question, but how is the total orbital angular momentum equal to  . If there is one electron in the  and one in the  and one in the  shouldn't the total angular momentum be  ? Could someone please tell me what I'm missing!!!
coke_man
2008-08-01 13:01:11 |
Sorry, this should have been categorized under HELP :D
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note
2008-08-23 16:17:19 |
the s subshell refers to quantum angular momentum number l=0
the p subshell refers to quantum angular momentum number l=1
one electron is in s and two are in p, thus 0+1+1
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ebykl
2008-10-25 10:18:03 |
What I can't understand is how can 3electrons can go to third shell. 1s^2 2s^2 2p^6......so the 3electrons must be at must at the 2S shell?
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segfault
2009-09-06 07:43:35 |
 is just a given state. It doesn't necessarily have to be the ground state, 
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Moush
2010-09-28 18:58:26 |
I had the same problem as coke_man, and a lot of others did too (only 25% answered correctly) so this is worth mentioning.
Whenever I saw  =1, 2, 3, ... n-1, I incorrectly thought HAS TO BE n-1, but instead can be UP TO n-1. For some reason I never understood that an e- in orbital s has =0, and =1 for e- in orbital p, etc., REGARDLESS of the principal quantum number shell it's in. All examples I've seen use PQN + orbital combinations that give the same answer with both correct and incorrect assumptions...until ETS.
http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/hydcol.html#c2
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You are replying to:
We have l=0+1+1=2. Each electron has s=1/2. If we have two electrons, the possible values of s are 0 and 1. Therefore for three electrons, the possible values of s are 1/2 and 3/2. Hence if we have (l=2, s=1/2) the possible j values are 3/2 and 5/2. But if we have (l=2, s=3/2) the possible j values are 1/2, 3/2, 5/2, 7/2. Hence the largest possible j is 7/2.\r\n\r\nAlso note l=2 because m_l = 0 for the 1s state, but m_l = {-1,0,1} for the 2p and 3p states, and so the total m_l can take any value from {-2,-1,0,1,2} -> l=2.
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