Atomic Spectra
The Bohr atom is our starting point in understanding atomic spectra.
e Hydroge n Atom Z 2 p En = −13.6 2 eV Z = 1 n e
(Z-1)e Other Atoms Zp Z 2 E = −13.6 eff eV Zeff takes in to account screening of n n2 the nucleus by inner electron cloud. Total Energy Z 2 1 eff ! E = KE − + ΔESpin−Orbit Coupling total electron spin S will interact with 4πεo r L1 "! e L2 the total orbital angular momentum L S1 S2
e ! + ΔESpin−Spin Coupling electron spin S1 will interact with ! the spin angular momentum S2
Spin-Orbit Interaction and Total J=L+S
• The Spin-orbit interaction is the interaction between the atoms orbital magnetic moment L and spin magnetic moment S.
• The magnetic moment of a current loop is given by = iA = current x area.
!" q 2 q 2 q q !" µ L = iA = (πr ) = (πr ) = mvr = L orbital magnetic moment L T 2πr / v 2m 2m !" q " S µS == S spin magnetic moment N 2m r e 1 ∂V !!" !" !" " ESO ~ µL iµS ~ LiS Spin − Orbit Energy r ∂r
• The total angular momentum J = L + S of an atom is conserved and represents a good quantum number.
J Ψ (n l s j mj ) Jz n = principal quantum # L= orbital quantum # L s = spin quantum # S J = total quantum #
Jz projection of J on to z-axis |Jz| = mj h Values of Angular momentum
L=1 (px,py,pz) L=2 (dxy,dyz,dxz,x2-y2,dz2) • Orbital Angular momentum • 2L+1 states L=0 (s)
2 states
6 states 10 states
• Spin Angular momentum sz=+/- h/2 z • 2S+1 states Sz=+1/2h • S vector precesses about z-axis, Sx and Sy unknown y •
2 states Addition of Angular momentum
• We must add the angular momentum J=L+S of active orbital electrons. • The value of J that an atom takes on is the sum of total L and S for the atom.
• First determine total L = L1 + L2+… 1,2 refer to valence electrons L = |L1+L2| ……. |L1-L2| integer steps • Then total S = S1 + S2+ .. S = |S1+S2| ……. |S1-S2| integer steps Hund’s Rules (minimize the • Then total J = L + S energy) J = |L+S| ……. |L-S| integer steps •Highest S lies lowest • Mj takes on values of +J ….. -J integer steps •Highest L lies lowest • Spectroscopic notation Ψ = 2S+1 L J •Lowest J lies lowest < half filled •Highest J lies loest > half filled
Case of L1=1 S1=1/2: one electron in p orbital L = 1 S= 1/2 J=|1+1/2| … |1-1/2|= 3/2 and 1/2 J S L S Mj = +/-3/2, +/-1/2 and +/-1/2 L 2 J Ψ = P 3/2, 1/2
Term Diagram 1S 0 Hg Spectra 1 P1 30 cm-1 408 nm (3.04 eV) 3 -1 P2 3 40 cm P 3 1 P 0 Electric Dipole Transitions 50 cm-1 185 nm (6.70 eV) 60 cm-1 ΔL=1 Δm =0,+/-1 254 nm (4.88 eV) L ΔJ=0,+/-1 70 cm-1 J=0 <->J=0 not allowed -80 cm-1
1 S0
14 10 2 80 Hg Ground State (Xe) 4f 5d 6s 6p.. 1 1 L = 0 and S = 2 + 2 = 1 , 0 2S+1 3 J = L + S = 0 + 1 = 1 LJ = S1 not allowed by Pauli Excusion Principle ↑↑ 2S+1 1 J = L + S = 0 + 0 = 0 LJ = S0 ground state ↑↓
1st Excited State (Xe) 4f14 5d10 6s1 6p1 L 0 L 1 and S 1 S 1 1 = 2 = 1 = 2 2 = 2 L = 0 + 1 = 1 and S = 1 / 2 + 1 / 2 = 1, 0 2S+1 3 J = L + S = 1+ 1 = 2,1,0 LJ = P2,1,0 2S+1 1 J = L + S = 1+ 0 = 1 LJ = P1 2 2 6 1 Na-11 Spectrum Na−11 Ground State (Na) 1s 2s 6p 3s L = 0 and S = 1 2 2S+1 2 J = L+ S = 0+1/2= 1/2 LJ = S1/2 1st Excited State (Na) 1s22s26p6 3s03p1 L = 1 and S = 1 1 2 2S+1 2 J = L+ S = 1+1/2= 3/2,1/2 LJ = P3/2, 1/2
Electric Dipole Transitions ΔL=1 ΔmL=0,+/-1 ΔJ=0,+/-1 2 2 P3/2 J=0 <->J=0 not allowed P1/2
doublet
2 S1/2 Ground state of C-12
C − 12 : 1s2 2s2 2 p6
L = 1+ 1 = 0,1,2 S = 1 / 2 + 1 / 2 = 1,0 Hund 'rule L = 2 S = 1 lie lowest J = 2 + 1 = 3,2,1 J = 1 lies lowest 3 D1