Producing Leaders Since 1905www.um.edu.my

CHAPTER 1NANOPARTICLE

NANOPHYSICS 2008NANOPHYSICS 2008KOMPLEKS DEWAN KULIAH, FAKULTI SAINS, UNIVERSITI MALAYAKOMPLEKS DEWAN KULIAH, FAKULTI SAINS, UNIVERSITI MALAYA

27 27 -- 28 MAY 200828 MAY 2008

NANOSCOPE• A lot of interest to study crystals of the size of a few nm. • One nm =10 Å or 10-9 m.• Distance between atoms is of the order of a few Å. • X-ray scattering to measure this distance

• When the distance between atoms is of the order of integer multiples of the wave length, the x-rays are diffracted.

• Lattice periodicity to get sufficient scattering but the length of the periodicity of the thickness of the lattice is not measured.

• Surfaces measured by; • reflection of light, or • an optical microscope

• The smallest particles which we can see are then of the size of a few thousand Å.

• Scanning electron microscope can see much smaller distances because the de Broglie wave length of the electron can be made much smaller than optical wave length.

• Scanning tunneling microscope; a metallic tip is used. • The sample surface exerts a force on the electrons of the

tip so that the tunneling current maps the surface. • The sample surface is connected to the tip to measure

the tunneling current, even though the tip does not touch the surface.

• When the small changes in the current can be measured, it is possible to make an accurate map of the surface.

• When the sample is sufficiently isolated from the vibrations of the walls and the table, the force exerted by the sample on the tip is of the order of a few nanoNewton and the resulting tunneling current can resolve atoms separated by 1 Å.

• The size of a metal-oxide-semiconductor device is a few µm.

• At the present time, it is possible to make a ball mill which can produce particles of the size of a few nm, such as 1 or 2 nm.

• The physical properties of the bulk samples are very different from those of nanometer size particles.

• e.g. the melting temperature of nanometer size particles may be very low whereas for particles larger than 20 nm, much larger bulk value is found.

• Thick aluminium wires have much larger melting temperatures than thin wires. The superconducting transition temperature of bulk aluminium is 7 K but that of nanometer thick wires it is much less than 1 K.

• Large particles obey the Debye's T3 law of specific heat but smaller particles have exponential dependence.

• Bulk sodium is a solid at room temperature and it melts at 97.8 oC to a liquid state which evaporates at a higher temperature. However, a few thousand atoms of sodium show a condensed state of matter at temperatures of a few nK.

• Using electromagnetic force on atoms by laser diodes, produces temperatures of the order of nK at which Bose-Einstein condensation occurs.

• Attosecond spectroscopy to measure short lived energy levels.

• Femtosecond pulses helped understand short life times.

• Terawatt laser generate subattosecond pulses and still higher powers may lead to zeptosecond, zs, pulses.

•as = attoseconds•fs = femtoseconds•zs = zeptoseconds

• 1fs =10-15s, 1 as = 10-18s and 1zs = 10-21s.

• The atoms of a sample exert a force on a metallic tip through which the tunneling current can be measured.

• This tunneling current depends on the surface barrier and hence can be plotted in two dimensional graphs.

• The equipment used to make this type of graph is called an atomic force microscope (AFM).

• This differs from the scanning tunneling microscope only in the design of a tip through which the tunneling current is measured.

In Fig. 1.1, we show the design of the experimental setup of the AFM.

• Fig. 1.2 shows that the tunneling current makes the map of the surface.

• AFM image of TiO2 with Ti2O3 adsorbed on the surface, is shown in Fig. 1.3.

• The right hand side shows the simulation of Ti2O3 on TiO2surface.

• The force required to accelerate one kG of mass to an acceleration of one m/s2 is called one Newton.

• In the STM the beam which is connected to the tip moves by a force as small as 10-18 N.

• In the case of a profilometer, the radius of the stylus is about 1 µm and loading force extends from 10-2 to 10-5 N.

• In the ionic materials, the binding energy is often of the order of 10 eV.

• The energy is obtained from the force multiplied by the displacement.

• The binding energy of 1 eV requires a force of 10-8 N.• The interatomic forces therefore range from 10-7 N for ionic

bonds to 10-1 N for van der Waals bonds and can be as low as 10-12 N for weaker surfaces.

• The force required to move the tip in the atomic force microscope is thus of the order of a few nano Newton, nN.

• The mass of the spring in man made structures can be quite small but microfabrication methods are employed to fabricate a spring with a mass less than 10-10 kg and a resonant frequency larger than 2 kHz.

• A boron carbide nanospring has been grown on a silicon substrate by the technique of plasma enhanced chemical vapour deposition (PECVD) where Fe works as a catalyst.

• The nanospring is 60 nm in diameter and approximately 1 µm in length.

• The diameters of the boron carbide nanospringsrange from 20 to 30 nm with pitches from 31 to 43 nm and nanowire diameters from 20 to 30 nm.

• The atomic force microscope has been operated in various modes, which relate to feedback circuits.

• They serve to maintain a constant force, fo,between the sample and the diamond stylus while the latter follows the contours of the surface.

• The sample is modulated in the z-direction at its resonant frequency 5.8 kHz.

• The force between the sample and the diamond stylus deflects the lever holding the stylus.

• In turn, this modulates the tunneling current which is used to control the AFM feedback circuit and maintain the force fo at a constant level.

• Alternatively, the lever carrying the diamond stylus is driven at its resonant frequency in the z-direction with an amplitude of 0.01 to 1 nm.

• The force, fo, between the sample and stylus changes the resonant frequency of the lever.

• This changes both the amplitude and phase of the ac modulation of the tunneling current.

• Either of these can be used as a signal to drive the feedback circuits.

• Another alternative is to use a feedback circuit which is connected to the AFM and controlled by the tunneling current in the STM.

• It can be improved by reconnection of both feedback circuits in such a way that the AFM sample and the STM tip are driven in opposite directions with a factor of 10 to 1000 less in amplitude for the STM tip.

• The problem of statistics is a serious one. • When the number of atoms becomes small, it is not clear

whether the Bose-Einstein or the Fermi-Dirac statistics is applicable.

• The Maxwell-Boltzmann statistics tells us that the number of atoms in the upper states is less than those in lower states by a factor of exp(-E/kBT).

• This type of distribution is sufficient for a long column of a gas but not for a small system.

• In small systems, Bose-Einstein condensation occurs at ultra low temperatures such as ~ 20 nK.

• Large particles exhibit the phenomenon of ferromagnetism and the phase transition from ferromagnetic state to a paramagnetic state but the small particles need not exhibit ferromagnetism at the same temperature as large particles.

• The Curie temperature is much less in small particles than in large crystals.

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CRYSTALS & CLUSTERS

In table 1.1 we show the properties of the unit cells and in Fig. 1.4 we show the 14 Bravais unit cells.

• The crystals are subjected to a translational symmetry so that these are all of the possibilities, if unit cells have to be used.

• In recent years, a five fold symmetry is found in two dimensions.

• Although it is possible to fill the entire space by using kite and dart type figures, the usual periodicity is still missing.

• It is possible to make such symmetries in Al-Mn alloys.

It is possible to make quasi-crystals by using kite and dart combinations as shown in Fig. 1.5.

• The solutions of Schrödinger equation are in terms of Legendre's polynomials which determine the symmetries found in atoms but molecules need not be restricted by Legendres's polynomials.

• There is no restriction on the number of atoms in a unit cell.

• Large unit cells are found such as in YBa2Cu2O7 type systems.

• Molecules of interest in microbiology have very large cells.

Balls of C60, C70, C240, etc. have been made shown in Fig 1.6.

• If we put a crystal in a ball mill, it is possible to make clusters which are “smaller than unit cell”.

• Large cell can be broken into smaller “clusters”.

• We can solve the Schrödinger equation for a small cluster of atoms, such as 5 atoms of Au and determine the bond distances and the energy.

• The monoatomic layers of the metals are made easily in a laboratory.

i.e. a cluster of 5 or 20 Na atoms can be made which does not have the same symmetries as in a single crystal of Na metal as shown in Fig. 1.7.

NANOLITHOGRAPHY• The basic idea of the dip-pen nanolithography is that from a

tip single molecules can come out by means of activation energy so that the thickness of the pen becomes comparable to molecular size.

• The size of the dots made by the pen also depends on the time during which pen is held.

• The dot area increases by increasing the contact time called the “dwelve time”.

• The ink molecules can be doped with other molecules such as water molecules.

• These water molecules determine the humidity. • The dot diameter increases with increasing humidity for

small humidity. • At large humidity the dot diameter becomes independent of

humidity. • The faster is the tip speed, the narrower is the line drawn.

The line thickness depends on the inverse of the tip speed.

• The thickness of the pen is limited by quantum mechanics. For optical radiation, the wave length restricts the thickness of the pen. Otherwise, the distance between lines for printing is restricted by the Heisenberg's uncertainty relation, ∆x.∆p = ħ/2.

• The distance uncertainty is of the order of wave length due to Bragg's law,

so there is a large distance limitation. This limitation decides the resolution in lithography.

• One bit is a number 0 or 1. One megabit is one million bits, 1048576=220 bits. 15 gigabits per square inch can be achieved by a recording head by using extraordinary magnetoresistance of narrow gap Si doped InSb quantum well.

θλ

sin2nx ≅∆

X-RAY SCATTERING• x-ray scattering to measure the unit cell size in crystals. • It is found that the unit cell size is larger in powders of

smaller particle size. • In the case of α- Fe2O3 x-ray Rayleigh scattering data is

available. The kα x-rays from Co source are scattered by the (104) planes of α - Fe2O3 in powders of particle sizes 172 Å, 385 Å and 1250 Å.

• As the particle size reduces, the unit cell becomes longer and hence < sinθ > becomes smaller.

• There are particles of varying sizes so that an envelop of lines emerges.

• For particle size 172 Å, there is an expansion of the unit cell by about 1.2 %.

• This expansion is deduced by measuring the average scattering angle, which varies from sin 38o = 0.616 sin 39o = 0.6293.

The particles of Au of size 4 nm show thermal expansion below 125 K and contraction above this temperature as shown in Fig. 1.9.

In Fig. 1.8 such envelopes of lines are shown.

• Using Bragg's law,

so that the two angles correspond to about 1.3 per cent change in the lattice size.

• For particles of average size 385 Å, the angular width is 0.3o

which corresponds to 0.41 per cent expansion of the unit cell.

• Similarly, the average particle size 1250 Å shows only 0.27 per cent expansion of the unit cell with respect to the bulk value.

• Since there is scattering at the surface, the normal modes acquire the wave vector which corresponds to the size of the crystal. These wave vectors are excited at a particular temperature so that there is “thermal expansion” below a certain temperature and “thermal contraction” above that temperature.

2d1 0.6293 = λ2d2 0.6159 = λ

REFERENCESG. Binnig and H. Rohrer, Rev. Mod. Phys. 59, 615 (1987).G. Binnig, C. F. Quate and Ch. Gerber, Phys. Rev. Lett. 56, 930 (1986).D. N. Mcllroy, D. Zhang, Y. Kranov and M. Grant Norton, Appl. Phys. Lett. 79, 1540 (2001).M. Ashino, Y. Sugawara, S. Morita and M. Ishikawa, Phys. Rev. Lett. 86, 4334 (2001).W. A. Hofer, A. J. Fisher, R. A. Wolkow and P. Grütter, Phys. Rev. Lett. 87, 236104 (2001).K. N. Shrivastava, Nano Lett. 2, 21, 519 (2002).C. Kittel, Introduction to solid state physics, Wiley, New York.D. Shechtman, I. Blech, D. Gratias and J. W. Cahn, Phys. Rev. Lett. 53, 1951 (1984).D. Shechtman and I. A. Blech, Metallurgical Transactions A 16, 1005 (1985).R. Penrose, Bull. Inst. Math. Appl. 10, 266 (1974).A. L. Mackay, Sov. Phys. Crystallography. 26, 517 (1981).B. L. Weeks, A. Noy, A. E. Miller and J. J. De Yoreo, Phys. Rev. Lett. 88, 255505 (2002).S. A. Sonin, et al, Appl. Phys. Lett. 80, 4012 (2002).D. Schroeer and R. C. Nininger, Jr., Phys. Rev. Lett. 19, 632 (1967).W.-H. Li, S. Y. Wu, et al., Phys. Rev. Lett. 89, 135504 (2002).A. E. Kaplan and P. L. Shkolnikov, Phys. Rev. Lett. 88, 074801 (2002).

Thank You

CHAPTER 2MELTING

NANOPHYSICS 2008NANOPHYSICS 2008KOMPLEKS DEWAN KULIAH, FAKULTI SAINS, UNIVERSITI MALAYAKOMPLEKS DEWAN KULIAH, FAKULTI SAINS, UNIVERSITI MALAYA

27 27 -- 28 MAY 200828 MAY 2008

LINDERMANN’S CRITERION• The molecules in a solid oscillate around their equilibrium

position, due to the thermal energy. • Upon heating, this thermal energy of atoms increases to

such as extent that the solid melts. • When the amplitude of oscillations increases beyond a

critical value, the system is said to melt. • In 1910 Lindemann has found that when the amplitude of

oscillations becomes about 1/10 of the distance between atoms, the solid melts.

• This is called a solid to liquid phase transition. • The temperature at which this phase transition occurs is

called the “melting temperature”. • There is a discontinuity at the melting temperature because

the energy of the system with solid at the melting temperature is not equal to that of the liquid at the same temperature.

Producing Leaders Since 1905www.um.edu.my• This means that there is a latent heat required to convert

the solid into a liquid at the melting temperature. • The variation of energy as a function of temperature is not

a continuous function at the “melting temperature” so that it is called a first-order phase transition.

• We take the example of sodium chloride. It has a face centered cubic lattice with unit cell size, ao=0.563 nm.

• This is the distance between two Na+ ions along the cube edge.

• In this direction the distance between a Na+ and Cl- is half the unit cell size.

• The Na+ atoms oscillate due to thermal energy with amplitude of oscillations, u.

• Upon heating u increases with increasing temperature and when it becomes about 1/10 of ao, we say that the lattice melts and the temperature becomes the melting temperature, Tm.

• The melting temperature of NaCl is 777 K. • At this temperature u = 0.1ao. • Usually, a crystal in a bulk sample has very large number of

atoms of the order of Avogadro's number, ~ 6.02217 x 1023 per gram molecule, i.e. in 2 ( 11 + 17 ) = 56 grams of NaCl, where 11 is the atomic number of Na and 17 that of Cl. The amplitude of oscillations of the Na+ atoms is,

(2.1)

where m is the mass of one Na atom and ω its frequency.

( )122

2 +⎟⎠⎞

⎜⎝⎛= n

mu

ωh

Producing Leaders Since 1905www.um.edu.my• The number of vibrational quanta excited at a temperature

of T is given by the Bose-Einstein distribution,

(2.2)• At the melting temperature, Tm,

(2.3)

where c = 0.1. This equation can be solved to determine the melting temperature.

• This explains the Lindemann's criterion of melting. • The mass of the Na atoms is much smaller than that of Cl

atoms. • Therefore, the Na lattice melts at a lower temperature than

the Cl lattice. • However, only one melting temperature is sufficient for

NaCl.

/1

1B mk Tne ω=

−h

2 2/2 1

2 1B m ok T c am e ωω

⎛ ⎞ ⎛ ⎞= + =⎜ ⎟ ⎜ ⎟−⎝ ⎠ ⎝ ⎠h

h

• In some crystals such as AgF, the F lattice melts at much lower temperature than the Ag lattice so that there is a temperature range in which one sublattice is melted and the other is not.

• In such a case the melted sublattice can carry a lot of conductivity, and there are two melting temperatures.

• In bulk, it is assumed that all atoms have equal amplitude of oscillations, at a given temperature.

• However, the atoms at the surface have slightly different amplitude than those in the interior.

• To determine the melting temperature, we can take the amplitude of the atoms in the interior and ignore those of the surface.

• Usually, the number of atoms at the surface is very small compared with those inside so that ignoring the surface does not have a serious effect in the calculation of the melting temperature.

• If we take only two atoms, one Na and one Cl, then, the number of atoms on the surface is two and that in the interior is zero.

• Then the approximation of small number of atoms at the surface breaks down.

• If we take 4 atoms, two Na+ and one Cl-, then also the number of atoms in the surface is 4 and that inside is zero.

• Therefore, for a small number of atoms, the surface can not be ignored.

• In a large crystal, there are uniform phonon modes so that all atoms participate in the oscillations.

• However, in a small crystal, the phonons travel from one end to another so that the phonon modes are different.

• Therefore, not so much due to the surface atoms but due to phonon modes, the melting properties of small crystals are different from those in the bulk.

• In 1910, the quantum term was not considered in the Lindemann's criterion of melting.

• In the harmonic oscillator, there is a “zero-point” vibration which gives the minimum energy as (1/2)ħω.

• The linear momenta and coordinates, being conjugate variables, do not commute so that the minimum energy is explained to arise from the uncertainty principle,

(2.4)h

21. ≥∆∆ px

• In the amplitude of oscillations, the correlation function gives 2n + 1 which at zero number of excitations n = 0, is a finite number so that at T = 0, n = 0, u remains finite u(o). This finite value,

(2.5)

is independent of temperature but other parameters, such as frequency, may lead to interesting results.

• In small crystals, the frequency can be written in terms of the size of the crystal, d,

(2.6)

so that the vibrations in small crystals are different from those of the large crystals.

2/1

2)( ⎟

⎠⎞

⎜⎝⎛=

ωmou h

dk π

υω 2

==

• The zero-point oscillation, u(o) due to the quantum nature of the oscillator contributes to the melting criterion.

• We have written u2 = c2ao2 as the melting criterion which

is not changed by the quantum consideration but there is a zero-point additional term in u2.

• In the case of bulk material “temperature” is the only parameter so we can obtain the melting temperature.

• However, in the small crystals, temperature is not the only parameter.

• We can consider “size” of the crystal in addition to temperature as the parameter.

• For small crystals, the “melting temperature” then depends on the “size” of the crystal.

MELTING OF SMALL PARTICLES• The phonons in the bulk are determined from the normal

modes of oscillations of a unit cell. • In the small crystals, the vibrations are the result of

oscillations diffracted at the walls. • Therefore, normal modes in small crystals are different

from those of the bulk. • When phonon wave length is of the order of the size of the

crystal, important scattering effects occur. • Usually, we have the Bragg scattering from the unit cell,

2ao sinθ = nλ but in small crystals diffraction occurs from the size of the crystal.

• Therefore, the frequencies which are important are ω given by,

(2.7)where d is the size of the crystal.

ωυπυθ n

vnd 2sin2 ==

• For n = 1, sinθ = 1,

(2.8)

where υ is the sound velocity, υ = 5 x 106 cm/s. For a frequency, v = 1012 Hz, we get d = 5π nm.

• Therefore, particle size, the surface of which deflects sound waves is of the order of a few nanometers.

• In this size range, it is most likely that the material is a single crystal.

• Therefore, we call these samples as nanocrystals. • When, different angles of diffraction are considered, it is

clear that there is a divergence at θ = 0.

( )dπυω =1

Producing Leaders Since 1905www.um.edu.my• In order to keep the vibrational energy finite, we replace

the value of sinθ by its average value, < sinθ > = ½. • Thus the largest frequency scattered from the surface of

the nanocrystal is,

(2.9)• All frequencies, larger than ω(2) are cutoff so that the

lattice energy is finite. • Besides, the Bose-Einstein factor becomes very small at

large frequencies. • Therefore, it is reasonable to consider only the frequency

range from ω(1) to ω(2). • This frequency spectrum belongs to the phonon scattering

in nanocrystals when the phonon wave length is of the order of size of the crystal.

• There is the usual phonon spectrum and the surface scattering spectrum is superimposed on the phonon continuum.

.2)2(dπυω =

Producing Leaders Since 1905www.um.edu.my• In a harmonic oscillator, the displacement of an atom is

given by,

(2.10)• In a crystal there are N atoms so that the total mass is Nm =

M. • The frequency of oscillations is ωk with k as the wave

vector, within the Brillouin zone. • The ak (ak†) are the operators which annihilate (create)

phonons so that when ak is applied to a wave function, it reduces the number of phonons of the given wave vector by one.

• Similarly, the creation operator increases the number of phonons by one.

• The number operator is ak†ak =nk which gives the number of phonons of a given wave vector.

• The average value, < nk > is determined by the Bose-Einstein distribution, nk = [exp(ħω/kBT)-1] -1.

).()2/( 2/1kkk aaMR −+= ωδ h

• The square of δR gives the amplitude of oscillations when the Brillouin zone is defined and δR is summed over all wave vectors as,

(2.11)• where the summation over all the wave vector can be

changed to an integral for a given dimensionality. In one dimension the above amplitude is given by,

(2.12)

where L is the length of the one-dimensional system.

( )( )∑ +>=<k

kk nMu ,122/2 ωh

( )( ) dkLnMdu kk )2(122/)1(2

πω∫ +==>< h

• In two dimensions,

(2.13)where L2 is the area of the two dimensional sheet. In three dimensions, we obtain,

(2.14)where a factor of 3 has been introduced to account for two transverse and one longitudinal branch of the phonon spectrum. ρ = M / V is the mass density per unit volume.

• We use the phonon dispersion relation, k = ω/υ, where υ is the sound velocity to obtain,

( )( ) dkLnMdu kk )2(122/)2(

22

πω∫ +==>< h

( )( ) ( )[ ] dkkVnMdu kk232 32/4122/)3( ∫ +==>< ππωh

[ ] ( ) ωωρυππ dnu k∫ +>=< 12)2(/6 332 h

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which can be written into two terms, one of which is the zero-point vibrational contribution and the other is a temperature dependent part,

(2.15)

• We assume a continuum of phonon spectrum of frequencies from zero to the Debye cut off frequency, Ω, so that the above can be written as,

(2.16)we make the change of variables,

(2.17)so that,

(2.18)where X = ħΩ/kBT.

[ ] [ ][ ].1)/exp(2)2/(6 1332 ∫∫ −−+>=< ωωωωωρυππ dTkdu Bhh

[ ] [ ] [ ]∫Ω

−−×+Ω>=<o

B dTku ωωωρυπρυπ 1323222 1)/exp(2/316/3 hhh

( ) xdxTkdxTk BB2/ , / hh == ωωω

( ) ( ) [ ]∫−

−×⎥⎦

⎤⎢⎣

⎡+

Ω>=<

X

o

xB xdxeTku 12

3232

22 1/

23

163

hhh

ρυπρυπ

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Thus, there is a zero-point vibrational correction represented by the first term above and there is a temperature dependent term which varies as T2 at low temperatures. In a bulk crystal, the phonons are well represented by a continuous spectrum. However, in a small crystal, the problem is completely different. The phonons go from wall to wall. The standing waves from one end to another are not at all like continuous phonons. Therefore, the phonons largely have the frequency given by (2.8). Due to mismatch between the phonon wave length and the size of the crystal, there is an angle for the formation of standing waves. This angle is similar to Bragg angle so that we can use the Bragg's diffraction formula with the interlayer spacing replaced by the size of the crystal.

Most of the scattering will occur at (2.8) but we may include upto a maximum of (2.9) for the frequency of phonons which form the standing waves in nanocrystals. Therefore, we calculate the square of the amplitude of oscillations for a nanocrystal by using the crystal size based phonon frequencies. Accordingly, in a nanocrystal, the square of the amplitude of oscillations is given by,

(2.19)where ρn is the mass density per unit volume of the nanocrystal and υn is the sound velocity. This equation has two terms, one a zero-point vibrational contribution and the other a temperature dependent term.

[ ] ( )∫ +>=<)2(

)1(

322 12)2/(6ω

ωωωυρππ dnu nnh

In Fig. 2.1, we show the Bragg's law scattering from a bulk crystal with only two planes and a small crystal, the size of which is of the same order of magnitude as the wave length of the phonon.

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The zero-point contribution is easily integrable and its value is,

(2.20)We substitute the values of the two limiting frequencies from (2.8) and (2.9) in (2.20) to obtain,

(2.21)Therefore, for υ = υn, the above expression shows that the mean square displacement of atoms is proportional to the inverse square of the nanoparticle size, d. The temperature dependent part of (2.19) is given by,

(2.22)where x(1) = ħω(1)/kBT and x(2) = ħω(2)/kBT. From (2.8) and (2.9) we find x(1) = ħπυ/( kBTd) and x(2) = 2ħπυ/ kBTd. At very low temperatures, x >> 1, so that the integral in (2.22) can be treated as a constant and the displacement varies as T 2.

[ ] .)1()2(8/3 22322 ωωυρπ −=>< nnnanou h

).8/(9)( 22 dou nnnano υρh=><

( )[ ]( ) ( )∫−

−=><)2(

)1(

1222 1/4/3)(x

x

xBnnnano xdxeTkTu hh υρπ

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At high temperatures, x << 1, so that it is sufficient to write ex

= 1 + x, in the denominator of the integrand of (2.22) so that,

(2.23)and the displacement becomes,

(2.24)where,

(2.25)We use the Lindemann criterion to determine the melting temperature of a nanocrystal. It is assumed that the lattice melts when,

(2.26)

where ao is the usual lattice constant and c ≈ 0.1.

( ) ( )Tdkxdxe B

x

x

x /11)2(

)1(πυh≅−

−

∫

dTa

da

Tuou nanonano 42122 )()( +=><+><

( ).2/3 24 nnBka υπρ≅

22ocau >=<

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We apply (2.24) and (2.26) so that by ignoring the quantum mechanical zero-point term, we obtain the melting point as,

(2.27)which means that melting temperature is linearly proportional to the size of the crystal, i.e. larger nanocrystals have larger melting temperature. We apply the Lindemann criterion on the zero-point contribution to obtain,

(2.28)for melting temperature of the nanocrystal. Now the value of Tnm is slightly reduced compared with the value given by (2.27). Whereas Tnm versus d is linear in (2.27) it has acquired a negative d-1 term which means that Tnm(d) slightly bends,

(2.29)

42 /1.0 adaT onm =

( ) ( ) nmo Tdadaca // 42

12 =−

.1.0 212

4⎥⎦⎤

⎢⎣⎡ −=

da

aadT onm

EXPERIMENTS ON MELTING• The experimental values of the melting

temperatures of fine particles of Si are given by Goldstein.

• For small particle size, there is a linear region. • This experimental observation is in agreement with

the theory. • For larger particles, there is a bending so that the

melting temperature is not as large as expected from the linear extrapolation but the value is less.

• This bending is also in accord with the theory. • The experimental dependence is in agreement with

the theory.

In Fig. 2.2, we show the melting temperature of Si particles as a function of size of the particles.

The melting temperature of CdSas a function of size of the particles is given in Fig. 2.3.

DEBYE-WALLER FACTOR

• The recoilless fraction (Debye-Waller factor) is written as,

(2.30)where λ is the wave length of the γ ray. The recoilless fraction in a nanocrystal is given by,

(2.31)which we can write,

(2.32)where fc is the value in a crystal and fc in a nanocrystal. From eq.(2.24) the above can be written as,

(2.33)

( )222 /4exp λπ ><−= uf

( )[ ]2222 /4exp λπ nanouuf ><+><−=

nc fff =

)()( Tfofff nnc=

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• When the frequency increases by means of extra scattering at the surface of the nanocrystal and when ω1 and ω2 of equations (2.8) and (2.9) are larger than the Debye cutoff frequency of the bulk material, we expect u2 to increase compared with that in the crystal but the interpretation is nontrivial.

• The equation (2.14) has the bulk mass density while (2.19) has the mass density in a nanocrystal.

• In some experiments the surface area is very large because the recoil has to be absorbed by the entire sample, so the area becomes important.

• Therefore, we can write the volume as the area multiplied by the thickness, t so that V = t . A.

• The mass density in the nanocrystal is then, ρn = M / t . A so that for large area A, ρn becomes small.

• As ρn becomes small, large value of < u2 > results and the recoilless fraction reduces.

• This is like saying that there is a “negative pressure”. The material becomes less dense in going from the bulk to the nanocrystal.

• Let us take the largest term from (2.24) and substitute it in(2.32). The relevant term which depends on the crystal size is,

(2.34)• This predicts larger f for larger d. Experimentally in the

case of 197Au for particle sizes 60 Å and 200 Å, this is not the case. In fact, the experiment shows larger f for 60 Åsize particles than for 200 Å size.

• Usually, the material shows thermal expansion but we know that small Au particles show “thermal contraction”.

• Therefore, the recoilless fraction can be written as,

(2.35)in which a1 = 9ħ/8ρnυn has to be adjusted point-to-point.

.4

exp 221

2

⎥⎦

⎤⎢⎣

⎡−=

λπ

da

f d

⎟⎟⎠

⎞⎜⎜⎝

⎛ −⎟⎟⎠

⎞⎜⎜⎝

⎛ ><−== 22

12

2

22 4exp4exp

λπ

λπ

daufff dc

Fig. 2.4 Recoilless fraction of 197Au as a function of temperature for two particle sizes, 6 and 20 nm.

• In the study of specific heat it has been shown that there are new models due to nanometer size of the particles.

• The application of this mode gives the correct interpretation of the recoilless fraction in Au.

• Usually, the recoilless fraction is f = exp(-4π2u2 / λ2) and the mean square deviation in the displacement of an atom is

• u2 = (ħ/2mω )( 2n + 1).• Considering only the n = 0 term, the mean square deviation

becomes u2(o) = (ħ/2mω).• The mode found in the nanocrystal has frequency of ω1 = πυ/d.

• Substituting this value of the frequency in the mean square deviation gives u2(o) = (ħd / 2mπυ). The recoilless fraction now depends on d, the size of the crystal, as,

(2.36)( )./2exp 2υλπ mdf h−=

Producing Leaders Since 1905www.um.edu.myUsually a solid expands due to anharmonic force in the

solid. The size of the unit cell is measured by using X-ray diffraction or a high resolution microscope as a function of temperature. As the temperature is increased, the unit cell expands. In the nanocrystals there is a vibrational mode which depends on the size, d, of crystal. The frequency of this mode is predicted to be about πυ/d. It is of interest to study the effect of excitation of this mode on the unit cell size. As the system is warmed, at a particular temperature, the size dependent mode acquires sufficient population to dominate over the usual phonon continuum. Therefore, the thermal expansion of the unit cell switches from the phonon continuum to the size-dependent mode. This type of “switching temperature” can be detected by X-ray measurement.

Thermal expansion

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We assume there is no kinetic energy and the potential energy is given by,

(2.37)We suppose that the thermal expansion arises from the gx3

type anharmonicity and ignore the x4 type anharmonicity. The cx2 is the usual harmonic potential. The thermal expansion is determined by,

(2.38)

where β = 1/kBT. We assume that cx2 is much larger than the x3 and x4 type terms so that,

(2.39) (2.40)

which describes the linear thermal expansion of the unit cell size of a bulk crystal as,

(2.41)

.)( 432 fxgxcxxU −−=

∫∫

−

−

>=<U

U

dxe

xdxex

β

β

2/32/5

2/1

43~

βπβ

cgxdxe U∫ − 2/1)/( cdxe U βπβ ≅∫ −

Tkcgx Bbulk 24

3=><

As the temperature is increased the harmonic force in the frequency domain is described by c = mω2/2 with frequency ω = πυ/d where υ is the sound velocity and d is the size of the nanocrystal,

(2.42)If the nanograins are spherical, it is sufficient to use only one value of ω so that the above expression gives,

(2.43)which varies as d4. As temperature is increased the thermal expansion switches from < x >bulk to < x >nano so that there is a peak in < x > (T).

42

3ωm

Tgkx B>=<

244

43m

Tkgdx Bnano υπ

=><

At low temperatures < x > expands with increasing Twith slope given by < x >bulk.At a particular temperature the surface mode becomes important and the slope of < x >bulk becomes irrelevant and then the slope is determined by < x > nano. This means that dx/dT has a positive sign at low temperatures as at high temperatures but in the intermediate range of temperatures, it has negative sign so there is “thermal contraction” in addition to “thermal expansion” at low and high temperatures. This thermal contraction is characteristic of the nanocrystals.

This predicted behavior agrees with experimental observation of thermal contraction in Au nanoparticlesshown in Fig. 1.9.

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