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\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}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).

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Comments
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.
NEC
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_man
2008-08-01 13:01:11
Sorry, this should have been categorized under HELP :D
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
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?
segfault
2009-09-06 07:43:35
1s2p3p is just a given state. It doesn't necessarily have to be the ground state, 1s^22s
NEC

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