Lecture: P1_Wk3_L1

Tip-Sample Interactions: the Water Meniscus

Ron Reifenberger

Birck Nanotechnology Center

Purdue University

2012

1

2

AFM

Instrumentation

Topographic Imaging Theory

Piezos Cantilever Controller

Calibration

Contact

Elasticity

Controller Tip-Sample Interaction

Force Sensing and Control

Static Cantilever

Local Material Property Map

Adhesion

Force Spectroscopy

Deflection vs. z

Air

Week 3 Overview

P1_Wk3_L1

Liquid water (~3x1022 molecules/cm3)

Vapor (~2x1019 molecules/cm3)

The asymmetry in number density at a liquid-vapor surface gives rise to a pressure differential and a surface curvature

Liquid

Vapor Surface (~1015 molecules/cm2)

Solid substrate (~6x1022 atoms/cm3)

P1_Wk3_L1 3

Local radius of curvature R

4

The “Curvature” of a Liquid’s Surface has far reaching consequences

See Appendix for a general discussion on how to quantify the curvature of a surface using the radius of curvature. Two important consequences:

1. If the surface of a liquid is curved, there must be a pressure difference: Young-Laplace Equation

a) Fill tube with liquid:

Liquid Liquid Liquid

b) Liquid/vapor interface in equilibrium Curvature implies Pin < Pout Curvature implies Pin > Pout

( )

( )

2 1

1 2

1 1

,12

in out

in out

P PR R

for a spherical surface R R R and

P P PR

γ

γ

− = +

= =

∆ ≡ − =

see Appendix 2 for derivation

P1_Wk3_L1

Implications of the Young‐Laplace Equation

• If the shape of a liquid surface is known, its curvature is known and the pressure difference can be calculated. • In equilibrium, the pressure is the same everywhere in a liquid (otherwise there would be a flow of liquid to regions of lower pressure). Thus, Pin is constant and the Young‐Laplace equation tells us that the surface of the liquid must have the same curvature everywhere. Example: Calculate the pressure difference that develops for a spherical drop of water at STP with a radius of 1 μm? • Using the Young‐Laplace equation, the equilibrium shape of a liquid surface can be calculated if we know the pressure difference and boundary conditions (volume of liquid and its contact line).

( ) ( )3 26

5 525

1 12 2 75 10 /1 10

1 .1.5 10 / 1.5 101 10

1.5

in outP P P J mR m

atmN m PaPa

P atmospheres pressure drop across interface

γ −−

∆ = − = = × × ×

= × = × ××

∆

5 P1_Wk3_L1

2. It may not be so obvious, . . . but if the surface of a liquid is curved, the equilibrium vapor pressure of the liquid must also change: Kelvin Equation

The equilibrium vapor pressure measures a liquid's evaporation rate: smaller drops evaporate faster

Liquid film

Flat interface: drop bubbles condensation

into pores

Curved interfaces:

gasf

cvap

mol 1 2o

R T 1 1nV R RP

P = γ +

2

2

2

3

1 2

0.072

18 ,

8.314

H O

mol H O

gas

J T temperature in KmcmV R R radii of curvature of liquid surfacemoleJR K mole

γ = =

= =

=⋅

6 P1_Wk3_L1

foP

see Appendix 3 for derivation

Rtip

R1 R2

Sign convention for radius: R1 is negative, R2 is positive

Application to AFM

Rtip

In humid air, both tip and substrate will be coated with thin layer of adsorbed water. When the tip contacts the surface, the adsorbed

water will form a meniscus neck around the AFM tip.

7 P1_Wk3_L1

h

System in equilibrium

Temperature T

Pressure P water

meniscus neck

How thick is h? (measured thickness of water layer on a silicon oxide substrate)

Asay and Kim, J. Phys. Chem B 109, 16760 (2005)

How long does it take for this layer of water to form? The answer depends on relative humidity, but for a typical ambient humidity of say 30%, the time is very fast, much less than one second – see Appendix 3.

8 P1_Wk3_L1

~0.8 nm at RH=30%

• Vapor pressure in handbooks are measured for liquids having a flat surface • Vapor pressure depends on the curvature of the liquid’s surface • Use Kelvin equation - derived from the Young-Laplace Equation:

( )

1 2

1 2

2

1 1

,

10.07210

gas mol

mo

cvap

fo

c

cva

l

gas

pf

o

water

vapf

o

nR R

P vapor pressure of liquid having curved surface with radii R RP vapor pressure of liquid having flat sur

PP

PRelative Hum

R T

id

V

VR

ity

fac

m

P

e

TJ

m

γ

γ

= +

=

=

≡

= ⋅

33

2 7

9

1 20.5 1 12

1 2

11810 0.52

8.314 300

; ,Rco

Rfvap

nmcmmole cm nmJnm KK mole

P P e R R in nm

−

+

⋅ × = ⋅ ⋅

=

9 P1_Wk3_L1

for water:

At contact - what determines R1, R2?

0.0

2.0

4.0

6.0

8.0

10.0

0.0 0.2 0.4 0.6 0.8 1.0Equi

libriu

m |R

1| (in

nm

)

Relative Humidity

10

1 2

1 1AssumingR R

>>

Results for water meniscus

Will the liquid evaporate from the meniscus neck? No, the liquid will not evaporate due to the curvature of the

air/vapor interface.

P1_Wk3_L1

Rtip

R1R2

System in equilibrium

~0.5 nm at RH=30%

Conclusion: A capillary meniscus force binds a sphere to a flat plane

11 P1_Wk3_L1

Rtip

lift offF −

Rcap

Water meniscus forms: R1 is small (and negative) R2 is large (and positive) Curvature implies Pin < Pout

2

2

1 2

1

1 1

, ,

cgas vap

Laplace fmol o

cgas vap

capillary Laplace fmo

cap

c

l o

tipap

R T PP Laplace pressure n

R V P

R T PF P n

V P

will depend

A R

on h R R and RR

R

π

γ = ≡ + =

= =

Relative humidity

( ) 43

vdW tip

JKR tip

F Derjaguin R independentof RHF R

πγ

πγ

=

=

( )

( ) ( ) ( ) ( )

1

11

2

2

0.4

52

2

4

cvap

fo

cva

gasLaplace cap

mol

ga

capi

s

mo

pf

o

c c

ll

va

a

l

gas

mol

ry

tip

tip tipp vapf f

o o

R TP A n R

V

R Th

PP

PP

P P

R

h R

nV

R Tn

P Ph R

F

R

nR RV nm

π

πγγ

γ

π

π

=

= = =

+

+

+

Compare to:

Relative Humidity

DATA: Asay and Kim, J. Phys. Chem B 109, 16760 (2005) He et al., J. Chem. Phys. 114, 1355 (2001).

Estimate magnitude of capillary force (see Appendix 4 for geometric details)

12 P1_Wk3_L1

(1.3/0.5) x 1.2 ∼ 3

• Capillary force models range from simple to complex • Strong dependence on humidity

capillary neck breaks

Effect of capillary condensation – important modifications to the DMT model

13 P1_Wk3_L1

capillaryF

14

Next Lecture: An overview of VEDA – the online AFM simulation software

P1_Wk3_L1

15

Appendix 1: Defining the “Curvature” of a Surface in 3D?

Surprisingly, the “curvature” of any surface can be specified by two numbers, the two

radii of curvature at a point P.

P1_Wk3_L1

• Define a point P on surface qrst by the vector S.

• Define the orientation of the surface by the normal vector n.

• Define two tangent unit vectors u1 and u2 that are normal to each other and to n.

• This local coordinate system (u1,u2,n) can be used to specify any point ρ on the surface.

• How much a surface “bends and twists” along a line L can be defined by the largest circle tangent to the line (positive radius when located on the concave side of the surface).

• The reciprocal of the tangent circle radius R=PP’ is called the curvature (κ=1/R units: radians/m) of the surface along the line at the point P. For gently curving surfaces, the curvature is basically the second derivative of the line L at the point P. • The curvature of the surface will be a continuous function as the plane defining L is rotated through 180o .

• As the plane is rotated, there will be a maximum and a minimum radius of curvature. These two extrema are called the principal curvatures at the point P.

x y

z u1

u2

n

S

ρ

( )1 2

( , , )ˆˆ ˆ, ,

S f x y zr S u u n

=

=

q

r

s

t Surface qrst

P’

L P

16 P1_Wk3_L1

plane defining L

A B

C D

F E

H G

R1

R2

q

r

s

t A

B

C

D

E F

G

H

Defining the curvature of qrst requires finding the radii of two circles tangent to the surface.

L M

N O

L

M N

O

P Curvature: κ1=1/R1

Curvature: κ2=1/R2 Mean radius of curvature: Gaussian curvature: Κ = κ1 κ2

1 2

1 1 12

1mean R RR

+

=

17

Circle 1

Circle 2

P

P

18

1. An important theorem in differential geometry proves that so long as plane ABCD is perpendicular to plane FEHG (see previous slide), the mean radius of curvature is invariant, independent of the alignment of the two planes wrt the surface qrst. 2. Once the curvature is known at a point, it can be used to derive a differential equation that describes the equation of the surface at that point. 3. A few surfaces with their of radii of curvature:

Further Considerations

R1

R2=∞

+ = ⋅

=

1 2 1

1 1 1 1 12 2

1m R R RR

R1 Sphere, radius R

R2=R1

1 2

1 1 1 1 2 12 2

1 + = ⋅ =

=

m R R R RRP1_Wk3_L1

Cylinder

19

Appendix 2: Derivation of Young-Laplace and Kelvin Equations

P1_Wk3_L1

1. If the surface of a liquid is curved, there must be a pressure difference: Young-Laplace Equation

2. If the surface of a liquid is curved, the equilibrium vapor pressure of the liquid must also change: Kelvin Equation

20

( )( ) ( )

( )( ) ( )( )( )

1 1 2 2 1 2

1 11

2

2 12 1

11

1 1 1

2 2 22

2 2 2

1 2

1 2

1 2 2 1

1 2 2 1

in out in out

u du u du u u

u du u dnuR dn R R similar triangles

u du u dnuR dn R R

Work P P A dn P P u u d

Au du u du

A u du u du

du

du

dn dn

n

u uu uR R

γ γ

γ

= + + −

+= ⇒ = +

+ = ⇒ = +

= − = −

≡

= +

∆

+

∆ +

( )

( )

2 1

1 2

1 1

,12

in out

in out

P PR R

for a spherical surface R R R and

P P PR

γ

γ

− = +

= =

∆ = − =

u1

R2

u2

dn

R1

Young – Laplace Equation: Pressure difference across a curved interface

By convention, the radii of curvatures R1. R2 are assigned positive values if they fall inside the concave surface under consideration .

20

(Laplace Pressure)

P1_Wk3_L1

a liquid surface

see P1_Wk2_L2

21

The vapor pressure above a curved surface: the Kelvin equation

Let pressure of vapor above flat surface= Po

h

x

hvapPz

r2

liqρ

vapρ

r1

hliqP

The situation: Insert open tube into liquid. Assume the liquid does not wet the tube. Liquid will fill tube to a depth h as shown. In general, the vapor-liquid interface will be curved, and will be characterized by the radii of curvature r1 and r2. In equilibrium at temperature T, the pressure difference between the vapor at depth h and the liquid at depth h must be described by the Young – Laplace Equation. Let the vapor/liquid density be designated by respectively.

hvap(P ) h

liq(P )

lvap iq, ρρ

P1_Wk3_L1

Note: The end result is completely general and is identical to that obtained using more general arguments that are not so physically transparent.

liquid

22

Pressure of vapor above flat surface= Po

h

x

hvapPz

r2

( )

( )

o

hvap

vap vap gas n 1

vapvap o

gas

vap vap

gas

hvap vap

P 0vap gas

vapm

mvap

P

vap vap

vap

h

o

molar massmolar volume

for ideal gas : P P nR T

MP; now, P z

Mvapor density

P gzR T

dP z MPg g

dz R T

dP PM Mgdz

V

MV

nP R T P R

=

= =

=

=

=

ρ

ρ

ρ ρ

ρ

= = +

= =

= ⇒ =

∫ ∫

gas

o gas 1 2

g

hv

mol

as 1 2

gas molo 1 2

ap

hvap

gT

1 1MnP R T r rP

P

1 1R T r r

1 1

h

R T n VP r r

V

γ +

= γ +

= γ +

ρ

( )

hliq o

o

o

mgP PA

h A gP

AP gh

= +∆ρ ∆

= +∆

= + ρ

In equilibrium:

( )

( )( )

h hliq vap

1 2

ho vap

1 2

hvap o

1 2h

vap o

1 2

1 1P P (Young Laplace)r r

1 1P gh Pr r

1 1gh P Pr r

P P1 1g r r

hg

− = γ + −

+ ρ − = γ +

ρ = γ + + −

− γ

= + + ρ ρ

small - neglect

hvapP ?What is

see K.P. Galvin, Chem. Engin. Sci. 60, 4659 (2005) P1_Wk3_L1

A∆

ρ

vapρ

r1

Po

How to describe velocity of N gas molecules of mass m at

a temperature T: Maxwell Speed Probability

Distribution function (1852) ri

vi

23

Appendix 3: Kinetic Theory of Gases

2 2 2≡ + + =x y zv v v v speed of gas molecule

m T

( )2

32

224( )2π

− =

B

mvk T

B

mf v v ek T

P1_Wk3_L1

Note asymmetry – there are more ways to get a large speed than a small one.

Maxwell Speed Distribution Function

Shaded area represents probability that molecule will have a speed v ± dv/2

24

v

N2 molecule

P1_Wk3_L1

dv

25

Once you know f(v), you can answer many questions: 2

1

1 2 1 2

0

( ) ( )

( ) ( ) ( )∞

≤ ≤ ≡ ≤ ≤

< >≡

∫

∫

v

v

P v v v f v dv probability of finding molecule with velocity v v v

x v x v f v d v

Why Are Distribution Functions Useful?

where <x> is the expected (average) value for some quantity x that you define. For example, the average velocity will be

2

2

32

23

0 0

2 11

0

4( )2

! ; 1;22

8

π

π

∞ ∞−

∞+ −

+

= ≡ =

= = =

= =

∫ ∫

∫

Bmv

k Tavg

B

n axn

Bavg

mv v v f v dv v e dvk T

n mx e dx n akTa

k Tv v

mP1_Wk3_L1

x

y

z v θ

22 22

0 0

1

2

1 1

2 2

cos cos

1cos cos2

= = =

= = =∫ ∫πππ

θ θ

θ φ θ θ θ θπ

z avg

o

v v v v

d sin d sin

n = molecules/volume What is the mean value of velocity in the z-direction?

What is the molecular flux (number of molecules/sec) that will hit a unit area of the plate?

1 1/2 4z avgJ Collision rate unit area n v nv= = =

For n molecules in the volume, ½ will be moving up, the other half will be moving down

Assuming an ideal gas, how does n depend on T and P?

B

B

PV Nk TN PnV k T

=

≡ =26

P1_Wk3_L1

Standard Results

Assuming an ideal gas, what is the incident molecular flux in the –z direction?

1 281 1 :4 4 2

Bavg

B B

k TP PJ nv units s mk T m mk Tπ π

− − = = ⋅ ⋅ =

Assuming an ideal gas, how many molecules will strike a unit area of the surface in a time t?

2

0

:2

t

B

PJdt t units mmk Tπ

− Φ = = ∫

Assuming an ideal gas, what is the number of molecules that fill a 1cm3 volume at STP (0oC (273.15 K), 1 atm (100 kPa))?

23 319 3

3 6 3

1 6 10 1 1 2.7 1022.4 0.001 10

N mole molecules liter mn cmV liters mole m cm

−×≡ = ⋅ ⋅ ⋅ = ×

27 P1_Wk3_L1

28

x

y

z v θ

J= molecular flux (number/unit area) in unit time n = molecules/volume

surface

How many gas atoms strike a unit area of a surface in a given time t?

2

0

: /2

Φ = = ∫ π

t

B

PJdt t units number mmk T

The answer is found by integrating the molecular flux over the time t

If the gas has some fraction of water molecules, then it is not surprising that a thin layer of water will form on a surface.

mass m

Temperature T

Pressure P

P1_Wk3_L1

Estimate the length of time for a monolayer of gaseous atoms to accumulate on a surface?

The answer depends on the flux of gas atoms on the surface, the sticking probability of each gas atom, and the number of active adsorption sites on the surface. If the sticking probability is assumed to be 1, then a minimum estimate for the monolayer deposition time can be made. Furthermore, it is customary to assume there are ~1019 adsorption sites per square meter on a surface, roughly equal to the number of atoms per square meter that characterize most surfaces.

π

π−

× ≈ =

×≈ ⇒ = = × =

2

19

1926

2

2; 18 /

10

103 10 ; 300

monolayerB

Bmonolayer H O

PJ t where Jmk T

mk Tt assume gm mole m m kg T K

P

Vacuum P (Torr) P (Pa) n (m-3) tmonolayer (s)

atmospheric 760 1x105 2.4x1025 2.8x10-9

low ~1 130 3.1x1022 2.2x10-6

medium ~1x10-3 0.13 3.1x1019 2.2x10-3

high ~1x10-6 1.3x10-4 3.1x1016 2.2

ultra-high ~1x10-10 1.3x10-8 3.1x1012 2.2x104

29 P1_Wk3_L1

30

Appendix 4: Geometric details required to calculate Rcap Here, treat R1, R2 as positive numbers

( )( ) ( )( ) ( )( )( )

( )( )

( )( )

+ + = − + + +

+ + − + =

− ± − − +=

= + + −

+ = +

2 2 2 21 1 2 1

2 22 1 2 1 1

2 21 1 1 1

2

1 1

1 2 1

2 4 0

2 4 4 4

22

2

tip tip

tip

tip

ti

tipcap

p

R

R h R R h R R R

R R R R R h R

R R R R h RR

R h R R

R R R h R

h

R2

2R1

Rtip

Rtip –h-R1

Rcap=R2+R1

an ideal tip!

h~R1

P1_Wk3_L1

B. Bhushan, Springer handbook of nanotechnology, 2nd Edition, Vol. 2; p. 960 (2007).

Hydrophobic surfaces Hydrophilic surfaces

Appendix 5: Further Considerations

31 P1_Wk3_L1

A water molecule has a diameter do of approximately 0.3 nm.[1,2] The areal density of water is x/(do)2, where x is the adsorbed water layer thickness. A water “monolayer (i.e., x=1)” consists of about 11 molecules/nm2.

32 P1_Wk3_L1

Water Facts

[1]. Karis, T. E. “Water Adsorption on Thin Film Media,” Jour. Colloid Interface Sci., 225, 196-203, (2000). [2]. Adamson, A. W., Hu, P. and Tadros, M. E. “Adsorption and Contact Angle Studies,” Jour. Colloid Interface Sci., 49, 184-195 (1974) .