FUNDAMENTALS OF
SURFACE MODES:
WORK OF ADHESION
¥ WORK OF ADHESION AND COHESION
Ð
Ð
Ð
Ð
Ð
Ð
Measuring surface tension
1
1
We will define work of adhesion and cohesion
wetting
model studies
precipitation
capillary rise
Liquid helium climbing up the walls of a beaker
σ1
σ1/2
l
2
σ=
Wa = σ 1 0 + σ 2 0 − σ 1 2
1
WORK OF COHESION
¥ 0 indicates vacuum. We may skip this
index!
2
F ⋅ ∆x 2 ⋅ γ ⋅ l ⋅ ∆x
=
=γ
∆A
2 ⋅ ∆x ⋅ l
¥ Water: 73 erg/cm2
¥ Mercury (Hg) 490 erg/cm2
¥ (UNDER CONDITION THAT THE FILM IS THICK ENOUGH)
3
WETTING
¥ Work of adhesion in vacuum:
W12 = σ 1 + σ 2 − σ 1 2
1
θ
θ
¥ Work of adhesion in a 3rd medium:
σ1
1
W123 = σ 1 3 + σ 2 3 − σ 1 2
σ1
= [σ 1 + σ 3 − W13 ] + [σ 2 + σ 3 − W23 ] − [σ 1 + σ 2 − W12 ]
1
θ
= σ 3 − W13 + σ 3 − W23 + W12 = W12 − W13 − W23 + Wc3
Wc = 2σ 1 0 = 2σ 1
σ S = σ S / L + σ L ⋅ cos(θ )
W113 = W11 + Wc3 − 2W13 = W11 + W33 − 2W13 = Wc1 + Wc3 − 2W13
4
¥ In vacuum we have:
5
¥ i.e. work of adhesion always positive.
L2
not wetting
partly wetting
fully wetting
64447448 64447448 64447444
L1
σS
σS/L- σL
σS/L
σS/L+σL
144424431444444424444444
¥ if σL2< σL1 this will never happen.
σ L1 L 2 > σ L1 + σ L 2 ; Will never happen.
¥ If σS/L< σL the left limit is 0.
7
6
¥ not until σL1/L2< σL2 - σL1 will droplet let go
of surface and enter interior;
Complete wetting.A layer of
σ L1 > σ L 2 + σ L1 L 2 ; 
L2 covers thesurface of L1.
σ L 2 > σ L1 + σ L1 L 2 ; {L2 will be engulfed by L1.
Lyophilic solid
¥ The liquid is said to wet the solid if the
contact angle, θ, is sharp, i.e. < 90°.
¥ Interesting to notice that if σL1/L2< σL2
energetically favorable if droplet entirely
submerged as compared to fully above
surface, but even more favorable if droplet
penetrates surface;
Liquid drop on a liquid surface:
σS/ L < σS +σ L ⇒ σS/ L −σ L < σS
L
S
¥ Work of cohesion in a 3rd medium:
Lyophobic solid
F = 2 γ. l
σ2
2
8
9
Modelling of adhesion, cohesion
and wetting.
Gravitational effects:
¥ bulk modes
¥ Let both materials be represented by a
dielectric function of the form:
2a ω
ε i (ω ) = 1 − 2 i i 2 ; i = 1, 2
ω − ωi
ε i (ω ) = 0 ; ω ibulk = ω 02i + 2 a
¥ surface modes
ε i (ω ) + 1
= 0 ; ω i = ω 02i + a
2
¥ interface modes
¥ density of carriers in the two substances the
same
2a
ε i (ω ) = 1 − 2
; i = 1, 2
ω − ω 02i
10
ω
2
(
(
(
0
0
1
2
3
ω 02
13
)
)
(
)
(
1
(
(
)
) (
12
)
)
¥ Assumed that there are N modes (q-values)
and all modes give the same contribution.
II
III
fully wetting
spreading
0.2
14
0.1
IV
0.2
s1
s2
s1/2
σ
0.1
0
0
1
2
3
ω 02
¥ four characteristic regions: I : σ 2 > σ 1 2 > σ 1
II : σ 2 > σ 1 > σ 1 2
IV : σ 1 > σ 1 2 > σ 2
15
Surfactants have this property.
not wetting
s2
s1/2
s1+s1/2
s1/2-s1
III
¥ By reducing the surface energy of the liquid
we may extend the region of spreading.
IV
partly wetting
σ
II
III : σ 1 > σ 2 > σ 1 2
0.3
I
I
)
Criterion for wetting
¥ Surface energy of the interface smallest
when the two media have the same
dielectric properties; in that case it is zero.
¥ Also always smaller than the largest of the
two surface energies for the free surfaces.
¥ The solid is metal like towards the left and
insulator like towards the right.
2
2
2
2
 ω 01

− ω 02
2

 +a
2


0.3


 σ = ω − ω − 1 ω bulk − ω = 2 − 1 − 1 3 − 1
1
1
01 2 1
01
2


2
2
1 ω bulk − ω
=
−
−
=
+
− ω 02 − 12  ω 02
+ 2 − ω 02 
σ
ω
ω
ω
1
 2
2
02 2
2
02
02


bulk
bulk
σ 1 / 2 = ω + + ω − − ω 01 − ω 02 − 12 ω1 − ω 01 − 12 ω 2 − ω 02

2
2

2
2
2 
2 
1 + ω 02
1 + ω 02
 1 − ω 02
 1 − ω 02

=
+1+ 
+1 +
+1− 
+1



2
2
 2 
 2 


2
− 12 3 + 1 − 12  ω 02
+ 2 + ω 02 


w01
w02
w1
w2
wp
wm
(ω 012 + ω 022 ) + a ±
11
Surface energies
¥ Let both a and ω01 be equal to unity.
3
ε1 (ω ) + ε 2 (ω )
= 0 ; ω± =
2
1: Liquid
2: Solid
They also reduce the tension at the interface
and extend the partly wetting region.
0
0
1
2
3
Surfactants have a tendency to accumulate
at the interface.
ω 02
16
¥ Rough indication that metals have a
tendency to be wet while wetting is less
likely for insulators like, teflon.
17
18
Variation the numerator
Quantities important for the
wetting properties:
Surface energies:
¥ We have now varied the frequency. Let us just
very briefly vary the numerator instead. We let a
in the numerator for material 2 now be b.
0.2
2.5
0.2
0.15
s2
2
σ
0.1
σ
s12
0.1
s2
s1/2
s1/2+s1
s1/2-s1
zero level
0.15
s1
0.05
1.5
ω
0.05
1
w1
w2
wp
wm
0.5
0
0
0.5
1
1.5
2
0
0
-0.05
0
0.5
1
1.5
2
2.5
3
0
0.5
1
1.5
b
2.5
3
b
2
2.5
3
b
19
20
Birds of a feather flock together.
(Lika barn leka bŠst)
21
¥ With a and ω01 both unity :
¥ The energy change is proportional to
They have both the same number of atoms.
¥ What happens with the energy when we
combine them into one material with the
same density as before?
E=
(ω 012 + ω 022 ) + a +
+
(ω 012 + ω 022 ) + a −
1
−
2
[
2
2
2
ω 01
2
2
2
 ω 01

− ω 02
2

 +a −
2


2
2
2
 ω 01

− ω 02
2

 +a −
2


+ 2 a − ω 01 +
2
ω 02
+ 2 a − ω 02
(ω 012 + ω 022 ) +
2
(ω 012 + ω 022 ) −
2
0.2
2
2
 ω 01

− ω 02


2


2
2
2
 ω 01

− ω 02


2


2
0.15
Energy
¥ We have two equal volumes of materials 1
and 2.
0.1
0.05
]
0
0
1
2
3
4
5
ω0 2/ω0 1
22
Precipitation and the need for
nucleation centers
23
Original system
¥ Energy should be lowered when the two
phases separate, i.e., when precipitation
occurs .
¥ System consisting of a host material with a
fraction of impurity atoms.
¥ Two types of atom have the same size; this
we do to avoid complications from strain
fields building up in the system.
24
¥ The eq
ε (ω ) = 1 −
a
b
−
=0
ω 2 − ω12 ω 2 − ω 22
has two solutions:
ω ± ( a, b,ω1,ω 2 ) =
¥ In this case we also create an interface
between the two materials and that, we
know, costs energy.
25
1
2
(
a + b + ω12 + ω 22 ± (a + b) + ω12 − ω 22
2
)
2
(
+ 2(a − b) ω12 − ω 22
)
¥ Energy per atom proportional to:
E1 = ω + (1 − x, x,ω1,ω 2 ) − ω + (0, 0,ω1,ω 2 )
+ω − (1 − x, x,ω1,ω 2 ) − ω − (0, 0,ω1,ω 2 )
26
27
¥ The energy change per impurity atom at
precipitation is:
¥ Energy per atom in the pure host system
proportional to:
E2 = ω + (1, 0,ω1,ω 2 ) − ω + (0, 0,ω1,ω 2 )
+ω − (1, 0,ω1,ω 2 ) − ω − (0, 0,ω1,ω 2 )
¥ and in the pure impurity system:
E3 = ω + (0,1,ω1,ω 2 ) − ω + (0, 0,ω1,ω 2 )
+ω − (0,1,ω1,ω 2 ) − ω − (0, 0,ω1,ω 2 )
¥ Energy per surface atom in the precipitated
system is proportional to:
E2,3 = ω + ( 12 , 12 , ω1, ω 2 ) − 12 [ω + (1, 0, ω1, ω 2 ) + ω + (0,1, ω1, ω 2 )]
+ω − ( 12 , 12 , ω1, ω 2 ) − 12 [ω − (1, 0, ω1, ω 2 ) + ω − (0,1, ω1, ω 2 )]
¥ All proportionality constants are the same.
π
η
≈ 0.91
 area =
2 3

ηvol ≈ 0.74
∆E E2 (1 − x )
surface atoms E1
=
+ E3 + E2,3
−
Ni
x
volume atoms x
=
¥ Typical result for the change in energy per
impurity atom when the impurities precipitate:
ηarea
E2 (1 − x )
E
+ E3 + E2,3
− 1
2
1
x
(ηvol ) 3 Ni 3 x
0
Energy change per atom
and
1
4πR3
4πr 3
R  N 3
→ = i 
ηvol = Ni
3
3
r  ηvol 
28
¥ where R is the radius of the precipitate, r the
radius of the atom and η the packing
fraction.
29
31
34
-0.008
-0.012
0
200
400
600
Number of atoms
800
1000
30
¥ These examples are the results from
minimization of the surface energies and
potential energies in the gravitational field
¥ This means that the results are quite
different in micro gravity.
¥ I will here not go into these effects.
¥ I have in the book made a more extensive
derivation of the capillary rise in thin tubes.
¥ Upper surface of a liquid in a beaker
¥ No gain until the number of atoms exceeds a
critical value, around 50 in our example.
Need for nucleation centers or seeds
¥ Energy gain per atom increases with the size of
the precipitate. This means that large
precipitates grow at the cost of smaller ones.
This is a typical behavior. It is called particle
coarsening, or sometimes Oswald ripening or
Ostwald ripening.
-0.004
-0.016
Capillary rise
Learn two things:
¥ In most cases the energies are just the areas
times the surface energies for flat surfaces.
¥ Only for very small objects or if for other
reasons interfaces are very close apart there
will be effects from the interference
between modes on different interfaces or on
parts of the same interface.
¥ An example where this interference occurs
is thin films. A spectacular result of such an
interference is, as I have demonstrated in
the book, that liquid Helium climbs the
walls of a beaker.
¥ For close packing we have:
¥ The rise of the liquid in a capillary tube
32
¥ Let us now determine the thickness, and its
variation with height, of a thin liquid film adhered
to a vertical solid surface.
¥ We are interested in corrections to the results
neglecting interference.
¥ In the case of a capillary tube we limit ourselves to
the region where the film is very thin, at large
heights.
Ð In this situation we may treat the film as having
constant thickness at each height and we may
disregard the change in surface energy with any
deformation of the film. We make use of the
theory developed in Sec 6.5 and the result in the
35
exercise E 4.15.
33
¥ Let us study a system where we have two
half spaces of two different media 2 and 3
separated by a thin film of medium 1.
¥ In our case medium 1 is the liquid, medium
2 the solid, and medium 3 air.
¥ Since we are only considering thin films we
can neglect retardation effects.
36
¥ We should also notice that if the middle medium
has dielectric properties in between the other two
the interaction is repulsive.
¥ The force per unit area, or pressure, is
¥ Then the interaction energy between the air and the solid is
∞∞

 ε ' (ω ) − ε1' (ω )   ε 3' (ω ) − ε1' (ω )  
∆E( d ) = 2 2 ∫ ∫ dωdxx ln 1 − e −2 x  2


4π d 0 0
 ε 2 ' (ω ) + ε1' (ω )   ε 3' (ω ) + ε1' (ω )  

h
∞∞
 ε ' (ω ) − ε1' (ω )   ε 3' (ω ) − ε1' (ω )  
1
= − 2 2 ∫ ∫ dωdxx ∑ e −2 xn  2


4π d 0 0
 ε 2 ' (ω ) + ε1' (ω )   ε 3' (ω ) + ε1' (ω )  
n n
h
∞
∞
1  ε ' (ω ) − ε1' (ω )   ε 3' (ω ) − ε1' (ω )  
= − 2 2 ∫ dω ∑ 3  2

4π d 0 n =1 4n  ε 2 ' (ω ) + ε1' (ω )   ε 3' (ω ) + ε1' (ω )  
h
n
n
P( d ) = −
∞
 ε ' (ω ) − ε1' (ω )   ε 3' (ω ) − ε1' (ω )  
h
≈−

∫ dω  2
16π 2 d 2 0  ε 2 ' (ω ) + ε1' (ω )   ε 3' (ω ) + ε1' (ω )  
¥ where in the last line we have assumed that the dielectric
properties of the three media are not very different from
each other; if this is the case it is enough to keep just the
first term of the expansion.
¥ The Hamaker constant for the system
considered is
A=
≈
∂
∆E( d )
∂d
=−
∞
h ∞
1  ε 2 ' (ω ) − ε1' (ω )   ε 3' (ω ) − ε1' (ω )  
dω ∑


2π 2 d 3 ∫0 n=1 4n3  ε 2 ' (ω ) + ε1' (ω )   ε 3' (ω ) + ε1' (ω )  
≈−
 ε ' (ω ) − ε1' (ω )   ε 3' (ω ) − ε1' (ω )  
dω  2

8π 2 d 3 ∫0  ε 2 ' (ω ) + ε1' (ω )   ε 3' (ω ) + ε1' (ω )  
h
∞
P( d ) = −
A
+ ρdgh
12πd 2
¥ The minimum defines the equilibrium
thickness:
∂
A
 −A 
+ ρgh ⇒ d0 = 
W (d ) =

 6πρgh 
∂d
6πd 3
13
if A negative
3π n α i (0)α j (0)hω 0,i hω 0, j
(
2 hω 0,i + hω 0, j
)
40
A
12πd 2
;
A
6πd 3
39
¥ If the Hamaker constant is positive there is an attractive
force between the free surface of the film and the solidliquid interface;
Ð this tends to thin the film and in this case one needs to
include other energy terms as well to get a non-zero
thickness.
¥ The dielectric constant of liquid He is very close to one.
¥ This means that its dielectric constant lies between most
solids and air with the effect that the Hamaker constant for
a film on a solid is negative.
¥ This has the effect that the film thickens and explains why
the liquid climbs the walls of a beaker.
2 2
Aij =
∆E( d ) = −
38
W (d ) = −
¥ With the London approximation these
constants are
3h ∞  ε 2 ' (ω ) − ε1' (ω )   ε 3' (ω ) − ε1' (ω )  
dω 

4π ∫0  ε 2 ' (ω ) + ε1' (ω )   ε 3' (ω ) + ε1' (ω )  
¥ and
¥ Let us now study the free energy of a unit
area of the film. It is
A = A11 + A 23 - A12 - A13
n
n
37
¥ If the three media are diluted or if we use
summation over pair interactions to
calculate the Hamaker constant for the
system this constant may be written as
∞
3h ∞
1  ε 2 ' (ω ) − ε1' (ω )   ε 3' (ω ) − ε1' (ω )  
dω ∑


π ∫0 n=1 4n3  ε 2 ' (ω ) + ε1' (ω )   ε 3' (ω ) + ε1' (ω )  
41
42