양자화학 (8주차 원자구조...
TRANSCRIPT
133
→ The hydrogen atom is the first example of a physical system that one encounters in the study of electronic structure.
→ The hydrogen atom orbitals and energy levels will be crucial for understanding both behavior of larger atoms and the basis sets that exist for the description of molecules.
→ Schrödinger equation for the hydrogen atom can be solved exactly.
• Differences between 3D rigid rotor and hydrogen atom
2 2 2
2 2
0
ˆ2 2 4r
ZeH V rr
21837 0.99946
1838e n e
ee n e
m m m mm m m
수소 원자
134
Hamiltonian in 3D rigid rotor
, , ,nlm nl lmr R r Y
2 2
2 2 2 20
2 ( 1) 2 2( ) ( )4
l l mZe mER r R rr r r r r
This suggest that the solution depends on the quantum number l.
슈뢰딩거 방정식
135
Radial Kinetic Energy
2 2 22
2 20
( 1)2 2 4
d dR Zer l l Er R dr dr r r
Rotational Kinetic Energy (Centrifugal)
CoulombicPotential
Nucleus
ElectronRadial motion
(θ and φ constant)
Angular motion(r constant)
에너지
136
2 2 22
2 20
( ) ( 1) ( ) ( )2 2 4
d dR r Zer l l R r ER rr dr dr r r
2 2
20
12 4eff
l l ZeVmr r
수소 원자의 유효 포텐셜
137
2 2
2 2 2 20
2 ( 1) 2 2( ) ( )4
l l mZe mER r R rr r r r r
Ground state energy
( ) rR r e
22 /
2 2 20
2 ( 1) 2 2 04
rl l mZe mE er r r
2 2 42
22 40
24
mE m Z e
2 4
2 204 2
mZ eE
0 0
1/23
/ /2 1, ,34 3
0 00
4 ( 1)! 2 2( )!
lZr na Zr nal l
n l n l n lZ n l Zr ZrR r e L N L r r e
na nan a n l
Principal quantum number (n): n = 0, 1, 2, …Orbital angular momentum (l): l = 0, 1, 2, …, n‐1
20
0 2 0.524 9ame
Bohr radius
Å
방사 방향 해
138
0 0
1/23
/ /2 1, ,34 3
0 00
4 ( 1)! 2 2( )!
lZr na Zr nal l
n l n l n lZ n l Zr ZrR r e L N L r r e
na nan a n l
0
0
0
10
220
0
221
12
Zra
Zra
Zra
R r e
ZrR r ea
R r re
The number of radial nodes: (n‐l‐1)
방사 방향 파동 함수
139
0 0
1/23
/ /2 1, ,34 3
0 00
4 ( 1)! 2 2( )!
lZr na Zr nal l
n l n l n lZ n l Zr ZrR r e L N L r r e
na nan a n l
방사 방향 파동 함수
140
n l Degeneracy
Name Energy(Eh)
1 0 1 1s -Z2/2
2 0 1 2s -Z2/8
2 1 3 2p -Z2/8
3 0 1 3s -Z2/18
3 1 3 3p -Z2/18
3 2 5 3d -Z2/18
Quantum numbers, names, and energies of lowest states of H ato
m
2 2 2 2
2 20
ˆ1ˆ2 2 4
J ZeH rm r r mr r
' ' '
' ' ' '
* 2
* 2 *
' ' '
' ' '
sin
sin
n l m
n l l m
nlm
nl lm
n n l l m m
n l m nlm
r drd d
R R r dr Y Y d d
100
200
21 1 , 210 , 211
300
31 1 , 310 , 311
32 2 , 32 1 , 320 321 , 322
2
2
, , ,
2
nlm nl lm
hn
r R r Y
nlm
E ZEn
1, 2,3,0,1,2,3, , 1
nl n
수소의 파동 함수
141
* 2
0
( ) ( ) 1nl nlR r R r r dr
2
2 *
0 0 0
sin 1nlm nlmr dr d d
2 2ˆ ( 1)ˆ
nlm nlm
z nlm l nlm
L l l
L m
수소 원자 파동 함수
142
2 4
22 20
2 2 2
2 20 0
12 4
1,2,3, 4,...4 2 2
n
hn
mZ eEn
e Z Z na n n
E E
2
0 0
27.211 eV 627.51 kcal/mol4h
eEa
' "n nE E E
2 22
2 22
1 1 1 12 ' "
1 1" '
h
H
E ZEhc hc n n
Z Rn n
21
0 0
1 109737.31 2 2 4
hH
E eR cmhc hc a
수소 원자 에너지
143
22
2 2
12
hn H
E ZE Z hcRn n
수소 원자 에너지
144
The number of radial nodes: (n-l-1)
방사 방향 파동 함수
145
방사 방향 확률 밀도 (R(r)2)
146
2 0, 2
111 12n l
l l ar nn Z
What is the most probable distance of finding electron in 1s orbital?
방사 방향 확률 밀도 (R(r)2)
147
22 2 2( ) sin ( )nl nlm nlP r r d d R r r the probability of finding the electron between r and r + dr
확률 밀도 함수
148
확률 밀도 함수
149
Number of Radial Node: n – l – 1
방사 방향 밀도 및 마디
150
각 방향 파동 함수
151
Number of Angular Node Node: l
각 방향 파동 함수 및 마디
152
• For l = 0, s orbitals (spherically symmetric, nodal surfaces are spherical)• For l ≠ 0 and m = 0, z‐type orbitals• For l ≠ 0 and m ≠ 0, the total number of nodes is n ‐ 1
Radial nodes: (n‐l‐1)Angular nodes: l
전체 파동 함수
153
1/20
20
0 2
2
0 0
2 30 17
4
1 (action unit)1 (mass unit)
/ 4 1 (charge unit)
4 1 bohr (length unit)
27.2114 1 hartree (energy unit)4
42.41888 10 (time unit)
e
e
e
h
e
m
q e
am e
eE eVa
sm e
2 2
2 2
02 2
2 2
1ˆ2 4 2
2 2h
n
Ze ZHm r r
E Z ZEn n
원자 단위
154
•Wavefunctions and Energies for H Atom
, , ,nlm nl lmr R r Y nlm
2
22h
nE ZE
n
요약 (CH 9 -1)