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GR9277 #76
Problem
 GREPhysics.NET Official Solution Alternate Solutions
\prob{76}
The configuration of three electrons $1s2p3p$ has which of the following as the value of its maximum possible total angular momentum quantum number?

1. 7/2
2. 3
3. 5/2
4. 2
5. 3/2

Atomic$\Rightarrow$}Orbitals

The total angular momentum is given by $j=l+s$ where l is the orbital angular momentum and s is the spin angular momentum. (Note that, to an extent, l and s can be viewed as magnitudes, while $m_l$ and $m_s$ as directions.)

The total orbital angular momentum is just $0+1+1$, since one should recall that $(s,p,d,f) \in (0,1,2,3)$.

The spin angular momentum is just $1/2+1/2+1/2$ because one has three electrons.(Electrons are fermions that have spin $1/2$.)

Thus, the total angular momentum is $j=2+3/2=7/2$, as in choice (A).

Alternate Solutions
 Mall-Saint2016-10-23 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.Reply to this comment
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$^2$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.
 flyboy6212010-10-22 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.
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 $0+1+1$. If there is one electron in the $1s$ and one in the $2s$ and one in the $3s$ shouldn't the total angular momentum be $0+1+2$? Could someone please tell me what I'm missing!!!
 coke_man2008-08-01 13:01:11 Sorry, this should have been categorized under HELP :D
 note2008-08-23 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
 ebykl2008-10-25 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?
 segfault2009-09-06 07:43:35 $1s2p3p$ is just a given state. It doesn't necessarily have to be the ground state, $1s^22s$
 Moush2010-09-28 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 $\.l$=1, 2, 3, ... n-1, I incorrectly thought $\.l$ HAS TO BE n-1, but instead $\.l$ can be UP TO n-1. For some reason I never understood that an e- in orbital s has $\.l$=0, and $\.l$=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

If the state were instead 1s$^2$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?

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