《结构化学 Structural Chemistry》课程教学资源（PPT课件）第二章 原子结构 Atomic Structure

函数Φm()复数解实数解m0D(g)= /1/2元D(g)= /1/2元1D(g)= /1/2元els0 (g)= //元 cos@-(0)= //2元e-ig-1[()=/1/元 sin2D2(g)= /1/2元ei2g0()=/1/元cos2gD-2(0)= /1/2元e-12g-2①()=/1/元 sin203Φ,(g)= /1/2元e13g①0()= /1/元cos3g-3D_;()= /1/2元e-3g ()=/1/元 sin3g

m 复数解 实数解 0 1 −1 2 −2 3 −3 i2 2 ( ) 1 2πe    = i2 2 ( ) 1 2πe    − − = i3 3 ( ) 1 2πe    = i3 3 ( ) 1 2πe    − − = 0  ( ) 1 2 = π 0  ( ) 1 2 = π i 1( ) 1 2πe    = 1 i ( ) 1 2πe    − − = cos 1 sin 1 ( ) 1 π cos ( ) 1 π sin           =   =  cos 2 sin 2 ( ) 1 π cos 2 ( ) 1 π sin 2           =   =  cos 3 sin 3 ( ) 1 π cos3 ( ) 1 π sin3           =   =  函数 m( )

2.1.4 ①方程的解dosingdsine+βsindedeO化为连带勒让德(AssociatedLegendre)方程，有已知解对于给定的[(1 = 0,1,2,3.有满足合格条β = l(I+1)件的解(/ ≥|m|m=0,±1,...±1与量子数l,m有关O(0)= CP" (cos0)Imld'+ndcosg+ (cos*0-1)-cos0)Dml(cosO):2/1连带勒让德函数为角量子数（angularmomentumquantumnumber）

 0,1,2,3. ( 1) | | l m l l l  = = +  化为连带勒让德(Associated Legendre) 方程，有已知解 2.1.4 ()方程的解 有满足合格条 件的解 对于给定的l m=0, ±1, .±l | | | | | | 2 2 2 | | 1 d (cos ) (1 cos ) (cos 1) 2 ! dcos m l m m l l l l m P l     + + = − − 连带勒让德函数 与量子数 l, m 有关 sin d d 2 2 sin sin d d m             + =   ( ) (cos ) l m    = CP l 为角量子数 (angular momentum quantum number)

连带Legendra函数m）3/143cos05cosAV6142cosesin 0(5cos2 0-1)28V3V105sinsincos24V1070sin"e(cos?0-1)48V15sincosO注意：归一化条件2V15sin"@'@sinodo=14

注意：归一化条件 π * 0    sin d =1  3 3, 3 70 sin 8    = 0,0  =1 2 1,0 6 cos 2   = 1, 1 3 sin 2    = 2 2,0 10 (cos 1) 4   = − 2, 1 15 sin cos 2     = 2 2, 2 15 sin 4    = 3 3,0 3 14 (5cos 3cos ) 4    = − 2 3, 1 42 sin (5cos 1) 8     = − 2 3, 2 105 sin cos 4     = 连带Legendra函数 lm( )

2.1.5 R(r)方程的解Ze2ur2 dR1dE+=1(1+1)连带拉盖尔（AssociatedLaguerre）方程h?R drdr4元00rn为主量子数n≥/+1n = 1,2,3,..有收敛解条件(principalquantumnumber)Z2le4E8c,h?nR'Rr?dr =l归一化条件：R函数：Rn(r)与量子数n,1有关Rn(r)= Cep'Llt(p)2Z4元6d2/+1n+1P=aonaom.edoBohrradius连带拉盖尔函数

2.1.5 R(r)方程的解 有收敛解条件 连带拉盖尔(Associated Laguerre)方程 R函数：Rn,l(r) 与量子数 n, l 有关 归一化条件： * 0 2 R R r r d 1  =  2 2 2 2 0 1 d d 2 ( 1) d d 4π R r Ze r E l l R r r r         + + = +       4 2 2 2 0 2 8 n e Z E h n   = −  n  l + 1 n = 1,2,3,. 2 2 1 ( ) ( ) l n l R r Ce L l l n    − + = + 2 1 1 2 1 1 2 1 1 d d ( ) e (e ) d d n n n l l l l n L       + + + − + + + +   =     0   2Z r na  = 连带拉盖尔函数 2 0 0 2 e 4π a m e  = Bohr radius n为主量子数 (principal quantum number)

类氢原子径向波函数Rm(r)13/(Z/ao)R.o(r) =2(Z/ao)"2 e-012(24 -36p+12p2 - p )e-0/2R4.o(r) =96)3/2(Z/ao)e-p/2(Z/a.)"2R20(r)2/2(20p-10p+p)e-p/2R4,(r)32V15(Z/a. )3/2pe-pl2R.(r):(Z/a.)"22V6e-p/2R4.2(r):3/96V15(Z/ao)Je-pl2R3,o(r)(6-6p+p9V303/2(Z/a.)e-p/203/2R4.3(r)(Z/ao)96V35p)e-plR(r)402Z9V6O(Z/a.)32naope-p12R3.2(r):9V30

( ) ( ) 3/2 0 / 2 2,0 ( ) 2 e 2 2 Z a R r   − = − ( ) 3/2 / 2 1,0 0 R r Z a ( ) 2 e− = ( ) 3/2 0 / 2 2,1( ) e 2 6 Z a R r   − = ( ) ( ) 3/2 0 2 /2 3,0 ( ) 6 6 e 9 3 Z a R r    − = − + ( ) ( ) 3/2 0 2 /2 3,1( ) 4 e 9 6 Z a R r    − = − ( ) 3/2 0 2 /2 3,2 ( ) e 9 30 Z a R r   − = 0 2Z r na  = ( ) ( ) 3/2 0 2 3 /2 4,0 ( ) 24 36 12 e 96 Z a R r     − = − + − ( ) ( ) 3/2 0 2 3 /2 4,1( ) 20 10 e 32 15 Z a R r     − = − + ( ) ( ) 3/2 0 2 3 /2 4,2 ( ) 6 e 96 15 Z a R r    − = − ( ) 3/2 0 3 /2 4,3 ( ) e 96 35 Z a R r   − = 类氢原子径向波函数Rnl (r)