Hydrogen Atom and One Electron Ions

Transcription

1 Hydrogn Atom and On Elctron Ions Th Schrödingr quation for this two-body problm starts out th sam as th gnral two-body Schrödingr quation. First w sparat out th motion of th cntr of mass. Th intrnal potntial function, V(r), is th lctrostatic attraction of two point chargs, V(r) = Z r whr Z is th intgr nuclar charg, +1 for H, + for H +,..., and is th magnitud of th fundamntal charg unit. V r in Å

2 Th potntial function for th hydrogn atom is only dpndnt on th sphrical polar coordinat r. That mans w immdiatly know that th wavfunctions for th hydrogn atom will consist of som typ of radial function, R n (r), tims a sphrical harmonic function, Y m (θ,φ). Th radial function w must solv is again dr n, d R n, Z ( +1) R n, (r) = E R n, (r) μ r dr dr r μr whr μ is th rducd mass of th two-body systm. Notic that with th sizabl diffrnc in mass of th two particls, th rducd mass is vry narly qual to th mass of th lightr particl, th lctron. μ = m M m m + M = P P m P M P is th proton mass and m is th lctron mass. m

3 Th diffrntial quation has known solutions. Thr ar an infinit numbr of ths solutions for ach particular valu of th quantum numbr, and w introduc a nw quantum numbr, n, to distinguish ths solutions. Both and n must labl th diffrnt ignfunctions,.g., R n (r). A condition on this nw quantum numbr, n, that coms about in solving th diffrntial quation is that for som choic of, n may only tak on th valus +1, +, +, and so on. This condition can b invrtd so as to rlat th valu of to n. Thn, w hav that for a givn choic of n, can b only 0, 1,,..., up to n-1. Also, n must b a positiv intgr. From th sparation of variabls, w now hav that th wavfunctions for th hydrogn atom ar R n (r) Y m (θ,φ) Th quantum numbrs that distinguish th possibl stats must satisfy th following conditions: n = 1,,,... = 0,1,,..., n-1 m = -,- +1,..., -1, 1 Th radial functions, R n, (r), that ar th ignfunctions may b constructd from a st of polynomials call th Lagurr polynomials. Som of th radial functions ar givn blow. W will also find numrical vrsions with MathCad. What about nods??

4 Z R ( r) = 1,0 a 0 Zr a 0 Z Zr Zr 0 = R,0(r) a 0 a a 0 0 a o m R (r) =,1 1 Z Zr a a 0 0 Zr a 0 R,0 Zr Zr a0 a Z Zr (r) = a a R,1 1 Z Zr Zr () (r) = 6 81 a a a Zr a 0

5 100(, r θ, φ) = R10() r Y 0, 0(, θ φ) 00 ( r, θ, φ ) = R 0 ( r ) Y 0, 0 ( θ, φ ) 10 ( r, θ, φ ) = R 1 ( r ) Y 1, 0 ( θ, φ ) 11 ( r, θ, φ ) = R 1 ( r ) Y 1, 1 ( θ, φ ) ( r, θ, φ ) = R ( r ) Y ( θ, φ ) , 1 00(, r θ, φ) = R0() r Y 0, 0(, θ φ) n,, m and =0 is calld s, =1 is calld p, tc.

6 Th constants ar for normalization ovr th rang from r =0 to infinity. Rcall that in sphrical polar coordinats th volum lmnt is dv = r dr sinθ dθ dφ, and so th normalization of th radial quations obys th following: z * R () r R,, () r r dr = n l n l 0 1 Th nrgy ignvalu associatd with a particular R n (r) function is found to b E n = μ 4 Z n 16. Z = n V with n = 1,,, Thus, th nrgis of th stats of th hydrogn atom dpnd only on th quantum numbr n. Now 1 V = cm -1. Th lowst nrgy stat is with n = 1, and, for this stat, th nrgy is -109,678 cm -1 ( V). Th nxt nrgy lvl occurs with n =, and this nrgy is on fourth ( - ) of th lowst nrgy or -7,40 cm -1 (-.4 V). Nxt is th nrgy lvl for stats with n =. This nrgy is on ninth ( - ) of th lowst nrgy or -1, cm -1 ( V). At n = 100, th nrgy of th hydrogn atom is -11 cm -1 ( V)

7 Clarly, as n approachs infinity, th nrgy approachs zro. This limiting situation corrsponds to th ionization of th atom; th lctron is thn compltly sparatd from th nuclus, and thr is zro nrgy of intraction. Sinc th nrgy of th hydrogn atom dpnds only on n, and thr ar svral stats with th sam n, th stats of th hydrogn atom xhibit dgnracy. W can us th quantum numbrs to s that th dgnracy of ach lvl is n. With lctron spin, it will b n. Thus th lowst nrgy ignfunction is 100(, r θ, φ) = R10() r Y 0, 0(, θ φ) with nrgy E 1. And th nxt ons ar ( r, θφ, ) = R ( r ) Y ( θφ, ) ,0 10(, r θ, φ) = R1() r Y 1, 0(, θ φ) 11(, r θ, φ) = R1() r Y 1, 1(, θ φ) ( r, θ, φ ) = R ( r ) Y ( θ, φ ) , 1 with nrgy E (, r θ, φ) = R () r Y, (, θ φ) and tc W rfr to ths stats as and =0 is calld s, =1 is calld p, tc. n,, m What do th ignfunctions look lik?? W must tak * to obtain th thr dimnsional i probability bilit distribution ib ti for th lctron. Whn w intgrat to rmov th angular dpndnc, w obtain th radial probability distribution, P radial (r) = r * dr

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9 Ral Hydrognic Wavfunctions Notic that our nic pictur of p z orbital was rtaind, but th othr p 1 and p -1 orbitals had idntical doughnut shaps, not at all what w xpctd. W can undrstand ths orbitals in trms of our arlir physical pictur of th m = +1 and m = -1 angular momntum ignfunctions! Ths two stats with idntical shaps corrspond to an lctron circulating in two diffrnt snss. This dscription will prov vry usful whn w talk about linar molculs, but for th momnt th complx - valud p 1 and p -1 orbitals ar not so usful. Ĥ Considr th following n,,m : 10, 11, and 1,-1. Ths functions ar all ignfunctions i of ĤH and L with ih th sam ignvalus. Thus any linar combination i of thm is still an ignfunction of and L, although not L z. Th functions involving m = ± 1 ar complx. Considr, howvr, th following linar combinations of thm. 1 p = 11 1, 1,1 y i [ ] R (r) sin θsin φ 1 p = 11 1, 1,1 x [ + ] R (r) sin θcosφ Ths linar combinations ar no longr ignfunctions of L z, but thy ar ral and thy hav dirctional proprtis. p Similar things can b don with th = (d) angular functions. Why might thy b usful? Ths ral hydrognic functions ar th ons commonly apparing in txts.

10 1 Z 1 s ( σθφ,, ) = π ao Ral Hydrognic Wav Functions σ 1 Z s ( σθφ,, ) = ( σ) 4 π ao σ σ 1 Z p ( σθφ,, ) = cos z σ θ 4 π ao σ 1 Z p (,, ) = sin cos x 4 π ao σθφ σ θ φ σ 1 Z p y (,, ) = sin sin 4 π ao σθφ σ θ φ 1 Z s ( σθφ,, ) = 7 18σ + σ 81 π a o σ Z p z (,, ) = ( 6 ) cos 81 π ao σ σθφ σ σ θ σ Z p ( σθφ,, ) = σ( 6 σ) sinθcosφ x 81 π ao σ Z p (,, ) = ( 6 ) sin sin y 81 π ao σθφ σ σ θ φ d z 1 Z σ ( σθφ,, ) = σ cos θ π a o Zr whr σ = ao o

11 Visualizing th ignfunctions

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14 Hydrognic Proprtis What do w know about hydrognic wavfunctions? Ĥ L L z n,,m = n,,m = n,,m = n,,m n,,m = L z,,,, pz pz = L z 100 ĤH 00 = Considr th z= (H + ) hydrognic systm. What is n m? = Ĥ H +