Crystal Structure of Graphite, Graphene and Silicon€¦ · graphite and carbon nanotubes. (:: Say something about interesting band structure here) In the following, we will examine

Crystal Structure of Graphite Graphene and Silicon

Dodd Gray Adam McCaughan Bhaskar Mookerjilowast

6730mdashPhysics for Solid State Applications

(Dated March 13 2009)

We analyze graphene and some of the carbon allotropes for which graphene sheets form thebasis The real-space and reciprocal crystalline structures are analyzed Theoretical X-ray diffrac-tion (XRD) patterns are obtained from this analysis and compared with experimental results Weshow that staggered two-dimensional hexagonal lattices of graphite have XRD patterns that differsignificantly from silicon standards

The wide-variety of carbon allotropes and their associ-ated physical properties are largely due to the flexibilityof carbonrsquos valence electrons and resulting dimensionalityof its bonding structures Amongst carbon-only systemstwo-dimensional hexagonal sheetsmdashgraphenemdashforms ofthe basis of other important carbon structures such asgraphite and carbon nanotubes ( Say something aboutinteresting band structure here)

In the following we will examine the planar lat-tice structure of graphene and its extension to higher-dimensional lattice structures such as hexagonalgraphite We first analyze the lattice and reciprocal-space structures of two-dimensional hexagonal lattices ofcarbon and use the resulting structure factors to esti-mate the x-ray diffraction (XRD) intensities of graphiteWe conclude by comparing its calculated XRD spectra toexperimental spectra of graphene and crystalline silicon

1 PRELIMINARY QUESTIONS

11 Lattice Structure

Our discussion of the crystal structure of graphite fol-lows partially from DDL Chungrsquos review of graphite [1]When multiple graphene sheets are layered on top of eachother van der Walls bonding occurs and the three di-mensional structure of graphite is formed with a lattice

FIG 1 In-plane structure of graphite and reciprocal latticevectors [1]

lowastElectronic address doddmitedu amccmitedu mook-

erjimitedu

spacing between sheets c = 671A The sheets align suchthat their two-dimensional hexagonal lattices are stag-gered either in an ABAB pattern or an ABCABC pat-tern The ABAB alignment is shown in Figure 1 whichindicates four atoms per unit cell labeled A B Arsquo andBrsquo respectively The primed atoms AndashB on one graphenelayer are separated by half the orthogonal lattice spacingfrom the ArsquondashBrsquo layer BBrsquo atomic pairs differ from theircorresponding AArsquo pairs in their absence of neighboringatoms in adjacent layering planes The coordinates ofthese atoms forming the basis are given by

ρA = (0 0 0) ρAprime =(

0 0c

2

)

ρB =a

2

(

1radic3 1 0

)

ρBprime =

( minusa

2radic

3minusa

2c

2

)

(11)

With respect to an orthonormal basis the primitivelattice vectors are given by

a1 = a(radic

32minus12 0)

|a1| = a = 246A

a2 = a(radic

32 12 0)

|a2| = a = 246A

a3 = c (0 0 1) |a3| = c = 671A (12)

The magnitudes of the primitive lattice vectors corre-spond to the lattice constants parallel and perpendicu-lar to the graphene sheet The corresponding ABCABClayer forms a rhombohedral structure with identical lat-tice spacing parallel and orthogonal to the layer

12 Reciprocal Lattice Structure

Recall that the reciprocal lattice vectors bi are definedas a function of the primitive lattice vectors ai such that

b1 = 2πa2 times a3

a1 middot a3 times a3

b2 = 2πa3 times a1

a2 middot a3 times a1

b3 = 2πa1 times a2

a3 middot a1 times a2(13)

The reciprocal lattice vectors for graphite are then

2

FIG 2 Reciprocal lattice planes [001] [110] and [111]

given by

b1 =2π

a

(

1radic3minus1 0

)

|b1| =2π

a

2radic3

b2 =2π

a

(

1radic3 1 0

)

|b2| =2π

a

2radic3

b3 =2π

c(0 0 1) |b3| =

2π

c (14)

The reciprocal lattice plane generated by the b1 and b2

vectors forms the outline of the first Brillouin zone asdepicted in Figure 1 The intersection of the the planeskz = plusmn2πc with the plane forms a hexagonal prism ofheight 4πc

13 Atomic form factors

As carbon is the only element present in graphene andgraphite the atomic form factor is uniform across theentire crystal and thus can be factored out when calcu-lating the structure factor Thus the atomic form factorhas no effect on the relative intensities of x-ray diffrac-tion occuring in different planes of graphite Accordingto the NIST Physics Laboratory the atomic form factorof carbon varies from 600 to 615 eatom with incidentradiation ranging from 2 to 433 KeV [2]

2 X-RAY DIFFRACTION

21 Planes in the Reciprocal Lattice

Provide pictures of the crystal and of thereciprocal lattice in the [100] [110] and [111]planes Indicate the vertical positions ofatoms with respect to the plane

Pictures of the crystal and of the reciprocal lattice inthe [100] [110] and [111] planes are included in Fig-ure 2 In MATLAB the crystal was represented as a set

of points in space using the specified lattice vectors andatom bases Normals generated from lattice vector sumswere used to extract planes and display them Varyingcolors were used to depict vertical spacing between adja-cent planes

22 Structure Factors and X-Ray Diffraction

Intensities

Calculate the structure factor for all thereciprocal lattice vectors Kl lt 16 (2πa)

2

The structure factor is calculated as

Mp (Ki) = fc

nsum

j=1

(Ki) eminusiKimiddotρi

where fc is the structure factor of Carbon and ρi are thebasis vectors of our lattice We find that only four uniquenon-zero values of Mp (Ki) occur in the reciprocal latticeEach of these corresponds to the height of a diffractionintensity peak and their relative values are referenced inTable I

Calculate the ratio of the intensities ex-pected for the following lines of the diffractionpattern with respect to the [111] line [100][200] [220] [311] and [400]

Including the structure factor there are other severalfactors contributing to the intensities of the diffractionpeaks [3]

1 The Lorenz correction is a geometric relation al-tering the intensity of an x-ray beam for differentscattering angles θ

2 The multiplicity factor p is defined as the num-ber of different planes having the same spacingthrough the unit cell Systems with high symme-tries will have different planes contributing to thesame diffraction thereby increasing the measuredintensity

3 Temperature absorption polarization each con-tribute higher-order corrections ultimately ignoredin our calculation These include Doppler broad-ening from thermal vibrations in the material ab-sorption of x-rays and the polarization of initiallyunpolarized x-rays upon elastic scattering

These factors all contribute to the relative intensity of a[hkl] diffraction peak given by

I[hkl] (θ) = p|Ma (Km) |2(

1 + cos2 2θ

sin2 θ cos θ

)

(21)

Using the formula from the previous question to calcu-late the ratios of the structure factors in the given planes

3

FIG 3 Crystal structure of silicon

TABLE I X-Ray diffraction intensities for graphite and sili-con [4 5] Structure factors are included in parentheses

Si C (Graphite)

2θ () Exp 2θ () Calc Exp

[111] 2847 10000 mdash (3) mdash

[100] mdash mdash 4277 4 (3) 345

[200] mdash mdash mdash (1) mdash

[002] mdash mdash 2674 106 (16) 100

[220] 4735 6431 mdash (1) mdash

[311] 3731 3731 mdash (3) mdash

[400] 6921 958 mdash (1) mdash

we obtain the following results presented in Table I Cal-culations show that planes [100] [200] [220] and [400] ex-hibit relative diffraction intensities 13 that of the [111]and [311] planes The [002] plane exhibits the highest in-tensity diffraction 163 that of the [111] and [311] planesWe also found several non-zero structure factors that arenot present in the experimental data

23 Crystal Structure of Silicon

What are the ratios if the material wereSi How could you use this information todistinguish Si from your material by x-ray

diffraction

The crystal structure of another common semiconduc-tor material silicon (Si) is featured in Figure 3 Siliconforms a diamond cubic crystal structure with a latticespacing of 542A This crystal structure corresponds to aface-centered cubic Bravais lattice whose unit-cell basiscontains 8 atoms located at vector positions

d0 = ~0 d4 =a

4(1 3 3)

d1 =a

4(1 1 1) d5 =

a

4(2 2 0)

d2 =a

4(3 3 1) d6 =

a

4(2 0 2)

d3 =a

4(3 1 3) d7 =

a

4(0 2 2) (22)

The structure factor contributing to its X-ray diffrac-tion pattern is given by

Ma (Km) =

nsum

j=1

f (j)a (Km) eminusiKmmiddotdj

= f(1 + (minus1)h+k

+ (minus1)k+l

+ (minus1)h+l

+(minus1)h+k+l

+ (minusi)3h+k+l

+ (minusi)3h+3k+1

+(minusi)h+3k+1

)

= f(

1 + (minus1)h+k

+ (minus1)k+l

+ (minus1)h+l

)

middot(

1 + (minusi)h+k+l

)

(23)

This term undergoes a number of simplifications basedon the parity of its Miller indices If [hkl] are all evenand are divisible by 4 then Ma (Km) = 8f If they arenot divisible by 4 or have mixed even and odd valuesthen Ma (Km) = 8f Lastly if [hkl] are all odd thenMa (Km) = 4f (1 plusmn i)

The experimental X-ray diffraction intensities fromthese contributions are listed in Table I The intensityvalues for silicon were measured with respect to a ref-erence value II0 = 47 which is a direct ratio of thestrongest line of the sample to the strongest line of a ref-erence sample αndashAl2O3 The number of visible peaks andthe relative intensities between them suggest that siliconand graphite can be easily distinguished from each otherusing an x-ray diffraction experiment

[1] D Chung Journal of Material Science 37 1475 (2002)[2] () URL httpphysicsnistgovPhysRefData[3] B Cullity Elements of X-Ray Diffraction (Addison-

Wesley 1978)[4] E E B P J d G C H S C M Morris H Mc-

Murdie Standard X-ray Diffraction Powder Patterns volMonograph 25 Section 13 (National Bureau of Standards1976)

[5] () URL httprruffinfographitedisplay=

default

Phonon Spectra of Graphene

Dodd Gray Adam McCaughan Bhaskar Mookerjilowast

6730mdashPhysics for Solid State Applications

(Dated April 17 2009)

We use a Born model to calculate the phonon dispersion of graphene by accounting for stretching(αs) and bending (αφ) interactions between nearest neighbors Our model describes four in-planevibrational modes to whose dispersion relations we fit experimental lattice mode frequencies yieldingforce constants αs = 445Nm and αφ = 102Nm Our model also reasonably accounts for graphenersquosmacroscopic properties particularly sound speeds and elastic constants

The lattice dynamical properties of graphene form thebasis of understanding the vibrational spectra of carbon-based allotropes of various geometries such as graphiteor carbon nanotubes We can use an analytical descrip-tions of micromechanical behavior to better understandthe acoustic and optical properties of these materials

In the following we calculate the in-plane vibrationalspectrum of graphene and its contributions to macro-scopic elastic and thermodynamic quantities We firstdiscuss the force parameters of our Born model and de-rive a general potential to calculate the dynamical ma-trix of a primitive cell Using our dispersion relations athigh-symmetry points we can determine the vibrationaldensity of states in-plane sound velocity and elastic con-stants We will briefly touch upon weak out-of-planevibrational modes and its contributions to graphenersquosmacroscopic properties

1 BACKGROUND FOR THE BOHR MODEL

11 Parameters of Bohr Model

How many force constants are required foreach bond Why

Two force constants αs and αφ are required to modeleach bond Let αs represent the restoring force seen whena bond is stretched and let αφ represent the restoringforce seen when a bond is bent away from the axis alongwhich it is normally aligned

What is the energy of a single bond in theBorn model

The energy in a single bond is the sum of the stretchingand bending energies Es and Eφ If p is a vector alongwhich the bond is aligned in equilibrium and R is a latticevector then the energy contained in a bond between the

lowastElectronic address doddmitedu amccmitedu mook-erjimitedu

atom at R and the atom at R + p is

E[RR + p] = Es + Eφ

=1

2αs|p middot (u[R + p] minus u[R]) |2

+1

2αφ

(

|u[R + p] minus u[R]|2 minus |p middot (u[R + a] minus u[R])|2)

(11)

The model assumes that the bond is onlyslightly displaced from equilibrium Howwould you modify the model to make thebond energy more realistically dependent ondisplacement from equilibrium - what or-der would the corrections be and of whatsign Justify your answer physically includesketches if appropriate

The energy in the bond between the atom at latticevector Ri and the atom at lattice vector Ri+a is generallyestimated using a Taylor expansion to the second orderas

V (u[Ri t])) =

V0 +sum

n

sum

m

(

part2V

partun[Ri t]partum[Ri + a t]

)

eq

un[Ri t]

(12)

where the position of the atom at Ri + a is fixed n andm index all dimensions being considered V0 is the bondenergy seen with zero displacement and V is the potentialdefined as the sum of all bond energies in the entire lattice

V = middot middot middot+sum

n

(E[RR + an] + E[RR minus an])+middot middot middot (13)

where this is the slice of the lattice potential related toan atom at R and the number of different vectors an

between the atom at R and atoms coupled to it dependson which nth nearest neighbor model is used to modelthe lattice

Note that there is no first order term in this expan-sion A first derivative of potential energy would implya net force so this term must equal zero as the Taylorcoefficients are evaluated at equilibrium

To make the Born model more accurate we would needto take into account higher order terms from the Taylor

2

expansion of the potential We know that the third orderterm is non-zero because the curve we are attempting tofit is not even around the equilibrium point Based onthe fact that the function has higher curvature to the leftof the equilibrium point than it does to the right it seemsthat the sign of the third order term would be negative

12 Nearest Neighbor Couplings

If you use the only nearest neighbor cou-plings how many force constants will yourmodel require for your material How largewill the dynamical matrix be What if youused nearest neighbor and next-nearest neigh-bor couplings

Two force constants are needed to model graphenewhether a nearest neighbor coupling or nearest and nextnearest neighbor model is used as long as the model istwo-dimensional As mentioned before these are αs andαφ representing restoring forces due to bond stretchingand bond bending respectively As graphene is a twodimensional material with a two atom basis the dynam-ical matrix will be four-by-four A third force constantαz is required to account for out-of-plane phonon modesThe size of the dynamical matrix is not affected by thenumber of force constants used in our model however ifa third dimension is added to account for out of planevibrations then the dynamical matrix will be six-by-six

13 Elastic Properties

How many independent elastic constantsdoes your material possess What are they(give numbers) Why will a nearest neighborapproach not provide the most general solu-tion for a cubic material

Graphene has two elastic constants λ and micro Themeasured values of transverse and longitudinal phononvelocities in graphene are vt = 14 middot 103 msminus1 and vl =217 middot 103 msminus1 respectively [1] Given the density ρ =

2MC [ 3radic

3a2

2]minus1 = 76 middot 10minus7kgmminus2 where the mass of

carbon MC = 199 middot 10minus26 kg and the lattice constanta = 142 middot 10minus10 m the Lame coefficients are calculatedusing that vt =

radic

microρ and vl =radic

(λ + 2micro)ρ to be micro ≃293Nm and λ ≃ 722Nm It is worth noting that thesemacroscopic elastic constants are not usually calculatedor measured for graphene because it is a two-dimensionalmaterial with single atom thickness

2 CONSTRUCTION OF THE DYNAMICAL

MATRIX

We will first determine the phonon dispersion relationsof the in-plane vibrational modes in the context of the

FIG 1 Hexagonal crystal structure of a graphene primitivecell and its neighbors First Brillouin zone of graphene andits symmetry points

Born force model and and then briefly discuss the out-of-plane vibrational modes

21 Lattice and Reciprocal Space Structures

[Label] all the atoms in the basis and all theirnearest neighbors For each atom labeled A-H [verify] the lattice vectors Rp to each unitcell

A graphene sheet forms a two-dimensional hexagonalcrystal lattice with a primitive cell containing two atoms(A and B) depicted in Figure 1 Based on our previ-ous discussion of the Born model let us assume thatthe primitive cell interacts with nearest neighbors andnext-nearest neighbors with spring constants αs and αφrespectively and that atoms of the primitive cell have alattice spacing given by a = 142A and a primitive lat-tice constant 246A Figure 1 also depicts graphenersquos firstBrillouin zone and its symmetry points located at kΓ =(0 0) kM =

(

2πaradic

3 0)

and kK =(

2πaradic

3 2π3a)

The lattice contains two sublattices 0 and 1 which

differ by their bond orientations The first atom A inthe primitive cell has three first neighbors in the othersublattice 1 with relative unit vectors

eB = (1 0) eC =

(

minus1

2

radic3

2

)

eD =

(

minus1

2minus

radic3

2

)

(21)

and six next-nearest neighbors in the same sublattice 0with relative unit vectors

eE =

(radic3

21

2

)

eF =

(radic3

2minus1

2

)

eG = (0 1)

eH =

(radic3

21

2

)

eI =

(

minusradic

3

2minus1

2

)

eJ = (0minus1)

(22)

22 Born Force Model

[Using a Born force model find] a generalexpression for the potential energy of all the

3

TABLE I Calculated nearest neighbor and next-nearest neighbor forces for graphene The first (second) row of a given planarindex specifies the force interaction between the A (B) atom and one of its neighbors

A B C D E F G H I J

xx 3αs2 + 3αφ minusαs minusαs4 minusαs4 minus3αs4 minus3αs4 0 minus3αφ4 minus3αφ4 0

minusαs 3αs2 + 3αφ mdash mdash minusαs4 minusαs4 mdash mdash mdash mdash

xy yx 0 0radic

3αs4 minusradic

3αs4 minusradic

3αφ4radic

3αφ4 0radic

3αφ4 minusradic

3αφ4 0

0 0 mdash mdash minusradic

3αs4radic

3αs4 mdash mdash mdash mdash

yy 3αs2 + 3αφ 0 minus3αs4 minus3αs4 minusαφ4 minusαφ4 minusαφ minusαφ4 minusαφ4 minusαφ

0 3αs2 + 3αφ mdash mdash minus3αs4 minus3αs4 mdash mdash mdash mdash

atoms in the crystal in terms of their displace-ment from equilibrium

The total elastic potential energy for the all the atomsin the crystal in terms of their displacements fromequilibrium is given by summing the nearest-neighborstretching interactions

Vs (R) =1

2αs|eB middot (u1 [R] minus u2 [R + eB]) |2

+1

2αs|eC middot (u1 [R] minus u2 [R + eC]) |2

+1

2αs|eD middot (u1 [R] minus u2 [R + eD]) |2 (23)

with the nearest neighbor bending interactions

Vφ

=1

2αφ

(

(u1 [R] minus u2 [R + eB]) |2 minus |eB middot (u1 [R] minus u2 [R + eB]) |2)

+1

2αφ

(

| (u1 [R] minus u2 [R + eC]) |2 minus |eC middot (u1 [R] minus u2 [R + eC]) |2)

+1

2αφ

(

| (u1 [R] minus u2 [R + eD]) | minus |eD middot (u1 [R] minus u2 [R + eD]) |2)

(24)

23 Dynamical Matrix

Use the expression for the potential energyto determine the force on a given atom inthe crystal in terms of its displacement andits neighbors displacements Check your an-swer by directly calculating the force from thespring constants and displacements

Verify from the potential by explicitly takingthe derivatives the factors in the matrix

From the potential energy the force on a given atomin the crystal is given in terms of a harmonic oscillatorforce expression

fi = Dij (k) uj (25)

where Dij is the dynamical matrix is given by

Dij (k) =sum

Rp

Dij (Rp) eminusikmiddotRp =sum

Rp

Vprimeprime

ij (Rp) eminusikmiddotRp

(26)

In this explanation Vprimeprime

ij are second-partial derivatives ofour Born model potential at equilibrium Rp are the rel-ative lattice vectors described in Equations 21 and 22

and u =(

u1x u2

y u1x u2

y

)T The force constants from tak-

ing the appropriate second derivatives is given in Table I

The dynamical matrix is then given by

D (k) =

A0 B0 C D

B0 A1 D B1

Clowast Dlowast A0 B0

Dlowast Blowast1 B0 A1

(27)

If we let (AA) and (BB) represent the 0 and 1 sublatticesrespectively these matrix elements represent

A0 = Dxx (AA) = Dxx (BB)

B0 = Dxy (AA) = Dyx (AA) = Dxy (BB) = Dyx (BB)

C0 = Dxx (AB) = Dlowastxx (BA)

D0 = Dxy (AB) = Dyx (AB) = Dlowastxy (BA) = Dlowast

yx (BA)

A1 = Dyy (AA) = Dyy (BB)

B1 = Dyy (AB) = Dlowastyy (BA) (28)

As a sample calculation consider the element Dxx givenby A0

A0 = Vprimeprime

xx (AA) + Vprimeprime

xx (AE) eminusi

ldquo radic3

2kxa+

kya

2

rdquo

+ Vprimeprime

xx (AF ) eminusi

ldquo radic3

2kxaminus

kya

2

rdquo

+ Vprimeprime

xx (AG) eminusikya

+ Vprimeprime

xx (AH) eildquo radic

3

2kxaminus

kya

2

rdquo

+ Vprimeprime

xx (AI) eildquo radic

3

2kxa+

kya

2

rdquo

+ Vprimeprime

xx (AJ) eikya

=3αs

2+ 3αφ minus 3αs

4eminusi

ldquo radic3

2kxa+

kya

2

rdquo

minus 3αs

4eminusi

ldquo radic3

2kxaminus

kya

2

rdquo

+ 0 minus 3αφ

4eildquo radic

3

2kxaminus

kya

2

rdquo

minus 3αφ

4eildquo radic

3

2kxa+

kya

2

rdquo

+ 0

=3

2αs + 3αφ

[

1 minus cos

(radic3

2kxa

)

cos

(

1

2kya

)

]

(29)

4

TABLE II Calculated and measured [2] macroscopic prop-erties of graphene

Quantity Measured Calculated

Lattice Mode Frequencies (meV)

Γ M K Γ M K

ωLO 1948 1751 15536 1703 1829 1974

ωLA 0 16984 15536 0 1515 1564

ωTO 1948 16984 15536 1703 1806 1564

ωTA 0 5003 12507 0 665 998

ωZO 0 75 60 mdash mdash mdash

ωZA 110 52 60 mdash mdash mdash

Sound Velocities (kms)

vLA 217 1312

vTA 14 621

vZA mdash mdash

Elastic Constants (10 GPa)

C11 106 plusmn 2 1310

C12 28 plusmn 2 723

C44 043 plusmn 005 mdash

A similar calculation for the remainder of the matrixelements gives us

A0 =3

2αs + 3αφ

[

1 minus cos

(radic3

2kxa

)

cos

(

1

2kya

)

]

B0 = minusαφ

radic3 sin

(radic3

2kxa

)

sin

(

kya

2

)

C = minusαs

[

eminusi

ldquo

kxaradic3

rdquo

+1

2eildquo

kxa

2radic

3

rdquo

cos

(

kya

2

)]

D = minusi

radic3

2αse

ildquo

kxa

2radic

3

rdquo

sin

(

kya

2

)

A1 =3

2αs + αφ

[

3 minus 2 cos (kya) minus cos

(radic3

2kxa

)

cos

(

1

2kya

)

]

B1 = minus3

2αse

ildquo

kxa

2radic

3

rdquo

cos

(

kya

2

)

(210)

3 MODEL OPTIMIZATION AND

COMPARISON TO MACROSCOPIC

PROPERTIES

31 Comparison with Published Theoretical and

Experimental Data

Plot the phonon dispersion in appropriateunits along the Γ minus X X minus L and Γ minus Ldirections

For these values of force constants andmasses determine the atomic displacementsfor all the modes at Γ and for the highest

FIG 2 Calculated and experimental phonon dispersioncurves for graphene (a) Fitted experimental dispersioncurves using inelastic x-ray scattering in graphite [2] (b) (c)calculated dispersion relations using our fitted force constants(αs = 445Nm and αφ = 102Nm) and suggested force con-stants (αs = 1 and αφ = 025)

optic and lowest acoustic modes at X andL Provide drawings of the atomic motionof these modes How many modes are thereat Γ

5

32 Phonon Dispersion Relations

The phonon dispersion relation ω = ω (k) is deter-mined from the eigenvalue equation

(

Mminus1

D (k))

ǫ = ω2ǫ (31)

Note additionally that while graphene is two-dimensional it possesses miniscule elastic movement inthe vertical direction leading to a very small but non-zero elastic constant C44 We can define force constantsβs and βφ for the nearest and next-nearest interactions ofthe out-of-plane vibrational modes It has been shown [3]that the phonon dispersion relations for the out-of-planeoptical and acoustic vibrational modes are

ωZOZA =radic

u plusmn v (32)

where

u = 2βφ

[

cos(radic

3kya)

+ 2 cos

(

3kxa

2

)

cos

(radic3kya

2

)]

minus 3βs

v = βs

[

1 + 4 cos2

(radic3kya

2

)

+ 4 cos

(

3kxa

2

)

cos

radic3kya

2

]12

(33)

where the out-of-plane spring constants are given byβs = minus1176 and βs = 0190 These dispersion relationsare decoupled from the in-plane vibrational modes Bydiagonalizing the Hermitian dynamical matrix in Equa-tion 27 we can add these dispersion relations a sub-space From empirical data (see II) we know that elasticout-of-plane motion is several orders of magnitude lessthan the in-plane motion Our own plots of Equation 33yielded high lattice mode frequencies contradicting thisintuition As such we have omitted its inclusion in ourdetermination of macroscopic elastic constants

Figure 2 compares the measured dispersion relationof graphene against our calculated dispersion relationsusing the suggested force constants (αs = 1 and αφ =025) and our fitted force constants (αs = 445Nm andαφ = 102Nm) The optimum force constants are foundfor our model by parameterizing the force constants andcomparing the results of ω at the relevant zone edgesTwo force constants are chosen and ω is calculated foreach of the three zone edges Γ M and K An RMSerror is generated by subtracting the experimental valuesfrom the calculated values and iteratively the optimumvalues for the force constants are found by searching forthe lowest RMS error value

The relative atomic displacements for the acoustic andoptical modes at the Γ K and M symmetry pointsis shown in Figure 5There are a total of four in-planemodes (acoustical and optical transverse and longitudi-nal) present at Γ however both the optical and acousti-cal pairs are degenerate The atomic displacement pat-terns associated with the various eigenmodes at differentpoints of high symmetry in K space are shown in Figure

4 Many qualitative features of our calculated disper-sion relations are consistent with the experimental andab initio data in Figure 2 [4] Note degeneracies at theΓ point and linear dispersion relations for small displace-ments from Γ Hexagonal symmetry of the graphene lat-tice also accounts for the measured and calculated de-generacies at the M and K points The largest discrep-ancies between experimental data and our Born modelare for the TO and LO modes which require accountingof electron-phonon interactions in ab initio models foraccurate prediction [5]

Theoretical sound velocities for both longitudinal andoptical polarizations are estimated by calculating theslopes of the acoustical phonon dispersion curves of bothpolarizations near Γ and taking c = δω

δk The elasticconstants C11 and C12 are determined from the phononsound velocities as

vLA =radic

C11ρ vTA =radic

(C11 minus C12) ρ (34)

where ρ is the mass density of graphene

33 Density of States

Plot the total density of states (histogrammethod include all modes) versus frequency

Figure 3 shows the overall density of states and thosefor individual modes Experimental data for graphene isunavailable although the our calculation seemed qualita-tively consistent with experimental data for graphite [4]

34 Specific Heat

Calculate the specific heat of your materialversus temperature using (a) your calculateddensity of states (b) a Debye model and (c)a combined Debye-Einstein model (Debye foracoustic modes Einstein for optical modes)Plot your results for temperatures between 0Kand 3ΘD Comment on the strengths andweaknesses of your model

The specific heat for graphene is calculated using thecombined density of states (3) Debye model and com-bined Density of States (optical branches) and Debyemodel (acoustic branches) The two models result in rad-ically different specific heats perhaps because the De-bye model does not account for the optical modes uponwhich the specific heat significantly depends The plotsfor the specific heats in these models is given in Figure 4

In the Debye model calculation the longitudinal acous-tic velocity is used as the speed of sound instead of theinverse addition of longitudinal and tranverse velocitiesbecause in the calculated phonon dispersion relation thetransverse acoustic line is by far the most deviant fromexperimental value In addition it should be noted that