the electrons. Expanding the exponential and neglecting the constant term Ze 2 λ, we have

Transcription

1 LECTURE.8 Prof.R.Parthasarathy Atomic Structur - Elmntary Tratmnt Th ground stat of hydrogn atom has bn solvd xactly in th nonrlativistic tratmnt. Th ground stat of hlium atom has bn handld in variational schm. What about havir atoms? Hr w assum that ach of th atomic lctrons movs in a sphrically symmtric potntial V (r) producd by th nuclus and all th othr lctrons. This is cntral fild approximation. W hav sn in hydrogn-lik atom, th potntial Z r producd infinit numbr of nrgy lvls. W xpct similar situation for V (r) as wll. Howvr, thr is going to b a diffrnc! W hav sn in hydrogn-lik atoms, th nrgy is proportional to n and indpndnt of l, so that for givn n, th stats having diffrnt l s ar dgnrat. For xampl, th stats m and 00 hav th sam nrgy. This dgnracy is rmovd in a non-coulomb cntral fild. How? Whn many lctrons ar prsnt, th nuclus is scrnd by lctrons. W want to show that for givn n, th stats of lowr l hav lowr nrgy. Lt us introduc th scrning by th scrnd Coulomb potntial V (r) = Z λr, whr λ is a positiv r constant. Thn, H = p Z λr. This is th hamiltonian flt by on of m r th lctrons. Expanding th xponntial and nglcting th constant trm Z λ, w hav whr H = p m Z r Z λ r + 6 Z λ 3 r = H 0 + H + H +, () H = Z λ r ; H = 6 Z λ 3 r. () In (), H 0 is th hamiltonian for hydrogn-lik atom whos Schrödingr quation has bn solvd xactly. Th nrgy ignvalus ar E n = Z a 0 and n th ignfunctions ar ψ nlm (r, θ, φ). For small λ, th trm H is rlvant and can b considrd as a prturbation. Th first ordr corrction to th nrgy is < H > nlm. For a givn

2 n, this prturbation cannot connct diffrnt l stats. So non-dgnrat prturbation thory is a good approximation! W hav calculatd th xpctation valu of r in our tratmnt of hydrogn atom. So, From (3), w gt < H > nlm = Z λ < r > nlm, = λ a 0 n ( 3 l(l + ) ). (3) n E 00 = E 0 3 λ a 0, E m = E 0.5 λ a 0, (4) whr E 0 is th unprturbd nrgy in n = stat Z 8a 0. Clarly th 00 stat has lowr nrgy that m stat. Thus, th first ordr corrction not only rmovs th l dgnracy but also givs th rsult that lowr angular momntum stats hav lowr nrgy. Idntical Particls W hav sn th quantum mchanics of a singl particl in a potntial. Whn w hav mor than on particl, say lctron, bcaus of th absns of dtrminism in quantum mchanics, it is impossibl to labl th lctrons. Thy ar indistinguishabl, whn both th lctrons ar having sam quantum numbrs. So, physical obsrvation must b invariant undr thir xchang. As an xampl, considr hlium atom. Thr ar two lctrons. Lt us nglct thir mutual intraction. What is th ground stat? Lt us dnot th lowst singl particl stat by s >. Thn th amplitud for finding lctron (labl it tmporarily!) and in this stat is < s > and < s >. Knowing th lctron has spin, w hav < s + >, < s > ; < s + >, < s >. Thn th possibl two-particl amplituds ar: < s + >< s + >, < s + >< s >, < s >< s + >, < s >< s >,(5) or thir linar combinations. Exprimntally, only on linar combination appars! i.., < s + >< s > < s >< s + > (6)

3 which changs sign whn. This is a rsult of th xclusion principl of Pauli- th amplitud for a systm containing svral lctrons must b antisymmtric undr th xchang of any two lctrons. Equivalntly, no two lctrons ar in th sam singl particl stat. For if any two lctrons ar in th sam stat, th totally antisymmtric amplitud vanishs. Givn no two lctrons can b in th sam singl particl stat, can w gt th antisymmtry? Lt us procd furthr to a gnral situation. Lt Φ I = < a >< b >: Φ II = < b >< a >. (7) Sinc and ar indistigushabl, w hav two linar combinations; Φ S = Φ I + Φ II, Bosons, Φ AS = Φ I Φ II. F rmions. (8) Considr a systm of two idntical particls and govrnd by a hamiltonian H(). Lt thr b a stationary stat I >, which can b charactrizd by lablling by an indx α and by β, whr α, β rfr to position, spin projction tc. Thn I >= α β > will b in th Hilbrt spac H, = H H. Sinc and ar idntical, th stat II >= α, β > will also b a stationary stat with sam nrgy as I > sinc H() = H(). Introducing P th prmutation oprator, P I >= II >, P II >= I >, [P, H] = 0. (9) So P and H hav simulatanous ignfunctions. Lt P ψ >= η ψ >. Thn P P ψ >= η ψ >= ψ > and so η = ±. Thn, th cas, P ψ >= ψ > corrsponds to symmtric stat rprsnting Bosons whil th cas with P ψ >= ψ > corrsponds to antisymmtric stat rprsnting Frmions. For a two lctron systm, th totally antisymmtric wav function will b α β > α β > = ψ α ()ψ β () ψ α ()ψ β (), = ψ α() ψ α () ψ β () ψ β () For a thr lctron systm, th totally antisymmtric wav function is ψ α () ψ α () ψ α (3) ψ β () ψ β () ψ β (3) ψ γ () ψ γ () ψ γ (3) (0) () 3

4 Antisymmtry in () undr, 3, 3 is vidnt using th proprty of dtrminants! Symmtric wav functions will rprsnt Bosons, particls of intrinsic spin 0,, tc and oby Bos-Einstin statistics. Antisymmtric wavfunctions rprsnt Frmions, particls of intrinsic spin, 3, tc and oby Frmi-Dirac statistics. This connction spin-statistics thorm has bn vrifid. But th sourc or proof is not wll undrstood. Th abov wav function is not normalizd. A normalizd antisymmtric wav function is givn by Slatr dtrminant. For th thr-frmion systm, ψ αβγ (,, 3) = ψ α () ψ α () ψ α (3) ψ β () ψ β () ψ β (3) 3! ψ γ () ψ γ () ψ γ (3) () Is this procdur always corrct? No. Whn th lctrons ar non-intracting, it is a corrct mthod. Whn th intraction among thm is wak, th abov stat can b usd for prturbativ calculations. Ground Stat Configurations Th quantum numbrs ar: n =,, 3, principal quantum numbr l = 0,,, (n ), orbital ang.momntum qtm numbr m = l to + l in stps of, z componnt of ang.mom qtm numbr s =, m s =,, Hydrogn Z =. Ground stat: n =, l = 0, m = 0, m s = ± Th singl lctron configuration is (s). (notation: (nl) occupants. Th chmical valncy is just th numbr of lctrons in th last shll which hr is incomplt and so th valncy is. Hlium 4

5 Z =. Th two lctrons occupy n =, l = 0 shll and hav m s = ± so as to rspct Pauli principl. Th lctronic configuration is (s). With two lctrons in s shll, it is compltly occupid. Th valncy is zro and this lmnt is chmically inrt. Hlium is an inrt gas. Lithium Z = 3. First th shll s is compltly filld by two of th thr lctrons. Th third lctron can go to n =, l = 0 with m s = or. Th lctronic configuration is (s) (s). Th valncy is du to th lctron in th s shll and so it is. This lmnt is chmically activ. Brylliium Z = 4. Th lctronic configuration is (s) (s). It has a closd sub-shll sinc th p stat is mpty. Th normal valncy is. W giv th lctronic configurations of som light lmnts. Elmnt(Z) ElctronicConf iguration Hydrogn() (s) Hlium() (s) Lithium(3) (s) (s) Bryllium(4) (s) (s) Boron(5) (s) (s) (p) Carbon(6) (s) (s) (p) Nitrogn(7) (s) (s) (p) 3 Oxygn(8) (s) (s) (p) 4 F luorin(9) (s) (s) (p) 5 Non(0) (s) (s) (p) 6 Sodium() (s) (s) (p) 6 (3s) Magnsium() (s) (s) (p) 6 (3s) Aluminum(3) (s) (s) (p) 6 (3s) (3p) Silicon(4) (s) (s) (p) 6 (3s)(3p) P hosphrous(5) (s) (s) (p) 6 (3s) (3p) 3 Sulphur(6) (s) (s) (p) 6 (3s) (3p) 4 Chlorin(7) (s) (s) (p) 6 (3s) (3p) 5 Argon(8) (s) (s) (p) 6 (3s) (3p) 8. 5

6 Upto Z = 8, th filling up of th nrgy lvls with th undrstanding that for a givn n, lowr l lvls ar filld first works. Th rmoval of th l dgnrcay bing xplaind by th scrning of th nuclar charg by th rmaining lctrons. This squnc changs now! Aftr Argon, w hav Potassium Z = 9. Th nintnth lctron should go to th 3d stat. But it gos to 4s stat! Aftr filling th 4s subshll, 3d starts gtting lctrons. For xampl, th lctronic configuration of potassium (K,Z=9) is (Ar) + (4s), that of Calcium (Ca, Z=0) is (Ar) + (4s), that of Scandium (Sc, Z=) is (Ar) + (4s) (3d) and so on till, for Zinc (Zn, Z=30), it is (Ar) + (4s) (3d) 0. Now that th 3d subshll is compltly filld, th nxt lctron in Galium gos to 4p stat, namly, Ga(Z=3) has (Ar) + (4s) (3d) 0 (4p). Can w undrstand this qualitativly? From th first ordr corrction (3), it follows E 30 = Z E 400 = Z 8a 0 5.5λ a 0, 3a 0 λ a 0. (3) If th scrning is strong, thn it is plausibl to undrstand that th 4s lvl has lowr nrgy than 3d lvl. Alkali atoms Th alkali atoms ar Li, Na, K tc. From th lctronic configuration tabl, w not that th ground stat configuration of an alkali atom consists of filld shlls followd by a singl s lctron. Th filld shll configuration corrsponds to inrt gas configuration and so stabl. Thn, th xcitd stats of th alkali atom involv only th valnc s lctron. Thus, th alkali atom can b tratd to a good approximation in trms of a modl in which a singl lctron movs in a sphrically symmtric non-coulomb potntial nrgy V (r), non-coulomb du to scrning. In th absns of xtrnal filds, th hamiltonian could b + V (r). To this w add th spin-orbit intraction. m First of all w considr hydrogn atom. Th nrgy lvls of hydrogn atom obtaind using non-rlativistic Schrödingr quation dpnd on th principal quantum numbr n and thr is dgnracy for l. Classically, a particl of mass m, charg moving with vlocity v in an lctric fild E 6

7 xprincs a magntic fild B E v = and if th lctric fild E is drivabl c from a potntial V (r), E = r dv so that r dr dv dr B = c r = c rm r v, dv dr l, whr l = r p. Th nrgy of th intraction btwn B and a magntic dipol m is m B so that H s.o = m B = dv m l. For an lctron, mc r dr m = s so that m H s.o = m c r A corrct calculation (Dirac) lads to dv dr l s. H s.o = m c r For an lctron in hydrogn-lik atom, V = Z r H s.o = dv dr l s. (4) and so Z m c r 3 l s. (5) W know from hydrogn atom wav functions, < nlm r nlm > = Z 3 3 n 3 a 3 0l(l + (6) )(l + ). From th tim-indpndnt prturbation thory, th first ordr corrction to th nrgy lvls of hydrogn-lik atom du to th spin-orbit trm is < nlm H s.o nlm > = Z 4 m c a 3 0l(l + )(l + ) < l s >. (7) Th xprssion < l s > can b asily valuatd in th coupld basis sinc l s = {j l s }. 7

8 Thus th first ordr corrction is E s.o = Z 4 m c a 3 0n 3 l(l + )(l + ) h {j(j + ) l(l + ) s(s + )}. (8) Th abov rsult can b xprssd also as E s.o = Z E 0 n α n(l + ) l+ for j = l + l for j = l (9) whr E 0 n is th unprturbd nrgy. This can b asily gnralizd to alkali atoms. With this, lt us valuat th nrgis for 3p 3 and 3p. so that E 3p 3 E 3p = E3 0 + Z α E3 0 8, = E3 0 + Z α E3 0 9, (0) E 3p 3 E 3p = Z E 0 3 α 6. () Lt us considr sodium atom. Exprimntally, thr ar two dominant yllow doublt. Thir wavlngths ar: 5890Å and 5896Å. Thy corrspond to th atomic transitions 3p 3 3s and 3p 3s. For Coulomb potntials, such transitions cannot occur as th nrgy dpnds only on n. For non-coulombic potntials, such transitions can occur. As th non-coulombic potntial is not known (at this lvl), w cannot calculat th abov transition nrgis. Howvr, w adopt a modl: Th spctrum of sodium atom is dtrmind by th valnc lctron in 3s stat. Th closd shll configuration upto 0 lctrons corrsponds to Argon th inrt gas. Hr th nuclar charg (+ ) is compltly shildd by th 0 atomic lctrons, so that th valnc lctron ss th nuclar charg of on unit. In this modl, th diffrnc of th nrgis of 3p 3 and 3p is givn by (6) with E0 3 bing 8

9 simply 3.6 V. Using th valu of th fin structur constant α as 9 Z =, this nrgy diffrnc (6) bcoms, E 3p 3 E 3p 37 and = E = V. () For th transition E 3p 3 E 3s, th wavlngth of th radiation is λ = hc hc and for th othr transition, λ E 3p 3 E 3s =. Hnc, E 3p E 3s λ = λ λ E hc. (3) Using th valus of h = J.s and c = cm/s, w find, λ = 4.54Å whil th xprimntal valu is 6Å. Th simpl modl is abl to xplain th sodium doublt. It is possibl to calculat th intnsitis of ths lins using tim-dpndnt prturbation thory. Doublt of Sodium Yllow Lins 5890Å 5896Å Zman Effct W ar considring an atom with on valnc lctron placd in a magntic fild. Th rsult is that th nrgy lvls ar split into svral componnts, Zman splitting. Th intraction nrgy rsponsibl for ths splittings consists of two parts: () du to th spin of th lctron and () du to th orbital motion. Th lctron spin angular momntum is h and has projctions ± h. So th ltron has a componnt of magntic momnt h in th dirction in mc which th componnt of spin angular momntum is ± h. As th nrgy 9

10 of a particl of magntic momnt m in a fild H (xtrnal) is m H, th intraction nrgy of th lctron is mc s H = mc H s. (4) (In 95 Uhlnbck and Goudsmit (Natur 7 (96) 64) discovrd lctron possssing an intrinsic angular momntum whos componnt is rstrictd to ± h. Pauli in 97, dscribd it by his matrics!) Th ffct of xtrnal magntic fild on th orbital motion of th lctron is found now. Th hamiltonian for a particl of charg in th.m vctor potntial A ( H = A) is H = m ( p + c A) = m p + In hr, p, A ar oprators. Thn, ( p A)ψ = i h ( Aψ), mc ( p A + A p) + = i h( A)ψ i h A ( ψ), mc A. (5) = ( A p)ψ, (6) whr th gaug A = 0 (Coulomb gaug) is usd in th pnultimat stp. So th nrgy du to th magntic fild is A p + A. For a mc mc uniform magntic fild, w know A = ( H r). Show this! Thn, w s A = ( H r ( H r) ). W first nglct A trm. Thn, th intraction 4 can b writtn as: mc A p = = = = = mc ( H r) p, mc ɛ ijk H j r k p i, mc H j ɛ jki r k p i, H mc ( r p), H mc L, L = r p. (7) 0

11 Combining with th spin contribution, th total ffct of th xtrnal magntic fild on th atomic lctron is H = = = H mc s + H mc L, H mc ( L + s), H mc ( J + S), J = L + s. (8) If th applid magntic fild is along th z-axis, thn th intraction is mc H(J z + s z ). In (8), w hav combind th orbital and spin angular momnta. So, th wav functions should b appropriatly constructd. From th angular momntum wav functions lm > and spin wav functions m s > w hav th following: jm > = m s C(l j; mm sm) lm > m s >, This for j = l +, bcoms = C(l j; m m + ) lm > > + C(l j; m,, m ) lm > >. (9) l + l + M + M > = l + l M + + l + lm >, > lm + >, >, (30) and for j = l l l M + M > = l + l + M + + l + Now it is asy to find < l + H, M mc (J z + s z ) l +, M > = hh mc l, M > > l, M + >, >. (3) M(l + + ), l +

12 < l H, M mc (J z + s z ) l, M > = hh mc < l ± H, M mc (J z + s z ) l ±, M > = hh mc M(l + ), so that l + M(l ± + ) δ MM.(3) l + Th factor l±+ l+ is known as Landé g-factor. Wak Fild Cas Whn th xtrnal magntic fild is wak, th spin-orbit intractrion cannot b nglctd. This intraction ncssitats th j-j coupling. In this stat, w hav valuatd th matrix lmnts of th magntic intraction in (3). To this w add th spin-orbit coupling. Dnoting th radial matrix lmnt as ζ nl, w hav this as ζ nl{j(j + ) l(l + ) s(s + )}. Strong Fild Cas Whn th xtrnal magntic fild is strong, th spin-orbit intraction is small and in this cas, w us (n, l, m l, m s ) functions (without angular momntum coupling). Thn, < nlm l m s H mc (L z + s z ) nlm l m s > = hh mc (m l + m s ) = hh mc (M + m s). (33) Each lvl is split symmtrically into l + 3, as m l + m s taks l +, l, (l + ) which totals to l + 3. This cas is known as Paschn-Back ffct. W now includ th spin-orbit intraction. Th stats ar lablld by n, l, m l, m s. Th diagonal matrix lmnt < nlm l m s ξ l s nlm l m s > cannot connct x, y componnts of angular momnta and so this is ζ nl m l m s. Transition Cas W will us th fundamntal systm of ign functions lablld by nlm l m s following E.U.Condon and G.H.Shortly, Th Thory of Atomic Spctra. In this schm, th matrix of ɛ(l z + s z ) is diagonal, whr ɛ = H. W shall mc considr th matrix for a givn nl, say p and s prtaining to alkali atom sodium. W shall dnot th radial intgral of ξ by ζ.

13 s Stats: Th ign functions ar Y0 0 χ ; Y 0 0χ. Th matrix lmnt of l s can b omittd sinc it dos not shift thm with rspct to ach othr. Th shift du to th magntic intraction ɛ(l z + s z ) is ±ɛ h for m l = 0; m s = ±. p Stats: Th ign functions Y χ ; Y additional nrgis ar χ hav m = ± 3. For thm th ζ {j(j + ) l(l + ) s(s + )} + ɛ < l z + s z >, (34) which for j = 3 and m = ± 3 givs ζ ± ɛ h., Y 0 χ combin in pairs accord- Th wav functions Y 0 χ, Y χ, Y ing to m l + m s = m = ±. Ths ar: {Y 0 χ, Y χ On such pair can b considrd. Lt χ ; Y χ, Y 0 ψ > = Y 0 χ, ψ > = Y χ χ }. (35), (36) corrsponding to m =. Now w us dgnrat prturbation thory, as ths stats ar dgnrat. W want to find th matrix lmnts of First, th asy on. It is sn that H = ξ(r) l s + ɛ(l z + s z ). (37) < ψ ɛ(l z + s z ) ψ > = ɛ h, < ψ ɛ(l z + s z ) ψ > = 0, < ψ ɛ(l z + s z ) ψ > = 0, < ψ ɛ(l z + s z ) ψ > = 0. (38) Nxt, w nd < ξ(r) l s > btwn ths stats. Writing l s = {l +s + l s + } + l z s z, (39) 3

14 and using w find l ± l, m l > = (l m l )(l ± m l + ) l, m l ± >, (40) < ψ ξ(r) l s ψ > = 0, < ψ ξ(r) l s ψ > = ζ h, < ψ ξ(r) l s ψ > = ζ h, < ψ ξ(r) l s ψ > = ζ h. (4) Thus in th spirit of dgnrat prturbation thory, w nd up with whos ignvalus ar M = ( ɛ h ζ h ζ h ζ h ), (4) λ = {ɛ h ζ h ± (ɛ h + ɛζ h ζ h 4 )}. (43) Hr w rcall that ζ rprsnts th spin-orbit trm contribution and ɛ th magntic prturbation. Thn, ζ h 0 corrsponds to th Strong Fild Cas. In this cas, you ɛ h can s th two ignvalus ar: Similarly, ɛ h ζ h λ + = ɛ h ; λ = ζ h. (44) 0 corrsponds to th Wak Fild Cas and thn in this cas, λ + = ζ h + 3 ɛ h ; λ = ζ h + ɛ h. (45) 3 W rcall in th Strong Fild Cas, w had th nrgy shift as ɛ h(m + m s ) + ζ h m l m s which for m =, m l = m m s, m s = ± givs: m s = ; ɛ h 4

15 and m s = ; ζ h agring with (44). In th Wak Fild Cas, w had th nrgy shift as: ɛ hm l±+ + l+ ζ h (j(j + ) l(l + ) s(s + )) which is ɛ hm l±+ + l+ ζ h {l or (l + )} for j = l + or j = l. This, for l = agrs with (45). 3 p 3 3 p s Wak Fild Zman splitting 5

16 p s 0 Strong Fild Paschn-Back splitting 6