Download Text book: Physical Chemistry 3rd edition, Mortimer, 2008 (物化（三

Text book: Physical Chemistry 3rd edition, Mortimer, 2008
(物化（三）text book for next semester)
Reference books:
Physical Chemistry 2nd edition, Engel and Reid, Pearson 2010
Physical Chemistry, McQuarrie and Simon, 1997
Atkins' Physical Chemistry 7th edition, Atkins and Paula, 2002
Homework： 20%
Midterm-I ： 15%
Midterm-II： 20%
Midterm-III：15%
Final：
30%
任課老師：朱超原 老師
助
教：孔令鈞 助教
Physical Chemistry I (PChemI)
●●●●●
●●●●●
●●●●●
Many molecules in box
State of Equation
P
PV = nRT 2
T2
Ideal gas equation
T1
T 2 >> T1
V
2 ⎞
⎛
an
⎜P +
⎟ (V − nb ) = nRT
2 ⎟
⎜
V
⎝
⎠
Van Der Waals equation
Why is V2, not V3
Physical Chemistry II (PChemII)
One molecule in box
●
Schrödinger equation
Temperature
Hˆ Ψ k = E k Ψ k
Energy is discrete
time
Infinite molecules
Classical ≈ Quantum
One molecule
Pure
quantum
Physical Chemistry III (PChemIII)
From one to infinite molecules
●●●●●
●●●●●
●●●●●
●
2
⎛
an
⎜P +
2
⎜
V
⎝
Hˆ Ψ k = E k Ψ k
Partition function Q =
∑
k
⎞
⎟ (V − nb ) = nRT
⎟
⎠
⎛
Ek ⎞
⎟⎟
exp ⎜⎜ −
⎝ k BT ⎠
I. Origins of quantum mechanics
• Introduction
• Classical harmonic oscillator
• Classical wave
• Maxwell electromagnetic wave
I-1. Introduction
Wave
Particle
d 2r
dv
F = ma = m
=m 2
dt
dt
Newton equation
∂2 f
2
=
1 ∂2 f
∂x
v 2 ∂t 2
Wave equation
Wave and particle
∂ψ
= 2 d 2ψ
i=
=−
+V ψ
2
∂t
2m dx
Schrödinger equation
Galileo once wrote
“The book of nature is written in the language of mathematics.”
Energy conservation x = x (t )
Potential energy
F =−
Newton equation
V
v
V0
v0
−
∂V ( x )
∂x
dv
dV ( x )
=m
dt
dx
− ∫ dV = ∫ mvdv
Upper limit represents time at any t
Lower limit 0 represents time at t0
1
E = mv
2
Total energy
2
One-dimensional example
−
dv
dV ( x )
dt = m dt
dt
dx
− dV ( x ) = mvdv
1 2
1 2
mv + V = mv0 + V0
2
2
+ V = const
Kinetic energy
Potential energy
SI -- Units
2
d x
dv
F = ma = m
=m 2
dt
dt
distance
x
Meter = m
t
time
Second = s
m
mass
Kilogram = kg
F
force
Energy
1
T = mv
2
2
Kg m/s2 = Newton=N
Kg m2 /s2 = Nm=Joule=J
Classical diatomic molecule
2
d x1
F1 = k ( x 2 − x1 − l 0 ) = m 1
> 0　2
dt
d 2 x2
F 2 = − k (x 2 − x1 − l 0 ) = m 2
< 0
2
dt
Coordinate transformation
Center-of-mass coordinate
Relative coordinate
d2X
=0
2
dt
dX
= const = 0
dt
set up center-of-mass at origin
X = v0 t + X 0 = 0
Solved
m1 x 1 + m 2 x 2
X =
m1 + m 2
x = x 2 − x1 − l 0
d 2x
k
+
x = 0
2
μ
dt
m1m 2
μ=
m1 + m 2
Reduced mass
m1m 2
Equivalent to mass μ =
m1 + m 2
Equilibrium position
←
Force direction
μ
→
F = μ
d
dt
2
x
2
F = − k ( l − l 0 ) = − kx
x = 0→ F = 0
x > 0→ F < 0
x < 0→ F > 0
= − kx
d 2x
k
+
x = 0
2
μ
dt
I-2. Classical harmonic oscillator
d 2x
k
+
x = 0
2
μ
dt
x ( t ) + ω
2
ω =
k
μ
x = 0
sin (ω t )
Solution
cos (ω t )
e iω t
e
− iω t
General solution x (t ) = A sin (ω t + φ 0
max x (t ) = A
)
A
τ =
x (t )
2π
τ = Period for one oscillation
ω
A
t
−A
x (t ) = A sin (ω t + φ 0
τ =
2π
ω
= 2π
μ ⎛⎜
k ⎜⎝
kg
=
N /m
=
)
φ0 = 0
(
kg
kg m / s 2
)
⎞
= s⎟
⎟
/m
⎠
Frequency = number of oscillations per second
1
1
ν =
=
τ
2π
⎞
⎛1
=
Hz
⎟
⎜
⎠
μ ⎝s
k
Example τ =0.1s
ν=10Hz
Example: Hydrogen molecule H2
force constant = 500 N/m
find vibrational frequency
mH = 1.008amu
μ=
mH mH
mH + mH
1
ν =
2π
1.008
1.008
=
amu =
× 1.6605 × 10 −27 kg
2
2
500 N / m
μ
= 1 . 23 × 10 14 Hz
Kinetic energy
2
1
1 ⎛ dx ⎞
1
=
μω 2 A 2 cos 2 (ω t + φ 0 )
T = μv 2 = μ ⎜
⎟
2
2 ⎝ dt ⎠
2
Potential energy
F =−
∂V ( x )
= − kx
∂x
V =
1 2
kx
2
1 2 1
V = kx = kA 2 sin 2 (ω t + φ 0 )
2
2
Total energy
1
E = T +V =
μω
2
2
A
2
1
= kA
2
2
A constant of the motion
PCIIweek1-1作業
Page 628
Problems
14.1
14.2
I-3. Classical wave
Waves carry momentum, energy, but not matter
z (x , t )
Oscillation amplitude
x
Travelling wave
Snapshot at t1
Snapshot at t2
z (x , t )
Standing wave
x
Nodes
Snapshot at t1
Snapshot at t2
Snapshot at t3
Snapshot at t4
Standing wave
2
∂ z
∂x 2
=
z (x , t )
x
2
1 ∂ z
c 2 ∂t 2
z (x = 0, t ) = 0
z (x , t = 0 ) = 0
z ( x , t ) = ϕ ( x )η (t )
d 2ϕ (x )
2
+ k ϕ (x ) = 0
2
dx
ϕ (0 ) = ϕ (L ) = 0
z (x = L , t ) = 0
Boundary conditions
⎛ nπ ⎞
x⎟
L
⎝
⎠
ϕ ( x ) = ϕ n ( x ) = D sin (k n x ) = D sin ⎜
k = kn =
nπ
,
L
n = 1,2,3 "
Wave number
d 2η (t )
2
+ k 2 c 2η (t ) = 0
dt
η (0 ) = 0
⎛ nπ ⎞
ct ⎟
⎝ L
⎠
η (t ) = η n (t ) = B sin (k n ct ) = B sin ⎜
z n ( x , t ) = ϕ n ( x )η n (t )
nth-mode wave solution
z n ( x , t ) = ϕ n ( x )η n (t ) = A sin (k n x )sin (k n ct )
Wavelength λ (at fixed time t2 )
Asin(kn x)sin(knct2 ) = Asin(kn (x + λ ))sin(knct2 )
nπ
kn =
,
L
n = 1,2,3 "
z (x , t 2 )
λ
2π
knλ = 2π → λ = λn =
kn
Period τ (at fixed space x2 )
Asin(kn x2 )sin(knct) = Asin(kn x2 )sin(knc(t + τ ))
kncτ = 2π →τ = τ n =
2π λn
=
ckn c
1
c
Frequency ν ν = ν n = τ = λ
n
n
x
z (x 2 , t )
τ
wave speed
t
z (x , t )
x=0
λ1 = 2L
n=1
Fundamental
z (x , t )
x=0
x=L
λ3 =
n=3
z (x , t )
x=0
λ2 = L
n=2
First overtone
z (x , t )
2
L
3
x=L
x=0
Standing wave
x=L
λ4 =
n=4
1
L
2
x=L
Wave number is quantized in classical wave mechanics
from Newton equation.
Schrödinger equation does not give anything quantized,
but its boundary condition does.
All
z n ( x , t ) = An sin (k n x )sin (k n ct )
∂ 2 zn
satisfy
∂x
2
=
1 ∂ 2 zn
c
2
∂t
2
∂2 (zn + zn′ )
∂x
2
=
1 ∂2 (zn + zn′ )
c2
∂t 2
So that general solution goes to
z (x , t ) =
∞
∞
n =1
n =1
∑ z n (x , t ) = ∑ An sin (k n ct )sin (k n x )
Fourier series
For example
⎧ z1(x,t1) + z2(x,t1) + z3(x,t1)
z(x,t) = ⎨
⎩z1(x,t2) + z2(x,t2) + z3(x,t2)
L
t1 =
4c
Figure 14.8 (a)
3L
t2 =
4c
Figure 14.8 (b)
z (x , t )
Travelling wave
∂2z
∂x 2
=
1 ∂2z
c 2 ∂t 2
Wave number
x
Boundary condition z ( x , t = 0 ) = sin( kx )
⎧ z R ( x, t ) = A sin (kx − kct )
z ( x, t ) = ⎨
⎩ z L ( x, t ) = A sin (kx + kct )
z R ( x, t ) = A sin (kx − kct ) = A sin (k ( x − ct )) x =ct = 0
node
z L ( x, t ) = A sin (kx + kct ) = A sin (k ( x + ct )) x = − ct = 0
z ( x, t ) = z L ( x, t ) + z R ( x, t ) = 2 A sin (kx ) cos(kct ) Standing wave
Wave number k is not discrete !
Wavelength λ (at fixed time t2 )
A sin (kx ± kct 2 ) = A sin (k ( x + λ ) ± kct 2 )
2π
kλ = 2π → λ =
k
Period τ (at fixed space x2 )
A sin (kx 2 ± kct ) = A sin (kx 2 ± kc(t + τ ))
2π λ
kcτ = 2π →τ =
=
ck c
Frequency ν
ν=
1
τ
=
c
λ
wave speed
Properties of wave
Interference
Reflection
Diffraction
Classical particle does not has interference property
I-4. Maxwell electromagnetic wave
We can define scalar potential
vector potential
2
∇ Φ=
Classical wave equations
∇2A =
Electric field
Magnetic field
∂A
E = −∇Φ −
∂t
B = ∇×A
Φ (r , t )
A (r , t )
1 ∂ 2Φ
c 2 ∂t 2
2
1 ∂ A
c 2 ∂t 2
c-Speed of light
Electric field
λ＝wavelength (Dis tan ce between successive peaks)
Distance
Magnetic field
ν＝Frequency
(Number of cycles per
second passing a fixed point)
δ
Velocity of light
ν/H Z
108
1010
1012
1014
106
104
102
1
10-2
10-4
10-6
1016
1018
1020
Ultraviolet
106
Visible
104
Infrared
102
10-8
1022
1024
γ rays
10-10
10-12
λ/m
ν=
c
λ
10-14 10-16
PCIIweek1-2作業
Pages 639~640
Problems
14.9
14.10
14.11