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Unformatted text preview: Â§ gig egg QghitalAngulag Mgmentum in Atoms Electron spin angular momentum of s=â€˜/z has a projection on the zâ€”axis of
ms=+1l2 or â€”1l2 (can align with or against a ï¬eld), electron spin has a magnetic moment, but when paired the moments cancel Electron orbital angular momentum for l=1 (p orbital) has projections on the z
axis of m=1,0,+1, when mÂ¢0 there is a magnetic moment, but when there is an
electron in each of the three p orbitals, the moments cancel. Filled subshells have zero net orbital angular momentum. Paired electrons
have no net spin angular momentum. Only partially ï¬lled subshells can have magnetic moments. Magnetic
materials must have unpaired electrons. Terms mbolsLa llin " ' i States 23+1 L\J S is the total spin angular momentum
L is the total orbital angular momentum
J is the total angular momentum â€˜go read â€œsinglet ess zero"
2?â€ read â€œdoublet pee threehalvesâ€
392 read â€œtriplet dee twoâ€ in general capital letters indicate quantities for the whole atom. whereas lower
case symbols are used for the individual electrons in the system. The total orbital angular momentum is L i i i LMAxalMAXâ€”v"LLEAST vector sum of the orbital angular momenta of each electron where the result must
be quantized, Le. a whole number. Use only open shell electrons because
closed shells have zero angular momentum. LMAX = iâ€˜ and LLEAST = IMAX " i! iatMAX L 0 1 2 3
Letter Code 8 P D F 4
G
In a manner analogous to the hydrogen atom M2 2 L(L + 1M2 The spin angular momentum is S :2 i = SMAX,SMAX_1u"SLEAST This is a vector sum of the spins of the electrons, use only open shell electrons. SWÂ»X = 21/2 and SLEAST 21/2 â€” 21/2 or zero (the larger)
i ' lâ€”"I The total angular momentum, RussellSaunders Coupling J = (L+ S),(L + 8â€”1),...L â€” s for example if L21, 8:1, then J=2,1,0 8:1 3:1 J=0
J=2
L=1 L=1 Q L=â€˜l s=1
J=1
3P2 3P1 3P0 Boron example: electronic conï¬guration is 1322322}:1 the two electrons in the K shell have L=0, 8:0, J=O, so we need only consider the valence electrons 2522p1 electrons. The 232 electrons have no orbital
angular momentum, 1=2=0, and the spins are paired cancelling their contribution to the spin angular momentum; so we need only consider the 2p1 electron
l3=1 so trivially L = Z I i = LMM,LW,O(_1,...LLEIE,â€˜ST = 53=1l2 so trivially 8 == 2â€”8.. = 1/2 J: (1+1/2),...l1â€”1/2I= 3/2,1/2 boron has 2PM and 2PW states These are called sginorbit states. They have energies which are slightly
different and small compared to the electronic energies of the orbitals. The spin
orbit splittings are smaller for the light atoms and get bigger and more important
as the atoms get heavier. You need relativistic quantum theory to calculate spin
orbit splittings. The spinorbit energy level splittings are described as E â€” lAn'LS[J(J + 1) r L(L +1) â€” 3(3 + 1)] spinâ€”orbit _ 2 and can be measured from the observed splittings in high resolution spectra of
atoms arising from the same 8 and L levels. Which spinorbit levels are the lowest in energy? You use Hunds Rules The determination and energy ordering of the Terms for the carbon atom are
illustrated in the example following Hunds Rules. Hund's Rules (work best for ground states) 1. For states from the same " J
electronic configuration (same n1),
the one with the highest S is
lowest in energy. 2. Of the states above with the same
S, the one with the highest L is
lowest in energy. 3. For multiplet J states: When a
shell or subshell is less than halfâ€”
full, states of low J are lower in
energy; when a shell or subshell is
more than halffull, states of
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 Fall '04
 DR.COE
 Quantum Chemistry, Angular Momentum, $2, $5, iL, Hund

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