Scaling in geometry. Scaling in rigid body dynamics. The Trimmer force scaling vector – scaling in electrostatic forces, electromagnetic forces, scaling in electricity and fluidic dynamics, scaling in heat conducting and heat convection. Force scaling vector: F=[l F ]=[l
Surface-to-volume ratio scales as A/V
→
L
2
/ L
3
= L
-1
; Mass (weight)
1 l 2 l 3 l 4 ]
Moment of inertia I=mR
2
~L
5 m=rV~L
3
Torque t=(ma)R~L
4
Diffusion D =kT/6πηr, η = viscosity r =is the hydrodynamic radius
Frequency: f = 1/2

(k/m) ½ Natural frequency
 n =
 n (EI/mL
4 ) ½ Equivalent Rigidity: Keq =
EI/L
3
where E : Elastic Modulus; I : The area moment of inertia; m : Mass per unit length; L :
Length of the beam; a n
: an appropriate coefficient for boundary conditions
Piezolectric effect: Hook’s Law D = 
E where D is the electric displacement,

is permittivity and
E is electric field strength. Constitutive equation for dielectrics S = sT where S is strain, s is compliance, T is stress. Coupled Equations S =s
E
T +d t
E; D = d
T
+

T
E
Electric potential in parallel plates :E = ½ CV
2
=

0
W idth
L ength
V
2
/d istance
Scaling of classical mechanical systems and electrical systems:’ displacement x
 charge q force F velocity dx/dt mass m spring K damping b




 voltage V current I(=dq/dt) inductance L capacitance 1/C resistance R
P
ROBLEM EXAMPLE
:
FREQUENCY SCALING derive the equation on the dependence of frequency and equivalent rigidity on cantilever density.
P ROBLEM EXAMPLE : S CALING IN ELECTROSTATICS
Consider charging up the capacitor plates through a large resistor R Now move the plates a small amount
 d. For a bias of a few volts and a spacing of 100 nm, and knowing that it is easy to measure microvolts, calculate the detectable displacements of the plates.
Intermolecular forces (Debye, Keesom and London interactions; Lennard-Jones potential, van der Waals forces).
Dipole-dipole interactions U d
= -2

The induction effect U i
= – [
 2 
1
2
1
2 
+

1
2
2

/ 3k T r
2
2
] / r
6
6
Where

is the dipole moment
where

is the polarisability
London dispersion forces: between two atoms where and are the dipole polarizabilities of the respective atoms. I
A
and I
B
and are the first ionization potentials of the atoms and R is the intermolecular distance.
Repulsion, short distances (F

r
12
, E

r
11
)

Casimir force F
C
= -
 2 ħc/240d 4
Cohesive energy = energy of free atoms – crystal energy. For nanoparticles E = E b
(1-d/D)
D=nanoparticle size, d=atomic size; E b
=cohesive energy of bulk
E
XAMPLES OF PROBLEMS
Sketch an effective interaction potential for van der Waals molecules, derive an expression for the maximum energy of attraction in terms of the interaction parameters of the potential. Answer:
E max
= C vdW
2 /2C rep
- Calculate the number of atoms in a 1.4 nm diameter Pt nanoparticle using the total number of unit cells present. Consider a FCC unit cell with a lattice constant of a = 0 : 391 nm.
- Calculate the number of atoms in a 1.4 nm diameter Pt nanoparticle using the bulk density of Pt.
Consider

= 21 .
5 g/cm 3.
- Estimate the number of surface atoms and percentage of surface atom in a 1.4 nm diameter Pt nanoparticle. One can calculate this through a unit cell approach but use whatever approach you like.
- Calculate the planar density of low-index planes (100), (110), and (111) for an fcc crystal
Answer: We assume that the atoms are hard spheres with the radius R. The volume of the unit cell is determined by the linear dimension a. Let’s start with [100] plane which is a plane parallel to a face of the unit cell and it looks like a square. We can easily calculate that a = sqrt (8R2) = 2R

2
There is one atom at the center of the square and a total of 4*1/4 atoms on the corners of the plane. Hence there are a net total of 2 atoms inside the square. Now the area of the unit cell is a
The area of 2 atoms is 2*

R 2 . Planar density = 2*

R 2 / a 2 = 2*

R 2 / 8R 2 = 0.785
2
- A drop of water is forced through an aspirator at 20 C. Calculate the amount of work required to increase the surface area of water from that of the 1 cm3 drop to aerosol particles 1 µm3 volune each. (Ignore any potential Joule-Thompson cooling effect). The surface tension of water at 20 C is 72.8 dyn/cm. Tips: calculate the new surface area, find the additional area, and apply the equation “Work = surface tension * additional area”
Scaling in micromechanisms. Dynamics of a cantilever. Deflection d
Fl
3

3 EI

4 Fl
Ebt
3
3
Compressed materials may withstand a maximal stress of σ max.
If they are submitted to their own weight, mg, σ max
~ mg/A (A – cross section ~ d 2 )
Maximum Stress σ max
= 12 FLb/2b
2
= 6F/l
2
Oscillating Frequency

=

2 sqrt (EI/

AL
4
)
Hall-Petch relationship for stress and hardness vs the grain size.

=

0
+ K/D
½
; H= H
0
+ K/D
½
.
Inverse Hall - Petch behavior.
P ROBLEM EXAMPLE A paper in Applied Physics Letters (Khaled M. Appl. Phys. Lett., Vol. 85,
929) reports “a unique way of using mechanical milling/in situ consolidation at both liquidnitrogen and room temperature to produce artifact-free nanocrystalline Cu (…. nm) with a narrow grain size distribution. This nanocrystalline Cu exhibits an extraordinarily high yield strength σ
=
770 MPa, as predicted from a Hall–Petch extrapolation. σ o
= 25 Mpa; k= 0.11 MPa m 1/2 . What was the size of nanocrystallites?
P ROBLEM EXAMPLE : S CALING IN C ANTILEVER B EAMS

Calculate a vibrational frequency characteristic of a cubic nanometer block made of diamond.
Diamond density = 3.5 ×10 3 kg /m 3; Young's modulus E = 10 12 N/m2; stretching stiffness is
1000 N/m.
Calculate the relative changes in maximum stress and oscillating frequency if the size of cantilever is reduced from 1 cm to 100 nm.

Adhesion; Friction; Lubrication; Wear
Scaling in Adhesive Forces. Attraction force between a sphere and surface F vdW
= HR/12

where
H- Hamaker constant, R- radius,

- distance
Fv dWr
= (

/

+

/2)
2
F vdW
where
 is surface roughness
Two atoms interact with E i
=

/r
6
, where

- London - van der Waals constant
E
XAMPLE OF PROBLEMS
Consider two planar planes. (a) Derive the total energy per unit area of the VdW interaction at distance D , considering that contact is where D = Do . (b) Based on (a) provide an expression for the adhesive energy, i.e., the energy needed to separate the two planes from each other.
Solution Assuming two planar surfaces in contact, the Van der Waals interaction energy per unit area is W
1
= HR/12D
2
. We have also to consider unsaturated "bonds" at the two surfaces, which in contact (i.e., D = Do ) are W
1
= HR/12D
0
2 at a distance D
≥
Do apart is W = W surfaces, i.e., D
= ∞, 
W =H/12

D
1
0
2
. Thus, the total energy of two planar surfaces
+W
2
= H (1-D
0
2
/D
2
) /12

D
0
2
. In the case of isolated
Viscosity of fluids. Flow patterns Reynolds number. Basic equation in continuum fluid dynamics.
Surface tension, capillary effect and micropumping. Fluid flow at the nanoscale.
Reynolds number = (Inertial Force)/(Viscous Force) = ρAv/ η = Re
Inertial Force = dp/dt ~ ρA²v² Viscous Force = F= ηAv, where η is the liquid viscosity.
Scaling in Surface Tension = (δG/ δA) p,T
where G is the free energy and A is the area
F surf tens
=
 d, where d is the object diameter,

is the surface tension coefficient
Drag force D g
= C d
* ½ (ρ A V 2
) where C d
= drag coefficient:describes the resistance to motion;
ρ = density of fluid; A = area of the body; V = velocity.
Non-Newtonian flow: Viscosity is a function of velocity gradient
Scaling in fluids mechanics : Navier-Stockes law
 p = -8
 xu/r
2
~ l
4
Hagen-Poisseuille law Q =
 r4
 p/8
 x where
 p is pressure which scales as l
-2
Surface-driving forces substituting pumping: surface-traction, surface-tension flow (capillary effects), piezoelectric, electro-osmotic, electro-hydrodynamic.
P
ROBLEM EXAMPLE
: Calculate the scaling of a relative force needed to empty liquids from a capillary tube compared to spilling coffee from a cup. Use the scaling relations for the mass of a liquid in a capillary tube, and the surface tension.
Heat conduction in multilayered thin films . Heat conduction in solids in submicrometer scale.
Thermal conductivity of thin film, heat conduction equation for thin films.
Scaling in Thermodynamics: Energy required to heat a system to temperature T is proportional to mass E th
~ L
3
. For conduction and radiation power dissipation is proportional to area P d
~ L
2
.
The time needed to homogenize the temperature in a system of given shape is t
(used to determine the time increments in a transient heat conduction analysis) h

~ L
=

C
2 p
.
Scaling in Heat Conduction Fourier law in heat conduction Q = -kA

T/
 x ~l; Fourier number
V/

A~l
Scaling in Heat Convection: Total heat flow primarily depends on the cross-sectional area A.
Newton’s cooling law
Q = hA

T~l
2
where h - Heat transfer coefficient; Q - Heat flow.
Microscopic Picture of Thermal Transport. Phonons in the 1-D Lattice with Diatomic Basis
  c
1


M
1
M
1

M
M
2





 c
1
2


M
≈ (
M
1
 nL
M
M
2
3
2 )


2
 , or
M
4
L
1
1
2
≈
M
2
( sin (
)
2 ka )



1 / 2
1
–2/3 n
–1/3

Definitions of temperature: E k
= < mvi
2
Phonon mean free path l ph
∝ T
–1
>/2 = 3/2 kT; <mv i
2
>/2 = Σ{ω(q)/[exp(ω(q)/kT) – 1]}
. Statistical collision parameter η = exp(–l ph
/ L ).
Violation of Fourier Law: size-dependent thermal conductivity. Phonon Transport in 1-D
Nanostructures. Implications for the Boltzmann Transport Equation.
Thermal Conductance Quantization. G
0
=
 2 k
B
2
T / 3ħ - quantum of thermal conductance
  
2 T o

/
 where ∆ 
=
Deviation of melting point from the bulk value; To= Bulk melting point;

= Surface tension coefficient for a liquid-solid interface;

= Particle density; r= Particle radius; L= Latent heat of fusion
The Liquid Drop Model for size-dependent phase transitions.
. The total cohesive energy E c
of a nanoparticle with N a atoms
E c
(N) =  c,v
N a
+ 4  r a
2
N a
2/3

The cohesive energy per cluster
Spherical case:
E c,a
(N) = + 4  r a
2
N a
N a
= (d/2r a
)
3
and  c,v,d
=  c,v
+ 6V
1/3
 =  c,v
+  c,s a
 / d
N a
1/3
The exponents

,

are called “critical exponents”.
Implications of the Liquid Drop model: Magic number clusters
Vacancies and disclocations in nanoparticles: n v
= N a
⋅ e xp (
−
Q f
/ kT)
Surface reconstruction.
Introduction to Thermoelectrics: Peltier Effect. Figure of merit for thermolelectric device Z =
S 2 
/k
P
ROBLEM EXAMPLES
:

In the case of relatively large systems, where (1) N is such that the thermodynamical arguments remain valid; (2) the surface of the particle may be characterized by a single value of the surface tension, derive the equation for the melting temperature, variation with the size.

Calculate the relative change in cohesive energy as the spherical golden nanoparticle decreases in diameter from 100 to 10 nm.

In material 1 the mean free path for phonons is 10 nm and for electrons 100 nm. Im material 2 the mean free path for phonons is 100 nm and for electrons 10 nm. Which material has higher figure of merit?
Diffusion D =kT/6πηr, x = (2D

)
½
. A molecule diffuses over 10 μm, 1 million times faster than over 1 cm.
Self-assembly principles . L c is the effective chain length the tails can achieve and is frequently called the critical chain length. For saturated hydrocarbon chains, L c
(nm) £ 0.154 + 0.1265 n
V is the chain volume which for hydrocarbons is, 10
3
V(nm
3
) = 27.4 + 26.9 n. For large n, V/L c
is
0.21 nm
2
or the effective chain radius is 0.26 nm. A
0
is optimum area. Spherical micelle
M – number of lipids = M = 4pR
2
/ A
0
= 4pR
3
/ 3V . Cylindrical micelle M = 2

RL / A
0
=

R2L /
V So, R = 2V/A
0
, or, 1/2 = V/A0R . Therefore for a micelle to form, V/A
0
L c
< 1/2
Generalized Euler relation , V
−
E + F = K, where F , E and V are the numbers of faces, edges and summits (vertices), respectively. K is the Euler characteristics K
= 2(1 − g ) .
Euler's polyhedron formula or theorem for convex polyhedron V
−
E + F = 2.
P
ROBLEM EXAMPLE
:
Explain mathematically, using Euler´s theorem, why chemical reactivity depends not only on size but also on the shape of a nanoparticle.

1. Confinement
Confinement arises when the actual physical length scale of the system is smaller than the exciton
Bohr radius or corresponding deBroglie wavelength the electron and the hole experience confinement. In turn, the energies of the carrier along that dimension of the material are no longer continuous as in the case where there is no confinement
Finite box:
Quantum well: E= E n
+ p x
2
/ 2m + p z
2
/ 2m
Quantum wire: E= E mn
+ p x
2
/ 2m
Quantum well: E= E mnl
2. Infinite box: depth equals to the workfunction
Lower level position in the box is compared to box (film) thickness a>
 ħ /(2  m*), where

is the workfunction, m* is the effective mass.
Example
The energy spacing between the energy levels for the quantum wells with thickness ~10 nm is a few 10’s to a few 100’s meV • At room temperature kT ~ 26 meV. This means only the first energy levels can be occupied by electrons under typical device operational conditions
3. Conditions at which the quantum effects are observed
1. E n+1
-E n
>>kT
2. E n+1
-E n
>>E
F
3. E n+1
-E n
> ħ/

= ħe/m*

, Mobility e

/ m* g
 
 dN
4. The electronic density of states dE
, where dN is the number of electron quantum states within the energy interval dE , changes drastically between systems of different dimensionalities. This variation of the density of states is very important for light emitting semiconductor devices.
5. Carrier distribution n(E) =

(E) f (E) = Density of states * Fermi distribution
P ROBLEM EXAMPLES :
Calculate the mean free time of an electron and mean free path having a mobility of 
=1000 cm2/V-s at 300 K. Assume m*= 0.26m
0
, where m
0
= electron rest mass = 9.1
 10 -31 kg. Thermal velosity 10 5 mean free path l = v/ 
= 14.8 nm
m/s; k = Boltzmann’s constant = 1.38 x 10 -23 J/K
Solution:

= m*

/e = (O.26

9.1

10
-31 kg)(1000

10
-4 m
2
V
-1 s
-1
) / 1.6

10
19
C= 0.148 ps;

Scaling in electromagnetism. R~l
-
1 ; W ~l ; power density~l -1 . Moore’s law, emerging devices: ballistic, mesoscopic, low-dimensional, single electronic, spintronic. Coulomb blockade :Ec = Q
2
/
C.
P
ROBLEM EXAMPLES
:

For a tunnel junction with C = 0.7 aF and RT = 100 k

, what is the time constant? atto = 10-18

For a tunnel junction with C = 0.7 aF and RT = 100 k

, what is the maximum temperature at which you expect to find Coulomb blockade?

For SET assume that C = C = 10 aF; Cg = 1.4 * 10 -16 F. V g
= 0.1 V. If initially there are 175 electrons on the island, then what is the condition on Vs for an electron to tunnel across the left junction and onto the island?
• Light can be described as a wavelength

or as a angular frequency

•
1 mkm

1.23 eV

300 THz (computer work on GHz)
• Electromagnetic energy at a particular wavelength λ (in vacuum) has an associated frequency

and photon energy. These quantities are related according to the equations:

= c/

• E = h

• c = 3×10 8
m/s
• h = 6.65 × 10
−34
J·s or, in alternative units, h = 4.1 μeV/GHz.
Visible range: from 0.4 to 0.7
 m. Energy is calculated from the equation eV
 m = 1.2359
Maxwell equations
 
E
 

1
 
P
 
H

0
0
 
 
H
E



0
 
0

E
 t


H
 t

P
 t

J



  2
E

Source-free Maxwell equations
∇ ⋅ D = 0
∇ ⋅ B = 0
1 c
2
 2 E
 t
2
 
0
 2 P
 t
2
 
0

J
 t
∇ × E = −∂ B /∂t
∇ × B = µ ∂ D/ ∂ t field

P
 
0


E
 
0
  
1

E

describes how a material is polarized in response to an electric
Under the influence of E, the oppositely directed motion of + and - charges gives rise to a dipole moment characterized by polarization P. D =ε
0
E + P
Magnetic field B leads to a magnetic polarization characterized by magnetization M . H = B /

0
M
:
1. m eh
* = optical effective mass : 1/ m eh
* = 1/ me* + 1/ mh *
2. E g
= E g solid +ħ 2

2 /2R 2 meh * -1.8e
2 /4

0
R

E g = bandgap energy of a quantum dot or bulk solid
R = quantum dot radius m e = effective mass of the electron in the solid m h = effective mass of the hole in the solid m eh
* = m eh
/ m
0
ε = dielectric constant of the solid
ε0 = permeability of a vacuum
3.
From Fermi golden rule: the absorption coefficient and joint density of states are proportional to each other.
4 . In a Qdot energy is quantized: En = ħ 2  2 / m eh
Exciton binding energy E b
= E g round
– E

8 R 2 e xited
= ħ 2 /8r 2 meh * - 1.8e
2 /4

0
r
Exciton energy in quantum wells
In order to correlate the calculated optical transition energies of a 1D quantum well to experimental data, one has to include the exciton (electron-hole pair) corrections
Bulk
The 3D bulk exciton binding energy can be calculated analytically
E ex,b
= - m eh
e
4
/ ( 32
 ²ħ² e r
² e
0
²) = - m eh
/ (m
0

²)

13.61 eV
where m eh
is the reduced mass of the electron-hole pair: 1/ m eh
= 1/m e
+ 1/m h
GaAs: 1/ m eh
= 1 / 0.067 + 1 / 0.5 ==> m eh
= 0.0591
CdTe: 1/ m eh
= 1 / 0.096 + 1 / 0.6 ==> m eh
= 0.0828
e is the electron charge

is the dielectric constant (GaAs: 12.93, CdTe: 10.6)

0
is the vacuum permeability
m
0
is the rest mass of the electron and
13.61 eV is the Rydberg energy.
In GaAs , the 3D bulk exciton binding energy is equal to -4.8 meV with a Bohr radius of

=
11.6 nm.
In CdTe it is equal to -10.0 meV with a Bohr radius of 6.8 nm.
Thus the energy of the exciton, i.e. band gap transition, reads
GaAs : E ex
= E gap
+ E ex,b
= 1.519 eV - 0.005 eV = 1.514 eV.
CdTe : E ex
= E gap
+ E ex,b
= 1.606 eV - 0.010 eV = 1.596 eV.
Both quantities are given in terms of the effective Rydberg energy and the Bohr radius for a 3D exciton in the same material.
P
ROBLEM EXAMPLE
:
Calculate the exciton Bohr radius for CdS with the following parameters:

= 5.6 m e
* = 0.21, m h
* = 0.8

0
= 8.85

10
-12 ħ= 1.034

10
-34 e = 1.6

10
-19 m
0
= 9.1

10
-31
kg a
0
= 0. 53

10
-10 m

where e the elementary charge, ħ the Planck constant divided by 2

and m
0
is the electron rest mass a
0
is the Bohr radius
Solution: 1.855 nm
Material response to electric field: (e.g., N.W. Ashcroft and N.D. Mermin “Solid state Physics”)
Instantaneous collisions with ions with a fixed probability per unit time dt: dt/τ . ( τ - relaxation time. Electrons move with constant velocity v; mean free path l.

= 1/
 m e
 2 r
 t
2
( t )
 m e

 r
 t
( t )
  e E ( t )
Collective long-range Coulomb interactions between free electrons displaced by the external field generate a force to restore the neutrality of the plasma (i.e., displaced electrons are pulled back to their equilibrium positions and oscillate with a characteristic frequency - plasma frequency. r (

)
 m e
(

2 e
 i
 
)
E (

Macroscopic polarization (dipole moment per unit volume):
Definition of the dielectric constant:
P
 
0
  
1

E
Drude permittivity function:
 p
2 = ne 2 /

0 m e

)
(

)

1

(

2


2 p i
 
)
P
  ne r

 ne
2
E m e
(

2  i
 
)
Plasmons in the bulk oscillate at determined by the free electron density and effective mass
Plasmons confined to surfaces that can interact with light to form propagating “surface plasmon polaritons (SPP)”
Confinement effects result in resonant localized (LSP) modes in nanoparticles
• ε(ω) = 1-(ωp/ω)2
• ω < ωp 
ε < 0 wavevector of light in the medium is imaginary

no propagating electromagnetic modes
• plasmon is the quantum of plasma oscillation with energy ħ ωp and lifetime  =2/γ
Surface plasmon modes
Boundary condition:
(a) transverse component of E is conserved, E
1x
= E 2x
(b) normal component of D is conserved D =

E =

0E + P
Localized plasmon modes
 p

4

0
R
3



 
 m
 m

E
0
2 m
Electric polarisability of the sphere

:
 
4

0
R 3
 
2
 m m
Where ε = ε
1
(ω)+i ε
2
(ω) = dielectric constant of the metal particle
ε m
= dielectric constant of the embedding medium usually real and taken independent of frequency
Resonant enhancement of p if

(

)

2
 m
 minimum
P ROBLEM EXAMPLES

Gold has an intense absorption of light with energy 2.3 eV. What is its colour in the reflected light. What is its colour of a thin film in the transmitted light?
- Calculate the plasma frequeny for gold
 p
= (ne
2
/
0 m ∗ ) with n and m ∗ being the density and effective mass of the conduction electrons, respectively. Calculate the electron relaxation time using

=
 m ∗ /ne2. The parameters are the following: n [m
− 3
] = 5.90

10
28
;

[
 m
− 1
]=4.9

10
7
; m ∗ /me=0.99. Answer:
 p /eV= 9.1;

/ fs = 29.
- Calculate the diameter of golden spherical nanoparticles which appear red in the Lycurgus cup.
Use the empirical formula for the localized surface plasmon resonance d(nm) = 215 (2.262 − E res
/ eV)
½
. Take the energy of the red light E = 2 eV. Answer: 100 nm.
- For crystallite sizes of 100 nm, scattering dominates and peaks in the red, with the nontransmitted light peaking at 610 nm. Which colour will be seen in the transmitted light (answer: blue color).
Optics of thin films, Bragg condition m

= 2 d sin

where m is integer(order of refraction);

is angle of incidence and d is the lattice constant.
Scaling in optical systems: The lattice constant: a; Frequency: c/a; Angular frequency: 2
π c/a;
Wavevector: 2
π
/a; Wavelength: a .
Plotting the dispersion relations in one dimession:

= c
 k

. plotting two-dimensional

=c(k x
2+k y
2)
½
.
Plotting the band structure of a 1d periodic system with a small periodicity of states in optical systems. Stop gap width
1)/(m
2

+1), m=n
1
/n
2
,

is the volume fraction.
=

/

0.

2
=

1
+

. Density
Concept of photonic strength

=3

(m
2
-
Band structure of 2D photonic crystals. Photonic effects inside the gap: spatial localization of light, waveguides, light trapped in a cavity, holey fibers. Effects outside the gap: Spectral dispersion: Photonic band gap • Slow light • Emission suppression and enhancement. Spatial
Dispersion• Super-prism •Self collimation• Invisbility•Negative-index media, also known as lefthanded media, have
 , μ that are simultaneously negative, 
< 0 and μ < 0. Metamaterials are periodic arrays of wires and split-ring resonators. Focusing with left-handed media.
P
ROBLEM EXAMPLES
:
- For a 1-D photonic crystal composed of multiple Si slices (100 nm thick, n=3.5) separated by air
(200 nm gap, n=1) calculate the wavelength of light which can be stopped.
- For a 2D photonic crystal, containing hexagonal lattices of high dielectric rods (ε = 12, r = 0.2a) in air, calculate the lattice parameter using the theoretically calculated band structure (picture) where the gap extends from 0.2837 c/a to 0.4183 c/a.
Answer: The mid gap frequency is at 0.3510 c/a. To design a crystal such that 1.55 micron light falls at the center of the gap, we have c/(1.55micron) = 0.3510 c/a, hence a = 0.3510 * 1.55 micron = 0.5440 micron
Energy : exchange energy, magnetocrystalline energy, magnetoelastic energy,
Magnetization dynamics. Barkhausen effect. Domains and domain walls in bulk materials and in thin films. Examples of magnetic configurations of nano-objects
Origin of magnetic moments in an isolated atom. - Influence of environment : crystal field effects.
Magnetic order. Magnetic phase transitions - Interactions between moments. Direct and indirect exchange (insulators and metals).- Types of interactions:(1) Exchange interaction (electrostatic)
(2) Dipolar interaction (spin-spin coupling) (3) Anisotropy interaction (spin-orbit coupling).

Antiferromagnetism. Ferrimagnetism. Magnetism in frustrated systems. Itinerant ferromagnetism.
Superparamagnetism. Superparamagnetic nanoparticles: Ea = KV sin
2 
, wkere K is the anisotropy constant, V is the particle volume. Effective magnetic moment of superparamagnetic nanoparticle .

 eff eff
=

0
[1-coth x] where x = N*

B / kT, N is the number of atoms in the cluster,
corresponds to the experimentally measured magnetic moment per atom , i.e.
 eff
=
 exper
.
P ROBLEM EXAMPLES :
- In sample 1, after H > Hc , <M> = Msat for H < < Hc, In sample 2, after H > Hc, <M> << Msat for H < < Hc, Which is the soft ferromagnet?
- Consider that the molar volume of magnetite is 4.4*10 5 m-3 and magnetic moment of Fe3O4 is
9.27*10
-24
, then calculate the magnetization of magnetite assuming that the magnetization os omly fue to the 4 3d electrons of the Fe2+ions and that only the spin angular momentum of the electrons contributes ti the magnetic moment. Solution: M = 4

B
N
A
/V m
= 5*10
5
JT
-1 m
-3
.
- If a magnetic domain, either a patterned ferromagnetic structure or single ferromagnet grain is too small, it becomes super-paramagnetic despite its anisotropy. Magnetic energy required to flip domain E mag
= ½ M sat
E mag
/k
B
T). τ = Γ −1
H c
V, where V is the particle volume. Average rate of flipping

=

0
-1 exp(-
is the mean lifetime or time spent in one of the possible magnetic states. τ
0
-1
= rate the magnet attempts to flip ~10
9
/sec. For data storage we need at least 10 years. Calculate the minimum diameter of a nanoparticle with M sat
= 100 emu/g, H c
= 1 Tesla.
Hybridization. Electron configuration of carbon. Pauli principle and Hund’s rule. Euler’s theorem for fullerenes V
−
E + F = 2 ⋅ (1− g ). Nanotube, chirality, notation. Zone-folding approximation for a SWNT C = n a 1 + m a 2 is a translation vector of graphene => y(C) = y(O) exp( ik.C
) => k.C
=

2p => discretization lines parallel to the nanotube axis in reciprocal plane with distance 1/ R.
Condition for band crossing at E
F.
Graphene.
P
ROBLEM EXAMPLE
Using the C-C bond distance in graphite and simple geometry, calculate the diameter, the perimeter, the specific surface area of the unit cell, and density of single-walled carbon nanotubes that have 12 hexagons engirdling its equator. The atomic weight of carbon is 12 g/mol, and the C-C bond distance is 0.122 nm. Is it zigzag or armchair, metallic or semiconducting? Calculate its diameter, density, specific surface area (the area of the cylindrical surface divided by the mass of the nanotube unit cell).
Hazards for nanosilver, nano-TiO2, nanoZnO, fullerenes, nanotubes, nanocosmetics.
Explosiveness and toxicity. Societal, health and environmental risks. Nanopollutants. Risk of privacy invasion: Virtually undetectable surveillance devices could dramatically increase spying on governments, corporations and private citizens. Risk of economic upheaval. Risk of nanotech weapons. Gray Goo scenario.
P ROBLEM EXAMPLE Using The field of nanotechnology is subdivided into incremental nanotechnology (improving the properties of many materials by controlling their nano-scale structure), evolutionary (nanotechnology microelectronic devices towards actual nano-scale devices), and radical nanotechnology (Assemblers, i.e. nanomachines that manipulate matter one atom at a time, and Replicators i.e. nanomachines that make copies of themselves). Which societal problems are related to evolutionary nanotechnology ?