x z y r but vary 0 - 180 o & 0 - 360 o r 2 = x 2 + y 2 + z 2 x = r sin cos y = r sin sin ...
Post on 19-Dec-2015
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x
z
y
q
f
r
but q & f vary0 - 180o & 0 - 360o
r2 = x2 + y2 + z2
x = r sin q cosfy = r sin q sinfz = r cosqr = (x2 + y2+ z2)½
q = Acos(z/r)f = Asin{y/(rsinq)}
dV = dt = r2 sinq dr df d q or …..dV(shell) = dx • dy • dz = 4pr2 sinq dr df dq
Polar Coordinates
CM - 2D – Particle on a ring“If a particle is confined to the xy plane, then it has angular momentum along the z axis.
angular momentum: Lz = mvr
Lz = x • py – y • px
E = K + V = K = ½mv2 • (m/m) = p2/2m = Lz2/2mr2 = Lz
2/2I
moment of inertia: I = mr2
QM - 2D – Particle on a ring“If a particle is confined to the xy plane, then it has angular momentum along the z axis.
Ĺz = x • py – y • px = -iħ d( )/df
ĤY = Ĺz2/2I Y = -ħ2/2I • d2Y/df2 = EY
E = m2ħ2/2I
y = Neimf
y = (2p)-1/2 eimf
N2 • ∫02p e-im f eim f d f = 1
quantum #, m (or ml ) = 0, ±1, ±2, ±3, ….
2D – Particle on a ring
y = (2p)-1/2 eimf
CM E = ½mv2 = p2/2m = J2/2I (I = mr2)
QME = L2/2I (I = mr2)L = hr/l = mlħl = hr/LE = (mlħ)2/2I
moon data m = 7.3 x 1022 kg v = 1,020 m s-1
r = 384000 km1 revolution = 27.3 days
What is I?What is E (CM)?
What is J?
What quantum state ml?
Goal - Separate problem into 2 distinct one-particle problems.
Two Particle Problems
Particle 1: Mass = m1 + m2
Position = center of mass XYZ e.g. X = (m1x1 + m2x2)/(m1 + m2)
example: translational motion of diatomic molecule (O2)
Particle 2: Mass = m = m1m2/(m1+ m2) Position = position of smaller particle relative to the center of mass
use x, y, z or (best) polar coordinates; r, , qf
example: rotation of 2 bodies attached at fixed distance (O2)
x -1 +1
e.g. H2 moleculem1 = m1 + m2 ~ 2 amu
Two Particle Problems
x = (m1x1 + m2x2)/(m1 + m2)
x1 = 0
m2 = m1m2/(m1 + m2) = ½amu
xcm = 0r2 = ½ bond distance = 0.60 Å
e.g. H atomm1 = m1 + m2 ~ 1 amu x = (m1x1 + m2x2)/(m1 + m2)
x -1 +1x1 ~ -1
m2 = m1m2/(m1 + m2) ~ me
xcm ~ -1 = -0.99891 r2 ~ r = 0.529 Å
m2 = 9.04 x 10-28 kgm2 = 1.67 x 1010-24 kg
r2 = x2 + y2 + z2
x = r sin q cosfy = r sin q sinfz = r cosqr = (x2 + y2+ z2)q = Acos(z/r)f = Asin{y/(rsinq)}
dt = r2 dr • sinq d q • df
Ĥ = -ħ2/2I {d2( )/dq2 + cotq d( )/dq + 1/sin2q d2( )/df2} & cot = q cos /q sinq
Ĥ = Ĺ2/2I + V̂
3D – rotations (fixed r) rigid rotor
0 - 360oq
fr
0 – 180o
Ĺ2 = -ħ2 (d2( )/df2 + cot q d( )/dq + 1/sin2dq d2( )/df2) L = {ℓ(ℓ+ 1)}½ħ
Ĺz = -iħ d( )/d f Lz = mℓħ
Fml = (2p)-1/2 exp(imlf)Ql,ml
associated Legendre polynomials
Q0,0 = 2-1/2
Q1,0 = (3/2)1/2 cos qQ1,1 = (3/4)1/2 sin qQ2,0 = (5/8)1/2 (3cos2 - 1)qQ2,1 = (15/4)1/2 sin q cos qQ2,2 = (15/16)1/2 sin2 q
Yrot = Fml Ql,ml After solving Ĥy = Eroty ....
where l = 0, 1, 2, …
degeneracy = 2l + 1
ml = -l, - l +1, - l +2, .... + l
Ĥ = -ħ2/2I {d2( )/dq2 + cotq d( )/dq + 1/sin2q d2( )/df2}
L = {ℓ(ℓ+ 1)}½ħĹ2 = -ħ2 (d2( )/df2 + cot q d( )/dq + 1/sin2dq d2( )/df2) L = {ℓ(ℓ+ 1)}½ħ
Lz = mℓħĹz = -iħ d( )/d f = mℓħ
Erot = Eq + Ef = {l(l + 1)-ml2}(ħ2/2I) + ml
2(ħ2/2I) = l(l + 1)ħ2/2I I = mr2
Fml = (2p)-1/2 exp(imlf)Ql,ml
associated Legendre polynomials
Q0,0 = 2-1/2
Q1,0 = (3/2)1/2 cos qQ1,1 = (3/4)1/2 sin qQ2,0 = (5/8)1/2 (3cos2 - 1)qQ2,1 = (15/4)1/2 sin q cos qQ2,2 = (15/16)1/2 sin2 q
Yrot = Fml Ql,ml After solving Ĥy = Eroty ....
where l = 0, 1, 2, …
degeneracy = 2l + 1
ml = -l, - l +1, - l +2, .... + l
Ĥ = -ħ2/2I {d2( )/dq2 + cotq d( )/dq + 1/sin2q d2( )/df2}
Erot = Eq + Ef = {l(l + 1)-ml2}(ħ2/2I) + ml
2(ħ2/2I) = l(l + 1)ħ2/2I I = mr2 Eq for E0,0 = 0 since Q0,0 = 2-1/2 and Erot = 0 + 0 = 0
Eq for E1,0 = Ĥ = q -ħ2/2I [d2( )/dq2 + cot q • d( )/dq]q = Eq q
= -ħ2/2I • (3/2)1/2 [d2(cos q)/dq2 + cot q • d(cos q)/dq] = Eq (q cot = q cos /q sin )q = -ħ2/2I • (3/2)1/2 [-1 - cos /q sin q • sinq] = (-ħ2/2I • -2) q
= ħ2/ I • qEq = ħ2/I and Erot = ħ2/I + 0 = ħ2/I =
(+1)ħ2/2I
angular momentum l = 0 mℓ = 0
L = 0 Lz = 0
Ĺ2 = -ħ2 (d2( )/df2 + cot q d( )/dq + 1/sin2dq d2( )/df2) L = {ℓ(ℓ+ 1)}½ħ
Ĺz = -iħ d( )/d f = mℓħ
Yrot = Q • F (spherical harmonics) … is an eigenfunction of the Ĺz and Ĺ2 operators but not Ĺ, Ĺx nor Ĺy.
The overall angular momentum scalar value is determined by (Ĺ2)½ .
Note that this is not a diagram indicating where the electron may be found.Rather it is a vector diagram representing the limitations on the values of the angular momentum of the electron. The electron with l = 0 (an s orbital)Has no z-component to its angular momentum – it is not confined to a circle.
angular momentum l = 0, 1
L = 21/2 ħ Lz = +1ħ
L = 0 Lz = 0
L = 21/2 ħ Lz = -1ħ
Ĺ2 = -ħ2 (d2( )/df2 + cot q d( )/dq + 1/sin2dq d2( )/df2) L = {ℓ(ℓ+ 1)}½ħ
Ĺz = -iħ d( )/d f = mℓħ
Yrot = Q • F (spherical harmonics) … is an eigenfunction of the Ĺz and Ĺ2 operators but not Ĺ, Ĺx nor Ĺy.
The overall angular momentum scalar value is determined by (Ĺ2)½ .
angular momentum l = 2
L = 61/2 ħ Lz = +2ħ
L = 61/2 ħ Lz = +1ħ
L = 61/2 ħ Lz = 0
L = 61/2 ħ Lz = -1ħ
L = 61/2 ħ Lz = -2ħ
The Hydrogen atom a 3D, 2 particle problem
.
Y = yN ye yN is translational motion of H atom
ye is electron motion relative to nucleus
e(r, , q f) = () () R(r)
“Radial function”
Spherical harmonics
Force between two charges in vacuum F = q1 • q2/(4peo • r
2) (N = kg m s-2)
V = F • r = q1 • q2/(4peor) (J = kg m2 s-2)
Ĥ = y R Q F
-ħ2/2m [1/r2 d(r2dy/dr)/dr + 1/(r2sinq) d(sinqdy/dq)/dq + 1/(r²sin²q) d²y/df²] - Ze2/(4peor)
= Ey
Hamiltonian: Ĥ = y Ky + Vy = Ey
e(r, , q f) = R(r) () ()
V = -Ze2/(4peor)Applies to H atom and any H-like ion with only 1 e-.
K (R(r) () () ) Vy
Spherical harmonics () () These solutions are identical to the rigid rotor model
Ĥ = y -ħ2/2m [1/r2 d(r2dy/dr)/dr + 1/(r2sinq) d(sinqdy/dq)/dq + 1/(r²sin²q) d²y/df²] - Ze2/(4peor)
reduces to ……
ESH = ħ2l(l + 1)/2mr2
ĤR = -ħ2/2m [1/r2•d(r2dR/dr)/dr + {ħ2l(l+1)/2mr2 - Ze2/(4peor)}R
Ĥ(r, , q f) = Ĥ(ℓ,mℓ() mℓ()) + Ĥ R(r)
Ĥ R = -ħ2/2m [1/r2•d(r2dR/dr)/dr + {ħ2l(l+1)/2mr2 - Ze2/(4peor)}R
As the solutions to the Q function already existed in the form of the associated Legendre Polynomials …..An exact solution to the Hamiltonian for R exists using pre-existing mathematics ….. the associated Laguerre Polynomials
n,ℓ,mℓ (r, , q f) = ℓ,mℓ() mℓ() Rn,ℓ(r)
These solutions contain the quantum numbers (n and ℓ) such that ….• n = 1, 2, 3, ….. (principle quantum #)• ℓ = 0, 1, … n – 1 (angular momentum q# as in Q)• mℓ = -ℓ, -ℓ+1, … +ℓ (magnetic quantum number as in F and )Q
R(r) = radial function
R = cst * (n-1)th order polynomial in r * e-Zr/2a
a = 4peoħ2/me2 (units = m) which happens to equal the Bohr orbit radii (0.529Å for H)
n,ℓ,mℓ (r, , q f) = ℓ,mℓ() mℓ() Rn,ℓ(r)
ĤR = -ħ2/2m [1/r2•d(r2dR/dr)/dr+ {ħ2l(l+1)/2mr2 - Ze2/(4peor)}R = ER
E = - Z2e4m/(8eo2h2n2) eq 11.66
What is ynlm?
y100 = 1s = (p-½) • (Z/a)3/2 • exp(-Zr/a)
y200 = 2s = (32 )p -½ • (Z/a)3/2 • {2-(Zr/a)} • exp(-Zr/a)
y210 = 2pz = (32p)-½ • (Z/a)5/2 • r • exp(-Zr/2a) • (cosq)
y211 = (64p)-½ • (Z/a)5/2 • r • exp(if) • exp(-Zr/2a) • sinq
y21-1 = (64p-½) • (Z/a)5/2 • r • exp(-if) • exp(-Zr/2a) • sinq
2px = 2-1/2(y211 + y21-1) 2py = -i2-1/2(y211 - y21-1)
If any 2 wave functions satisfy H and give the same E, then any linear combination of those 2 wave functions (renormalized) will also give the same E.
2px = (32 )p -½ (Z/a)5/2 • r • exp(-Zr/2a) • sincos
2py = (32 )p -½ (Z/a)5/2 • r • exp(-Zr/2a) • sinsin
2pz = (32 )p -½ (Z/a)5/2 • r • exp(-Zr/2a) • cos
a = 4peoħ2/me2 (units = m)
-0.500
0.000
0.500
1.000
1.500
2.000
0 2 4 6 8 10 12 14 16 18 20
R-10 R-20 R-30 R-10 R-20 R-30
1s
2s
3s
H-atom – Radial Functions
-0.10
-0.05
0.00
0.05
0.10
0.15
0.20
0 2 4 6 8 10 12 14 16 18 20
R-21 R-31 R-32 R-21 R-31 R-32
2p
3p
3d
H-atom – Radial Functions
dV = dx dy dz = d = 4pr2 dr sin d d
dV = dx dy dz
dV
dV(cube) = dvxdvydvz = ? dr
dV = 4p(r+dr)3/3 – 4pr3/3 = 4r2dr
4pr3/3 + 4pr2dr + 4prdr2 + 4pdr3/3
-0.100
0.000
0.100
0.200
0.300
0.400
0.500
0.600
0.700
0 2 4 6 8 10 12 14 16 18 20
R-10 R-20 R-30
r2R2
1s
2s
3s
Probability = r2R2 R2
Even though the radial function for s orbitals is maximal at r = 0 the r2 term in dt, drops the probability to 0 at the nucleus.
r2R2
r/a
-0.100
0.000
0.100
0.200
0.300
0.400
0.500
0.600
0.700
0 2 4 6 8 10 12 14 16 18 20
R-10 R-20 R-30
r2R2
1s
2s
3s
Y1s = (p-½) • (1/a)3/2 • exp(-r/a)
r/a
rmp = r when d (r2Y1s)dr = 0 = a
-0.05
0.00
0.05
0.10
0.15
0.20
0.25
0 2 4 6 8 10 12 14 16 18 20
R-21 R-31 R-32
r2R2
2p 3p
3d
1 2 3 4 5 6 7
-16.00
-14.00
-12.00
-10.00
-8.00
-6.00
-4.00
-2.00
0.00
Ĥ = y -ħ2/2m [1/r2 d(r2dy/dr)/dr + 1/(r2sinq) d(sinqdy/dq)/dq + 1/(r²sin²q) d²y/df²] - Ze2/(4peor)
= E y and = E = -Z2e4m/(8eo2h2n2) eq
11.66
eV
Probability = ∫e* e dt = 4 p ∫r
(r+dr) r2R dr • ∫q(q+dq) sin d • ∫f(f+df) d
e(r, , q f) = () () R(r)
1 = 4 p ∫0∞ r2R dr • ∫0p sin d • ∫02p d
= 1 1 1
prob = 4 p ∫r(r+dr) r2R dr (over all space)
Probability = ∫e* e dt
Prob = 4 p ∫00.1Å r2R2 dr = 4 p ∫0
0.189a r2R2 dr
51
R = 2 (1/a)3/2 exp(-r/a) R2 = 4/a3 exp(-2r/a)
Prob = 16 /p a3 ∫00.189a r2exp(-2r/a) dr
∫ r2exp(-2r/a) dr = (-ar2 • e-2r/a)/2 – (-a • ò r e-2r/a dr)
= (-ar2 • e-2r/a)/2 – (-a • a2/4 • e-2r/a • (-2r/a – 1) ]00.189a
ò xm ecx dx = (xm • ecx)/c – m/c • ò x(m-1) ecx dx
ò x2 ecx dx = (x2 • ecx)/c – 2/c • ò x ecx dx
ò x ecx dx = ecx)/c2 • (cx – 1) c = -2/a
Prob = 4 p • 4/a3 [(-ar2 • e-2r/a)/2 – (-a • a2/4 • e-2r/a • (-2r/a – 1) ]00.189a
a (Å) Z r -Zr/a A2 e-2zr/a r2/b -2r/b2 2/b3 e-zr/a Prob0.529 1 1.00E-01 -3.78E+00 27.0 0.68518 -2.65E-03 -1.40E-02 -3.70E-02 1 6.80E-030.529 10 1.00E-01 -3.78E+01 27020.5 0.02281 -2.65E-04 -1.40E-04 -3.70E-05 1 7.28E-01
What is an orbital?
the size of the orbital depends on confidence level desired. e.g. ....
0r r2R dr * 0
p sinq Q dq * 02p F df = 0.9… says that the probability is 90% of finding e-
within the distance r of the nucleus. r is different at different q & f angles for non s orbitals.
The shape of an orbital is the volume enclosed by a surface of constant probability density = |2| d
a one electron spatial wave function
y100/1s (p-1/2)(Z/a)3/2 exp(-Zr/a)
y200/2s (32 )p -1/2(Z/a)3/2 (2-Zr/a) exp(-Zr/2a)
y21-1 (64 )p -1/2 (Z/a)5/2 sin q exp(-if) r exp(-Zr/2a)
y211 (64 )p -1/2 (Z/a)5/2 sinq exp(if) r exp(-Zr/2a)
2px (32 )p -1/2(Z/a)5/2 r exp(-Zr/2a) sinqcosf
2py (32 )p -1/2(Z/a)5/2 r exp(-Zr/2a) sinqsinf
y210/2pz (32 )p -1/2(Z/a)5/2 r exp(-Zr/2a) cosq
(2px)2 (32 )p -1(Z/a)5 r2 exp(-Zr/a) sin2qcos2f
(2py)2 (32 )p -1(Z/a)5 r2 exp(-Zr/a) sin2qsin2f
(2pz)2 (32 )p -1(Z/a)5 r2 exp(-Zr/a) cos2q
Prob = |2| d
0r r2R dr * 0
p sinq Q dq * 02p F df = 0.9
0.00
0.02
0.04
0.06
0.08
0.10
0.12
0.14
0.16
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5
2s
2px
2py
2pz
r/a
Prob: = 90 & = 0
r222px
2s
2py & 2pz
0.00
0.02
0.04
0.06
0.08
0.10
0.12
0.14
0.16
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5
2s
2px
2py
2pz
r/a
Prob: = 90 & = 90
r222py
2s
2px& 2pz
0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.00.00
0.02
0.04
0.06
0.08
0.10
0.12
1s 2s 2px 2py 2pz
r/a
Prob: = 45 & = 35
r22
2py
2px
2pz
2s
1s
angular momentum l = 0 mℓ = 0
L = 0 Lz = 0
Ĺ2 = -ħ2 (d2( )/df2 + cot q d( )/dq + 1/sin2dq d2( )/df2) L = {ℓ(ℓ+ 1)}½ħ
Ĺz = -iħ d( )/d f = mℓħ
Yrot = Q • F (spherical harmonics) … is an eigenfunction of the Ĺz and Ĺ2 operators but not Ĺ, Ĺx nor Ĺy.
The overall angular momentum scalar value is determined by (Ĺ2)½ .
angular momentum l = 1 mℓ = 0
L = √2ħ Lz = 0
Ĺ2 = -ħ2 (d2( )/df2 + cot q d( )/dq + 1/sin2dq d2( )/df2) L = {ℓ(ℓ+ 1)}½ħ
Ĺz = -iħ d( )/d f = mℓħ
Note that this is not a diagram indicating where the electron may be found.Rather it is a vector diagram representing the limitations on the values of the angular momentum of the electron. The electron with l = 0 (an s orbital)Has no z-component to its angular momentum – it is not confined to a circle.
angular momentum l = 0, 1
L = 21/2 ħ Lz = +1ħ
L = 0 Lz = 0
L = 21/2 ħ Lz = -1ħ
Ĺ2 = -ħ2 (d2( )/df2 + cot q d( )/dq + 1/sin2dq d2( )/df2) L = {ℓ(ℓ+ 1)}½ħ
Ĺz = -iħ d( )/d f = mℓħ
Yrot = Q • F (spherical harmonics) … is an eigenfunction of the Ĺz and Ĺ2 operators but not Ĺ, Ĺx nor Ĺy.
The overall angular momentum scalar value is determined by (Ĺ2)½ .
angular momentum l = 2
L = 61/2 ħ Lz = +2ħ
L = 61/2 ħ Lz = +1ħ
L = 61/2 ħ Lz = 0
L = 61/2 ħ Lz = -1ħ
L = 61/2 ħ Lz = -2ħ