GREPhysics.NET: GR9277 #76

GR9277 #76

Problem

GREPhysics.NET Official Solution

\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?

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).

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Comments

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, $1s^22s$

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