Confinement of electrons in layered metals

Unit 2 Classification of MaterialsSolid materials have been conveniently grouped into three basic classifications: metals, ceramics, and polymers. This scheme is based primarily on chemical makeup and atomic structure, and most materials fall into one distinct grouping or another, although there are some intermediates. In addition, there are three other groups of important engineering materials —composites, semiconductors, and biomaterials.译文：译文：固体材料被便利的分为三个基本的类型：金属，陶瓷和聚合物。

固体材料被便利的分为三个基本的类型：金属，陶瓷和聚合物。

固体材料被便利的分为三个基本的类型：金属，陶瓷和聚合物。

这个分类是首先基于这个分类是首先基于化学组成和原子结构来分的，化学组成和原子结构来分的，大多数材料落在明显的一个类别里面，大多数材料落在明显的一个类别里面，大多数材料落在明显的一个类别里面，尽管有许多中间品。

尽管有许多中间品。

除此之外，此之外， 有三类其他重要的工程材料－复合材料，半导体材料和生物材料。

有三类其他重要的工程材料－复合材料，半导体材料和生物材料。

Composites consist of combinations of two or more different materials, whereas semiconductors are utilized because of their unusual electrical characteristics; biomaterials are implanted into the human body. A brief explanation of the material types and representative characteristics is offered next.译文：复合材料由两种或者两种以上不同的材料组成，然而半导体由于它们非同寻常的电学性质而得到使用；生物材料被移植进入人类的身体中。

1. The Elements and The Periodic Table元素和周期表The number of protons in the nucleus of an atom is referred to as the atomic number, or proton number, Z. The numbers of electrons in an electrically neutral atom is also equal to the atomic number, Z. The total mass of an atom is determined very nearly by the total number of protons and neutrons in its nucleus. This total is called the mass of the number, A. The number of neutrons in an atom, the neutron number, is given by the quantity A-Z.refer to sb. [sth.] as 称某人（物）为be determined by 由…确定原子核中质子的数目称为原子序数，或者质子数，以Z表示。

电中性原子中电子的数目也等于原子序数Z。

经测定，原子的总质量与原子核中质子与中子的总数差不多。

（几乎相同）（或者说原子的总质量几乎可以由原子核中质子与中子的总数确定。

）这个总数叫质量数，以A表示。

因此，原子中的质子的数目，质子数，可以定量地由A-Z给出。

即原子中质子数=A-ZThe term element refers to a pure substance with atoms all kinds of a single kind. To the chemist the “kind” of an atom is specified by its atomic number, since this is the property that determines its chemical behavior. At present all the atoms from Z=1 to Z=107 are known; there are 107 chemical elements. Each chemical element has been given a name and a distinctive symbol. For most elements the symbol is simply the abbreviated form of the English name consisting of one or two letters,for example:元素这个术语指的是仅仅由同一种类的原子组成的物质。

纳米技术的发展英语作文The Evolution of Nanotechnology: Transforming the World.In the realm of scientific advancements, nanotechnology has emerged as a transformative force, revolutionizing various industries and opening up unprecedented possibilities. Nanotechnology refers to the manipulationand engineering of matter at the nanoscale, typically ranging from 1 to 100 nanometers. At this microscopic scale, materials exhibit unique properties and phenomena that are distinct from their macroscopic counterparts, enabling scientists and engineers to design and fabricate materials with exceptional characteristics.Origins and Historical Development.The concept of manipulating matter at the nanoscale can be traced back to the early 20th century. However, it wasnot until the 1980s, with the advent of scanning tunneling microscopes and atomic force microscopes, that scientistsgained the ability to visualize and manipulate individual atoms and molecules. The field of nanotechnology gained significant momentum in the 1990s and early 2000s, as advancements in microscopy, synthesis techniques, and computational modeling laid the foundation for the rapid development of nano-enabled technologies.Key Principles and Applications.Nanotechnology encompasses a diverse range of disciplines, including physics, chemistry, biology, and materials science. Its key principles revolve around the manipulation of materials at the nanoscale, enabling the creation of materials with tailored properties. Some of the fundamental concepts in nanotechnology include:Size Dependence: The properties of materials change significantly at the nanoscale, as quantum effects become dominant. This size dependence allows for the creation of materials with enhanced strength, reactivity, andelectrical conductivity.Surface Area to Volume Ratio: Nanoparticles have a large surface area relative to their volume, providing increased reactivity and interaction with surrounding molecules.Quantum Confinement: The confinement of electrons in nanoparticles results in discrete energy levels, leading to unique optical and electronic properties.Industries Impacted by Nanotechnology.Nanotechnology has found applications in a wide range of industries, including:Electronics: Nano-enabled materials are used to create smaller, faster, and more efficient electronic devices, such as transistors, displays, and sensors.Healthcare: Nanoparticles are utilized for drug delivery, gene therapy, and tissue engineering, offering targeted treatment and improved patient outcomes.Energy: Nano-materials are employed in the development of more efficient solar cells, batteries, and fuel cells.Manufacturing: Nanotechnology enables the creation of new materials with enhanced properties, leading to advancements in lightweight materials, coatings, and manufacturing processes.Consumer Products: Nano-additives are incorporated into textiles, cosmetics, and other consumer products to enhance their properties and introduce new functionalities.Nanoscale Phenomena and Novel Properties.The manipulation of matter at the nanoscale has led to the discovery of novel phenomena and properties, such as:Nanoparticle Self-Assembly: Nanoparticles can self-assemble into ordered structures, forming periodic patterns and even complex architectures with tunable properties.Surface Plasmons: The interaction of light with metalnanoparticles can generate surface plasmons, which are collective oscillations of electrons that give rise to unique optical properties.Quantum Dots: Quantum dots are semiconductor nanoparticles that exhibit quantum confinement effects, resulting in tunable emission colors and improved quantum efficiency.Challenges and Future Prospects.While nanotechnology holds immense promise, it also presents several challenges:Toxicity and Safety: Concerns exist regarding the potential toxicity and environmental impact of nanomaterials, necessitating thorough risk assessment and regulation.Mass Production and Cost: Scaling up nanotechnology for mass production remains a challenge, as the synthesis and processing of nanomaterials can be complex and expensive.Ethical and Regulatory Issues: The ethical implications of nanotechnology, such as privacy concerns and potential misuse, require careful consideration and regulation.Despite these challenges, the future of nanotechnology looks promising. Continued advancements in microscopy, synthesis techniques, and computational modeling will enable the development of even more advanced and sophisticated nano-enabled technologies. Research in areas such as quantum computing, nano-biotechnology, and nano-medicine is expected to lead to groundbreaking discoveries and transformative applications that will shape the world for generations to come.Conclusion.Nanotechnology has emerged as a transformative force in the 21st century, empowering scientists and engineers to create materials with unprecedented properties and functionalities. By harnessing the unique phenomena and interactions that occur at the nanoscale, nanotechnology isrevolutionizing industries, improving healthcare, and opening up new frontiers in scientific research. As the field continues to advance, we can expect even more remarkable innovations that will shape our future and improve the human experience.。

a r X i v :q u a n t -p h /0005130v 1 31 M a y 2000MULTIPHOTON RESONANT TRANSITIONS OF ELECTRONS IN THE LASERFIELD IN A MEDIUMH.K.Avetissian,A.L.Khachatryan,G.F.MkrtchianPlasma Physics Laboratory,Department of Theoretical PhysicsYerevan State University1,A.Manukian,375049Yerevan,ArmeniaE-mail:Avetissian@ysu.amFax:(3742)151-087Within the scope of the relativistic quantum theory for electron-laser interaction in a medium andusing the resonant approximation for the two degenerated states of an electron in a monochromaticradiation field [1]a nonperturbative solution of the Dirac equation (nonlinear over field solution ofthe Hill type equation)are obtained.The multiphoton cross sections of electrons coherent scatteringon the plane monochromatic wave at the Cherenkov resonance are obtained taking into account thespecificity of induced Cherenkov process [1,2]and spin-laser interaction as well.In the result ofthis resonant scattering the electron beam quantum modulation at high frequencies occurs thatcorresponds to a quantity of an electron energy exchange at the coherent reflection from the ”phaselattice”of slowed plane wave in a medium.So,we can expect to have a coherent X-ray source ininduced Cherenkov process,since such beam is a potential source of coherent radiation itself.I.INTRODUCTION As is known the coherent interaction of electrons with a plane monochromatic wave in a dielectric medium can be described as a resonant scattering of a particle on the ”phase lattice”of a traveling wave similar to the Bregg scattering of the particle on the crystal lattice [1],[2].The latter is obvious in the frame of reference (FR)of rest of the wave.Since the index of refraction of a medium n >1(n (ω)≡n as the wave is monochromatic)in this FR there is only a static periodic magnetic field and an elastic scattering of a particle takes place.The law of conservation for Cherenkov process taking into account the quantum recoil translates into the Bragg resonance condition between the de Broglie wave of the particle and this static periodic structure.Hence,in induced Cherenkov process the interaction resonantly connects two states of the particle which are degenerated over the longitudinal momentum with respect to the direction of the wave propagation:the states with longitudinal momenta p x -of the incident particle and the states with longitudinal momenta p x +ℓ k -of scattered ”Bragg”particle,as far as the conservation law of this process is |p x |=|p x +ℓ k |(ℓ-number absorbed or radiated photons with a wave vector k =k x ).The latter is the same as the Bragg condition of coherent elastic scattering.Therefore,in stimulated Cherenkov process no matter how weak the wave field is the usual perturbation theory is not applicable because of such degeneration of the states.So,the interaction near the resonance is necessary to describe by the secular equation [1].The latter,in particular,reveals zone structure of the particle states in the field of transverse electromagnetic (EM)wave in a dielectric medium [1],[2].Note that the application of the perturbation theory ignoring the mentioned degeneration in this process has reduced to essentially incorrect results which have been elucidated in the paper [3].In the present work the case of strong radiation field is considered within the scope of the relativistic quantum theory for electron-laser interaction in a ing the resonance approximation for the above mentioned two degenerated states in a monochromatic radiation field [1]a nonperturbative solution of the Dirac equation (nonlinear over field solution of the Hill type equation)are obtained.The multiphoton probabilities of free electrons coherent scattering on a strong monochromatic wave at the Cherenkov resonance are counted,taking into account the above mentioned specificity of induced Cherenkov process [1],[2]and spin-laser interaction as well.In the result of this resonant scattering the electron beam quantum modulation at high frequencies occurs that corresponds to the electron energy exchange at the coherent reflection from the ”phase lattice”of slowed wave in a medium.So,we can expect to have,in principle,a coherent X-ray source in induced Cherenkov process,since such-quantum modulated beam is a potential source of coherent radiation itself.II.NONLINEAR SOLUTION OF THE DIRAC EQUATION FOR ELECTRON IN STRONG EMRADIATION FIELD IN A MEDIUMIn this section we shall solve Dirac equation for a spinor particle in the given radiation field in a medium1i∂Ψn2−1.(5) The wave function of a particle in R frame is connected with the wave function in laboratory frameΛby the Lorentz transformation of the bispinorsΨ= S(ϑ)ΨR,(6) whereS(ϑ)=chϑ2;thϑ=V=1∂t′= α( p′−e A R(x′))+ βm ΨR.(8) Since the interaction Hamiltonian does not depend on the time and transverse(to the direction of the wave propaga-tion)coordinates the eigenvalues of the operators H′, p′y, p′z are conserved:E′=const,p′y=const,p′z=const and the solution of Eq.(8)can be represented in the form of a linear combination of free solutions of the Dirac equation with amplitudes a i(x′)depending only on x′:ΨR(r′,t′)=4i=1a i(x′)Ψ(0)i.(9)HereΨ(0)1,2= E′+m2 ϕ1,2σx p′x+σy p′y 2E′ 1E′+mϕ1,2 exp i(−p′x x′+p′y y′−E′t) ,(10)wherep′x=(E′2−p′2y−m2)1The solution of Eq.(8)in the form (9)corresponds to an expansion of the wave function in a complete set of the wave functions of a electron with certain energy and transverse momentum p ′y (with longitudinal momenta ±(E ′2−p ′2y−m 2)12).The latter are normalized to one particle per unit volume.Since there is symmetrywith respect to the direction A R (the OY axis)we have taken,without loss of generality,the vector p ′in the XY plane (p ′z =0).Substituting Eq.(9)into Eq.(8)then multiplying by the Hermitian conjugate functions and taking into account (10)and (2)we obtain a set of differential equations for the unknown functions a i (x ′).The equations for a 1,a 3and a 2,a 4are separated and for these amplitudes we have the following set of equationsp ′xda 1(x ′)dx ′=−iep y A y (x ′)a 3(x ′)−eA y (x ′) p ′x +ip ′y ·a 1(x ′)exp(2ip ′x x ′).(12)A similar set of equations is also obtained for the amplitudes a 2(x ′)and a 4(x ′).For simplicity we shall assume that before the interaction there are only electrons with specified longitudinal momentum and spin state,i.e.|a 1(−∞)|2=1,|a 3(+∞)|2=0,|a 2(−∞)|2=0,|a 4(+∞)|2=0.(13)From the condition of conservation of the norm we have|a 1(x ′)|2−|a 3(x ′)|2=const(14)and the probability of reflection is |a 3,4(−∞)|2.The application of the following unitarian transformationa 1(x ′)=b 1(x ′)exp iep ′y 2 a 3(x ′)=b 3(x ′)exp −i ep ′y2 ,(15)simplifies Eq.(12).Here ϑ′is the angle between the momentum of electron and the direction of the wave propagation in the R frame.The new amplitudes b 1(x ′)and b 3(x ′)satisfy the same initial conditions:|b 1(−∞)|2=1,|b 3(+∞)|2=0,according to (13).From Eq.(12)and Eq.(15)for the b 1(x ′)and b 3(x ′)we obtain the following set of equationsdb 1(x ′)dx ′=−f ∗(x ′)b 3(x ′)(16)wheref (x ′)=eA y (t )p ′p ′x x ′−∞A y (η)dη ;p ′=db1(x′)dx′=−∞N=−∞f N exp[i(2p′x−Nk)x′]b1(x′)(18)wheref N=p′p′xp′y(n2−1)=sω(21)whereϑis the angle between the electron momentum and the wave propagation direction in theΛframe(the Cherenkov angle),v and E are the electron velocity and energy.So,we can utilize the resonant approximation keeping only resonant terms in the Eq.(18).Generally,in this approximation,at detuning of resonance|δs|=|2p′x−sk′|<<k′,we have the following set of equations for the certain s-photon transition amplitudes b(s)1(x′)and b(s)3(x′):db(s)1(x′)dx′=−f s exp[iδs x′]b(s)1(x′)(22) This resonant approximation is valid for the slow varying functions b(s)1(x′)and b(s)3(x′),i.e.at the following conditiondb(s)1,3(x′)ch ∞−∞f s dη ,b(s)3(x′)=sh∞x′f s dηwhere△x′is the coherent interaction length.The reflection coefficient in the laboratory frame of reference is the probability of absorption at v<1/n or emission at v>1/n.The latter can be obtained expressing the quantities f s and△x′by the quantities in this frame since the reflection coefficient is Lorentz invariant.SoR(s)=th2[F s△τ](26) whereF s= (1−nv cosϑ)22v sinϑJ s ξω(1−nv cosϑ) (27)and△τfor actual cases is the laser pulse duration in theΛframe.The condition of applicability of this-resonant approximation(23)is equivalent to the condition|F s|<<ω,(28) which restricts as the intensity of the wave as well as the Cherenkov angle.Besides,to satisfy the condition(28)we must take into account the very sensitivity of the parameter F s towards the argument of Bessel function,according to Eq.(27).For the wave intensities when F s△τ 1the reflection coefficient is in the order of unit which can occur for the large number of photons s>>1when the argument of Bessel function Z∼s≫1in Eq.(27)(according to asymptotic behavior of Bessel function J s(Z)at Z≃s≫1).Let us estimate the reflection coefficient of an electron from the laser pulse or the most probable number of absorbed/emitted photons due to resonance interaction in induced Cherenkov process.For the typical values of experimental parameters of this process in the gaseous medium with the index of refraction n−1∼10−4,at the initial electron energy E∼50Mev and Cherenkov angleϑ∼1mrad,during the”Bragg reflection”from Neodymium laser pulse(ω△τ∼102, ω=1.17eV)with an intensity1010W/cm2(ξ∼10−4)electron absorbs or emits about105 photons.For the offresonant solution,whenδs=0,but f2s>δ2s/4from Eq.(22)for R(s)the following expression we obtainR(s)=f2s1+f2sf2s−δ2s/4,which has the same behavior as in the case of exact resonance.In opposite case when f2s δ2s/4the reflection coefficient is a oscillating function on interaction length.During the coherent interaction with EM wave the quantum modulation of particles beam density occurs too which in difference to classic one after the interaction remains unlimitedly long(for the monochromatic beam).This is a result of coherent superposition of particle states with various energy and momentum due to absorbed and emitted photons in the radiationfield which is conserved after the interaction.The quantum modulated state of the particle leads to modulation of the beam density after the interaction at the frequency of the stimulating wave and its harmonics[4].The density modulated particles beam can be used to generate spontaneous superradiation[5]The various radiation mechanisms of quantum modulated beams are investigated in the works[6]-[8].In stimulated Cherenkov process the beam quantum modulation occurs if the particles wave packet size(∆x)is enough large:∆x>>λ(λ-is the radiation wavelength).In the opposite case the classic modulation or bunching of the beam takes place(klystron interaction scheme).From Eq.(9)and Eq.(26)for the electron wave function after the reflection from the wave pulse we have the superposition of incident and reflected electron waves(in the R frame)ΨR=a1(−∞)Ψ(0)1+a3(−∞)Ψ(0)3(30) and in the result the probability densityρR=Ψ+RΨR is modulated at the X-ray frequenciesρ(s)R=1+th2[f s△x′]+2 1−p′2xE′2 sinϕ0=sinϑ′.In the laboratory frame of the reference from Eq.(7)and Eq.(30)we have5ρ(s)≃1n2−1(1+th2[F s△τ]+2th[F s△τ]cos(sωτ−ϑ′)(32) where it is taken into account that in actual case|sω|<<E.As is seen from Eq.(30)the modulation depth is in the order of unit for the intensities when F s△τ∼1which can be satisfied for the moderate intensities of the laserradiation in the order of1010W/cm2.。

《材料科学与工程基础》英文习题及思考题及答案第二章习题和思考题Questions and Problems2.6 Allowed values for the quantum numbers ofelectrons are as follows:The relationships between n and the shell designationsare noted in Table 2.1.Relative tothe subshells,l ＝0 corresponds to an s subshelll ＝1 corresponds to a p subshelll ＝2 corresponds to a d subshelll ＝3 corresponds to an f subshellFor the K shell, the four quantum numbersfor each of the two electrons in the 1s state, inthe order of nlmlms , are 100(1/2 ) and 100(-1/2 ).Write the four quantum numbers for allof the electrons inthe L and M shells, and notewhich correspond to the s, p, and d subshells.2.7 Give the electron configurations for the followingions: Fe2+, Fe3+, Cu+, Ba2+,Br-, andS2-.2.17 (a) Briefly cite the main differences betweenionic, covalent, and metallicbonding.(b) State the Pauli exclusion principle.2.18 Offer an explanation as to why covalently bonded materials are generally lessdense than ionically or metallically bonded ones.2.19 Compute the percents ionic character of the interatomic bonds for the followingcompounds: TiO2 , ZnTe, CsCl, InSb, and MgCl2 .2.21 Using Table 2.2, determine the number of covalent bonds that are possible foratoms of the following elements: germanium, phosphorus, selenium, and chlorine.2.24 On the basis of the hydrogen bond, explain the anomalous behavior of waterwhen it freezes. That is, why is there volume expansion upon solidification?3.1 What is the difference between atomic structure and crystal structure?3.2 What is the difference between a crystal structure and a crystal system?3.4Show for the body-centered cubic crystal structure that the unit cell edge lengtha and the atomic radius R are related through a =4R/√3.3.6 Show that the atomic packing factor for BCC is 0.68. .3.27* Show that the minimum cation-to-anion radius ratio for a coordinationnumber of 6 is 0.414. Hint: Use the NaCl crystal structure (Figure 3.5), and assume that anions and cations are just touching along cube edges and across face diagonals.3.48 Draw an orthorhombic unit cell, and within that cell a [121] direction and a(210) plane.3.50 Here are unit cells for two hypothetical metals:(a)What are the indices for the directions indicated by the two vectors in sketch(a)?(b) What are the indices for the two planes drawn in sketch (b)?3.51* Within a cubic unit cell, sketch the following directions:.3.53 Determine the indices for the directions shown in the following cubic unit cell:3.57 Determine the Miller indices for the planesshown in the following unit cell:3.58Determine the Miller indices for the planes shown in the following unit cell: 3.61* Sketch within a cubic unit cell the following planes:3.62 Sketch the atomic packing of (a) the (100)plane for the FCC crystal structure, and (b) the (111) plane for the BCC crystal structure (similar to Figures 3.24b and 3.25b).3.77 Explain why the properties of polycrystalline materials are most oftenisotropic.5.1 Calculate the fraction of atom sites that are vacant for lead at its meltingtemperature of 327_C. Assume an energy for vacancy formation of 0.55eV/atom.5.7 If cupric oxide (CuO) is exposed to reducing atmospheres at elevatedtemperatures, some of the Cu2_ ions will become Cu_.(a) Under these conditions, name one crystalline defect that you would expect toform in order to maintain charge neutrality.(b) How many Cu_ ions are required for the creation of each defect?5.8 Below, atomic radius, crystal structure, electronegativity, and the most commonvalence are tabulated, for several elements; for those that are nonmetals, only atomic radii are indicated.Which of these elements would you expect to form the following with copper:(a) A substitutional solid solution having complete solubility?(b) A substitutional solid solution of incomplete solubility?(c) An interstitial solid solution?5.9 For both FCC and BCC crystal structures, there are two different types ofinterstitial sites. In each case, one site is larger than the other, which site isnormally occupied by impurity atoms. For FCC, this larger one is located at the center of each edge of the unit cell; it is termed an octahedral interstitial site. On the other hand, with BCC the larger site type is found at 0, __, __ positions—that is, lying on _100_ faces, and situated midway between two unit cell edges on this face and one-quarter of the distance between the other two unit cell edges; it is termed a tetrahedral interstitial site. For both FCC and BCC crystalstructures, compute the radius r of an impurity atom that will just fit into one of these sites in terms of the atomic radius R of the host atom.5.10 (a) Suppose that Li2O is added as an impurity to CaO. If the Li_ substitutes forCa2_, what kind of vacancies would you expect to form? How many of thesevacancies are created for every Li_ added?(b) Suppose that CaCl2 is added as an impurity to CaO. If the Cl_ substitutes forO2_, what kind of vacancies would you expect to form? How many of thevacancies are created for every Cl_ added?5.28 Copper and platinum both have the FCC crystal structure and Cu forms asubstitutional solid solution for concentrations up to approximately 6 wt% Cu at room temperature. Compute the unit cell edge length for a 95 wt% Pt-5 wt% Cu alloy.5.29 Cite the relative Burgers vector–dislocation line orientations for edge, screw, andmixed dislocations.6.1 Briefly explain the difference between selfdiffusion and interdiffusion.6.3 (a) Compare interstitial and vacancy atomic mechanisms for diffusion.(b) Cite two reasons why interstitial diffusion is normally more rapid thanvacancy diffusion.6.4 Briefly explain the concept of steady state as it applies to diffusion.6.5 (a) Briefly explain the concept of a driving force.(b) What is the driving force for steadystate diffusion?6.6Compute the number of kilograms of hydrogen that pass per hour through a5-mm thick sheet of palladium having an area of 0.20 m2at 500℃. Assume a diffusion coefficient of 1.0×10- 8 m2/s, that the concentrations at the high- and low-pressure sides of the plate are 2.4 and 0.6 kg of hydrogen per cubic meter of palladium, and that steady-state conditions have been attained.6.7 A sheet of steel 1.5 mm thick has nitrogen atmospheres on both sides at 1200℃and is permitted to achieve a steady-state diffusion condition. The diffusion coefficient for nitrogen in steel at this temperature is 6×10-11m2/s, and the diffusion flux is found to be 1.2×10- 7kg/m2-s. Also, it is known that the concentration of nitrogen in the steel at the high-pressure surface is 4 kg/m3. How far into the sheet from this high-pressure side will the concentration be 2.0 kg/m3?Assume a linear concentration profile.6.24. Carbon is allowed to diffuse through a steel plate 15 mm thick. Theconcentrations of carbon at the two faces are 0.65 and 0.30 kg C/m3 Fe, whichare maintained constant. If the preexponential and activation energy are 6.2 _10_7 m2/s and 80,000 J/mol, respectively, compute the temperature at which the diffusion flux is 1.43 _ 10_9 kg/m2-s.6.25 The steady-state diffusion flux through a metal plate is 5.4_10_10 kg/m2-s at atemperature of 727_C (1000 K) and when the concentration gradient is _350kg/m4. Calculate the diffusion flux at 1027_C (1300 K) for the sameconcentration gradient and assuming an activation energy for diffusion of125,000 J/mol.10.2 What thermodynamic condition must be met for a state of equilibrium to exist? 10.4 What is the difference between the states of phase equilibrium and metastability?10.5 Cite the phases that are present and the phase compositions for the followingalloys:(a) 90 wt% Zn–10 wt% Cu at 400℃(b) 75 wt% Sn–25wt%Pb at 175℃(c) 55 wt% Ag–45 wt% Cu at 900℃(d) 30 wt% Pb–70 wt% Mg at 425℃(e) 2.12 kg Zn and 1.88 kg Cu at 500℃(f ) 37 lbm Pb and 6.5 lbm Mg at 400℃(g) 8.2 mol Ni and 4.3 mol Cu at 1250℃.(h) 4.5 mol Sn and 0.45 mol Pb at 200℃10.6 For an alloy of composition 74 wt% Zn–26 wt% Cu, cite the phases presentand their compositions at the following temperatures: 850℃, 750℃, 680℃, 600℃, and 500℃.10.7 Determine the relative amounts (in terms of mass fractions) of the phases forthe alloys and temperatures given inProblem 10.5.10.9 Determine the relative amounts (interms of volume fractions) of the phases forthe alloys and temperatures given inProblem 10.5a, b, and c. Below are given theapproximate densities of the various metalsat the alloy temperatures:10.18 Is it possible to have a copper–silveralloy that, at equilibrium, consists of a _ phase of composition 92 wt% Ag–8wt% Cu, and also a liquid phase of composition 76 wt% Ag–24 wt% Cu? If so, what will be the approximate temperature of the alloy? If this is not possible,explain why.10.20 A copper–nickel alloy of composition 70 wt% Ni–30 wt% Cu is slowly heatedfrom a temperature of 1300_C .(a) At what temperature does the first liquid phase form?(b) What is the composition of this liquid phase?(c) At what temperature does complete melting of the alloy occur?(d) What is the composition of the last solid remaining prior to completemelting?10.28 .Is it possible to have a copper–silver alloy of composition 50 wt% Ag–50 wt%Cu, which, at equilibrium, consists of _ and _ phases having mass fractions W_ _0.60 and W_ _ 0.40? If so, what will be the approximate temperature of the alloy?If such an alloy is not possible, explain why.10.30 At 700_C , what is the maximum solubility (a) of Cu in Ag? (b) Of Ag in Cu?第三章习题和思考题3.3If the atomic radius of aluminum is 0.143nm, calculate the volume of its unitcell in cubic meters.3.8 Iron has a BCC crystal structure, an atomic radius of 0.124 nm, and an atomicweight of 55.85 g/mol. Compute and compare its density with the experimental value found inside the front cover.3.9 Calculate the radius of an iridium atom given that Ir has an FCC crystal structure,a density of 22.4 g/cm3, and an atomic weight of 192.2 g/mol.3.13 Using atomic weight, crystal structure, and atomic radius data tabulated insidethe front cover, compute the theoretical densities of lead, chromium, copper, and cobalt, and then compare these values with the measured densities listed in this same table. The c/a ratio for cobalt is 1.623.3.15 Below are listed the atomic weight, density, and atomic radius for threehypothetical alloys. For each determine whether its crystal structure is FCC,BCC, or simple cubic and then justify your determination. A simple cubic unitcell is shown in Figure 3.40.3.21 This is a unit cell for a hypotheticalmetal:(a) To which crystal system doesthis unit cell belong?(b) What would this crystal structure be called?(c) Calculate the density of the material, given that its atomic weight is 141g/mol.3.25 For a ceramic compound, what are the two characteristics of the component ionsthat determine the crystal structure?3.29 On the basis of ionic charge and ionic radii, predict the crystal structures for thefollowing materials: (a) CsI, (b) NiO, (c) KI, and (d) NiS. Justify your selections.3.35 Magnesium oxide has the rock salt crystal structure and a density of 3.58 g/cm3.(a) Determine the unit cell edge length. (b) How does this result compare withthe edge length as determined from the radii in Table 3.4, assuming that theMg2_ and O2_ ions just touch each other along the edges?3.36 Compute the theoretical density of diamond given that the CUC distance andbond angle are 0.154 nm and 109.5°, respectively. How does this value compare with the measured density?3.38 Cadmium sulfide (CdS) has a cubic unit cell, and from x-ray diffraction data it isknown that the cell edge length is 0.582 nm. If the measured density is 4.82 g/cm3 , how many Cd 2+ and S 2—ions are there per unit cell?3.41 A hypothetical AX type of ceramic material is known to have a density of 2.65g/cm 3 and a unit cell of cubic symmetry with a cell edge length of 0.43 nm. The atomic weights of the A and X elements are 86.6 and 40.3 g/mol, respectively.On the basis of this information, which of the following crystal structures is (are) possible for this material: rock salt, cesium chloride, or zinc blende? Justify your choice(s).3.42 The unit cell for Mg Fe2O3 (MgO-Fe2O3) has cubic symmetry with a unit celledge length of 0.836 nm. If the density of this material is 4.52 g/cm 3 , compute its atomic packing factor. For this computation, you will need to use ionic radii listed in Table 3.4.3.44 Compute the atomic packing factor for the diamond cubic crystal structure(Figure 3.16). Assume that bonding atoms touch one another, that the angle between adjacent bonds is 109.5°, and that each atom internal to the unit cell is positioned a/4 of the distance away from the two nearest cell faces (a is the unit cell edge length).3.45 Compute the atomic packing factor for cesium chloride using the ionic radii inTable 3.4 and assuming that the ions touch along the cube diagonals.3.46 In terms of bonding, explain why silicate materials have relatively low densities.3.47 Determine the angle between covalent bonds in an SiO44—tetrahedron.3.63 For each of the following crystal structures, represent the indicated plane in themanner of Figures 3.24 and 3.25, showing both anions and cations: (a) (100)plane for the rock salt crystal structure, (b) (110) plane for the cesium chloride crystal structure, (c) (111) plane for the zinc blende crystal structure, and (d) (110) plane for the perovskite crystal structure.3.66 The zinc blende crystal structure is one that may be generated from close-packedplanes of anions.(a) Will the stacking sequence for this structure be FCC or HCP? Why?(b) Will cations fill tetrahedral or octahedral positions? Why?(c) What fraction of the positions will be occupied?3.81* The metal iridium has an FCC crystal structure. If the angle of diffraction forthe (220) set of planes occurs at 69.22°(first-order reflection) when monochromatic x-radiation having a wavelength of 0.1542 nm is used, compute(a) the interplanar spacing for this set of planes, and (b) the atomic radius for aniridium atom.4.10 What is the difference between configuration and conformation in relation topolymer chains? vinyl chloride).4.22 (a) Determine the ratio of butadiene to styrene mers in a copolymer having aweight-average molecular weight of 350,000 g/mol and weight-average degree of polymerization of 4425.(b) Which type(s) of copolymer(s) will this copolymer be, considering thefollowing possibilities: random, alternating, graft, and block? Why?4.23 Crosslinked copolymers consisting of 60 wt% ethylene and 40 wt% propylenemay have elastic properties similar to those for natural rubber. For a copolymer of this composition, determine the fraction of both mer types.4.25 (a) Compare the crystalline state in metals and polymers.(b) Compare thenoncrystalline state as it applies to polymers and ceramic glasses.4.26 Explain briefly why the tendency of a polymer to crystallize decreases withincreasing molecular weight.4.27* For each of the following pairs of polymers, do the following: (1) state whetheror not it is possible to determine if one polymer is more likely to crystallize than the other; (2) if it is possible, note which is the more likely and then cite reason(s) for your choice; and (3) if it is not possible to decide, then state why.(a) Linear and syndiotactic polyvinyl chloride; linear and isotactic polystyrene.(b) Network phenol-formaldehyde; linear and heavily crosslinked ci s-isoprene.(c) Linear polyethylene; lightly branched isotactic polypropylene.(d) Alternating poly(styrene-ethylene) copolymer; randompoly(vinylchloride-tetrafluoroethylene) copolymer.4.28 Compute the density of totally crystalline polyethylene. The orthorhombic unitcell for polyethylene is shown in Figure 4.10; also, the equivalent of two ethylene mer units is contained within each unit cell.5.11 What point defects are possible for MgO as an impurity in Al2O3? How manyMg 2+ ions must be added to form each of these defects?5.13 What is the composition, in weight percent, of an alloy that consists of 6 at% Pband 94 at% Sn?5.14 Calculate the composition, in weight per-cent, of an alloy that contains 218.0 kgtitanium, 14.6 kg of aluminum, and 9.7 kg of vanadium.5.23 Gold forms a substitutional solid solution with silver. Compute the number ofgold atoms per cubic centimeter for a silver-gold alloy that contains 10 wt% Au and 90 wt% Ag. The densities of pure gold and silver are 19.32 and 10.49 g/cm3 , respectively.8.53 In terms of molecular structure, explain why phenol-formaldehyde (Bakelite)will not be an elastomer.10.50 Compute the mass fractions of αferrite and cementite in pearlite. assumingthat pressure is held constant.10.52 (a) What is the distinction between hypoeutectoid and hypereutectoid steels?(b) In a hypoeutectoid steel, both eutectoid and proeutectoid ferrite exist. Explainthe difference between them. What will be the carbon concentration in each?10.56 Consider 1.0 kg of austenite containing 1.15 wt% C, cooled to below 727_C(a) What is the proeutectoid phase?(b) How many kilograms each of total ferrite and cementite form?(c) How many kilograms each of pearlite and the proeutectoid phase form?(d) Schematically sketch and label the resulting microstructure.10.60 The mass fractions of total ferrite and total cementite in an iron–carbon alloyare 0.88 and 0.12, respectively. Is this a hypoeutectoid or hypereutectoid alloy?Why?10.64 Is it possible to have an iron–carbon alloy for which the mass fractions of totalferrite and proeutectoid cementite are 0.846 and 0.049, respectively? Why orwhy not?第四章习题和思考题7.3 A specimen of aluminum having a rectangular cross section 10 mm _ 12.7 mmis pulled in tension with 35,500 N force, producing only elastic deformation. 7.5 A steel bar 100 mm long and having a square cross section 20 mm on an edge ispulled in tension with a load of 89,000 N , and experiences an elongation of 0.10 mm . Assuming that the deformation is entirely elastic, calculate the elasticmodulus of the steel.7.7 For a bronze alloy, the stress at which plastic deformation begins is 275 MPa ,and the modulus of elasticity is 115 Gpa .(a) What is the maximum load that may be applied to a specimen with across-sectional area of 325mm, without plastic deformation?(b) If the original specimen length is 115 mm , what is the maximum length towhich it may be stretched without causing plastic deformation?7.8 A cylindrical rod of copper (E _ 110 GPa, Stress (MPa) ) having a yield strengthof 240Mpa is to be subjected to a load of 6660 N. If the length of the rod is 380 mm, what must be the diameter to allow an elongation of 0.50 mm?7.9 Consider a cylindrical specimen of a steel alloy (Figure 7.33) 10mm in diameterand 75 mm long that is pulled in tension. Determine its elongation when a load of 23,500 N is applied.7.16 A cylindrical specimen of some alloy 8 mm in diameter is stressed elasticallyin tension. A force of 15,700 N produces a reduction in specimen diameter of 5 _ 10_3 mm. Compute Poisson’s ratio for this material if its modulus of elasticity is 140 GPa .7.17 A cylindrical specimen of a hypothetical metal alloy is stressed in compression.If its original and final diameters are 20.000 and 20.025 mm, respectively, and its final length is 74.96 mm, compute its original length if the deformation is totally elastic. The elastic and shear moduli for this alloy are 105 Gpa and 39.7 GPa,respectively.7.19 A brass alloy is known to have a yield strength of 275 MPa, a tensile strength of380 MPa, and an elastic modulus of 103 GPa . A cylindrical specimen of thisalloy 12.7 mm in diameter and 250 mm long is stressed in tension and found to elongate 7.6 mm . On the basis of the information given, is it possible tocompute the magnitude of the load that is necessary to produce this change inlength? If so, calculate the load. If not, explain why.7.20A cylindrical metal specimen 15.0mmin diameter and 150mm long is to besubjected to a tensile stress of 50 Mpa; at this stress level the resulting deformation will be totally elastic.(a)If the elongation must be less than 0.072mm,which of the metals in Tabla7.1are suitable candidates? Why ?(b)If, in addition, the maximum permissible diameter decrease is 2.3×10-3mm,which of the metals in Table 7.1may be used ? Why?7.22 Cite the primary differences between elastic, anelastic, and plastic deformationbehaviors.7.23 diameter of 10.0 mm is to be deformed using a tensile load of 27,500 N. It mustnot experience either plastic deformation or a diameter reduction of more than7.5×10-3 mm. Of the materials listed as follows, which are possible candidates?Justify your choice(s).7.24 A cylindrical rod 380 mm long, having a diameter of 10.0 mm, is to besubjected to a tensile load. If the rod is to experience neither plastic deformationnor an elongation of more than 0.9 mm when the applied load is 24,500 N,which of the four metals or alloys listed below are possible candidates?7.25 Figure 7.33 shows the tensile engineering stress–strain behavior for a steel alloy.(a) What is the modulus of elasticity?(b) What is the proportional limit?(c) What is the yield strength at a strain offset of 0.002?(d) What is the tensile strength?7.27 A load of 44,500 N is applied to a cylindrical specimen of steel (displaying thestress–strain behavior shown in Figure 7.33) that has a cross-sectional diameter of 10 mm .(a) Will the specimen experience elastic or plastic deformation? Why?(b) If the original specimen length is 500 mm), how much will it increase inlength when t his load is applied?7.29 A cylindrical specimen of aluminumhaving a diameter of 12.8 mm and a gaugelength of 50.800 mm is pulled in tension. Usethe load–elongation characteristics tabulatedbelow to complete problems a through f.(a)Plot the data as engineering stressversusengineering strain.(b) Compute the modulus of elasticity.(c) Determine the yield strength at astrainoffset of 0.002.(d) Determine the tensile strength of thisalloy.(e) What is the approximate ductility, in percent elongation?(f ) Compute the modulus of resilience.7.35 (a) Make a schematic plot showing the tensile true stress–strain behavior for atypical metal alloy.(b) Superimpose on this plot a schematic curve for the compressive truestress–strain behavior for the same alloy. Explain any difference between thiscurve and the one in part a.(c) Now superimpose a schematic curve for the compressive engineeringstress–strain behavior for this same alloy, and explain any difference between this curve and the one in part b.7.39 A tensile test is performed on a metal specimen, and it is found that a true plasticstrain of 0.20 is produced when a true stress of 575 MPa is applied; for the same metal, the value of K in Equation 7.19 is 860 MPa. Calculate the true strain that results from the application of a true stress of 600 Mpa.7.40 For some metal alloy, a true stress of 415 MPa produces a plastic true strain of0.475. How much will a specimen of this material elongate when a true stress of325 MPa is applied if the original length is 300 mm ? Assume a value of 0.25 for the strain-hardening exponent n.7.43 Find the toughness (or energy to cause fracture) for a metal that experiences bothelastic and plastic deformation. Assume Equation 7.5 for elastic deformation,that the modulus of elasticity is 172 GPa , and that elastic deformation terminates at a strain of 0.01. For plastic deformation, assume that the relationship between stress and strain is described by Equation 7.19, in which the values for K and n are 6900 Mpa and 0.30, respectively. Furthermore, plastic deformation occurs between strain values of 0.01 and 0.75, at which point fracture occurs.7.47 A steel specimen having a rectangular cross section of dimensions 19 mm×3.2mm (0.75in×0.125in.) has the stress–strain behavior shown in Figure 7.33. If this specimen is subjected to a tensile force of 33,400 N (7,500lbf ), then(a) Determine the elastic and plastic strain values.(b) If its original length is 460 mm (18 in.), what will be its final length after theload in part a is applied and then released?7.50 A three-point bending test was performed on an aluminum oxide specimenhaving a circular cross section of radius 3.5 mm; the specimen fractured at a load of 950 N when the distance between the support points was 50 mm . Another test is to be performed on a specimen of this same material, but one that has a square cross section of 12 mm length on each edge. At what load would you expect this specimen to fracture if the support point separation is 40 mm ?7.51 (a) A three-point transverse bending test is conducted on a cylindrical specimenof aluminum oxide having a reported flexural strength of 390 MPa . If the speci- men radius is 2.5 mm and the support point separation distance is 30 mm ,predict whether or not you would expect the specimen to fracture when a load of 620 N is applied. Justify your prediction.(b) Would you be 100% certain of the prediction in part a? Why or why not?7.57 When citing the ductility as percent elongation for semicrystalline polymers, it isnot necessary to specify the specimen gauge length, as is the case with metals.Why is this so?7.66 Using the data represented in Figure 7.31, specify equations relating tensilestrength and Brinell hardness for brass and nodular cast iron, similar toEquations 7.25a and 7.25b for steels.8.4 For each of edge, screw, and mixed dislocations, cite the relationship between thedirection of the applied shear stress and the direction of dislocation line motion.8.5 (a) Define a slip system.(b) Do all metals have the same slip system? Why or why not?8.7. One slip system for theBCCcrystal structure is _110__111_. In a manner similarto Figure 8.6b sketch a _110_-type plane for the BCC structure, representingatom positions with circles. Now, using arrows, indicate two different _111_ slip directions within this plane.8.15* List four major differences between deformation by twinning and deformationby slip relative to mechanism, conditions of occurrence, and final result.8.18 Describe in your own words the three strengthening mechanisms discussed inthis chapter (i.e., grain size reduction, solid solution strengthening, and strainhardening). Be sure to explain how dislocations are involved in each of thestrengthening techniques.8.19 (a) From the plot of yield strength versus (grain diameter)_1/2 for a 70 Cu–30 Zncartridge brass, Figure 8.15, determine values for the constants _0 and ky inEquation 8.5.(b) Now predict the yield strength of this alloy when the average grain diameteris 1.0 _ 10_3 mm.8.20. The lower yield point for an iron that has an average grain diameter of 5 _ 10_2mm is 135 MPa . At a grain diameter of 8 _ 10_3 mm, the yield point increases to 260MPa. At what grain diameter will the lower yield point be 205 Mpa ?8.24 (a) Show, for a tensile test, thatif there is no change in specimen volume during the deformation process (i.e., A0 l0 _Ad ld).(b) Using the result of part a, compute the percent cold work experienced bynaval brass (the stress–strain behavior of which is shown in Figure 7.12) when a stress of 400 MPa is applied.8.25 Two previously undeformed cylindrical specimens of an alloy are to be strainhardened by reducing their cross-sectional areas (while maintaining their circular cross sections). For one specimen, the initial and deformed radii are 16 mm and11 mm, respectively. The second specimen, with an initial radius of 12 mm, musthave the same deformed hardness as the first specimen; compute the secondspecimen’s radius after deformation.8.26 Two previously undeformed specimens of the same metal are to be plasticallydeformed by reducing their cross-sectional areas. One has a circular cross section, and the other is rectangular is to remain as such. Their original and deformeddimensions are as follows:Which of these specimens will be the hardest after plastic deformation, and why?8.27 A cylindrical specimen of cold-worked copper has a ductility (%EL) of 25%. Ifits coldworked radius is 10 mm (0.40 in.), what was its radius beforedeformation?8.28 (a) What is the approximate ductility (%EL) of a brass that has a yield strengthof 275 MPa ?(b) What is the approximate Brinell hardness of a 1040 steel having a yieldstrength of 690 MPa?8.41 In your own words, describe the mechanisms by which semicrystalline polymers(a) elasticallydeform and (b) plastically deform, and (c) by which elastomerselastically deform.8.42 Briefly explain how each of the following influences the tensile modulus of asemicrystallinepolymer and why:(a) molecular weight;(b) degree of crystallinity;(c) deformation by drawing;(d) annealing of an undeformed material;(e) annealing of a drawn material.8.43* Briefly explain how each of the following influences the tensile or yieldstrength of a semicrystalline polymer and why:(a) molecular weight;。

DOI: 10.1126/science.1226419, (2013);340 Science et al.Valeria Nicolosi Liquid Exfoliation of Layered MaterialsThis copy is for your personal, non-commercial use only.clicking here.colleagues, clients, or customers by , you can order high-quality copies for your If you wish to distribute this article to othershere.following the guidelines can be obtained by Permission to republish or repurpose articles or portions of articles): July 3, 2013 (this information is current as of The following resources related to this article are available online at/content/340/6139/1226419.full.html version of this article at:including high-resolution figures, can be found in the online Updated information and services, /content/340/6139/1226419.full.html#ref-list-1, 8 of which can be accessed free:cites 136 articles This article/cgi/collection/chemistry Chemistrysubject collections:This article appears in the following registered trademark of AAAS.is a Science 2013 by the American Association for the Advancement of Science; all rights reserved. The title Copyright American Association for the Advancement of Science, 1200 New York Avenue NW, Washington, DC 20005. (print ISSN 0036-8075; online ISSN 1095-9203) is published weekly, except the last week in December, by the Science o n J u l y 3, 2013w w w .s c i e n c e m a g .o r g D o w n l o a d e d f r o mThe list of author affi liations is available in the full article online.*Corresponding author. E-mail: colemaj@tcd.ieLiquid Exfoliation of Layered MaterialsValeria Nicolosi, Manish Chhowalla, Mercouri G. Kanatzidis, Michael S. Strano, Jonathan N. Coleman*Background: Since at least 400 C.E., when the Mayans fi rst used layered clays to make dyes, people have been harnessing the properties of layered materials. This gradually developed into scientifi c research, leading to the elucidation of the laminar structure of layered materials, detailed understand-ing of their properties, and eventually experiments to exfoliate or delaminate them into individual, atomically thin nanosheets. This culminated in the discovery of graphene, resulting in a new explosion of interest in two-dimensional materials.Layered materials consist of two-dimensional platelets weakly stacked to form three-dimensional structures. The archetypal example is graphite, which consists of stacked graphene monolayers. How-ever, there are many others: from MoS 2 and layered clays to more exotic examples such as MoO 3, GaTe, and Bi 2Se 3. These materials display a wide range of electronic, optical, mechanical, and electrochemi-cal properties. Over the past decade, a number of methods have been developed to exfoliate layered materials in order to produce monolayer nanosheets. Such exfoliation creates extremely high-aspect-ratio nanosheets with enormous surface area, which are ideal for applications that require surface activity. More importantly, however, the two-dimensional confi nement of electrons upon exfoliation leads to unprecedented optical and electrical properties.Advances: An important advance has been the discovery that layered crystals can be exfoliated in liquids. There are a number of methods to do this that involve oxidation, ion intercalation/exchange, or surface passivation by solvents. However, all result in liquid dispersions containing large quantities of nanosheets. This brings considerable advantages: Liquid exfoliation allows the formation of thin fi lms and composites, is potentially scaleable, and may facilitate processing by using standard technologies such as reel-to-reel manufacturing.Although much work has focused on liquid exfoliation of graphene, such processes have also been demonstrated for a host of other materials, including MoS 2 and other related structures, lay-ered oxides, and clays. The resultant liquid dispersions have been formed into fi lms, hybrids, and composites for a range of applications.Outlook: There is little doubt that the main advances are in the future. Multifunctional composites based on metal and polymer matrices will be developed that will result in enhanced mechanical, electrical, and barrier properties. Applications in energy generation and storage will abound, with layered materials appearing as electrodes or active elements in devices such as displays, solar cells, and batteries. Particularly impor-tant will be the use of MoS 2 for water splitting and metal oxides as hydrogen evolution catalysts. In addition, two-dimensional materials will fi nd important roles in printed electronics as dielectrics, optoelectronic devices, and transistors.To achieve this, much needs to be done. Production rates need to be increased dramatically, the degree of exfoliation improved, and methods to control nanosheet properties devel-oped. The range of layered materials that can be exfoliated must be expanded, even as methods for chemical modifi cation must be developed. Success in these areas will lead to a family of materials that will dominate nanomaterials science in the 21st century.21 JUNE 2013 VOL 340 SCIENCE 1420ARTICLE OUTLINE Why Exfoliate?Large-Scale Exfoliation in Liquids?PioneersRecent Advances in Liquid Exfoliation Potential Applications of Liquid-Exfoliated Nanosheets OutlookADDITIONAL RESOURCESJ. N. Coleman et al ., Two-dimensional nanosheets produced by liquid exfoliation of layered materials. Science 331, 568–571 (2011). doi:10.1126/science.1194975K. Varoon et al ., Dispersible exfoliated zeolite nanosheets and their application as a selective membrane. Science 334, 72–75 (2011). doi:10.1126/science.1208891READ THE FULL ARTICLE ONLINE /10.1126/science.1226419Liquid exfoliation of layered crystals allows the production of suspensions of two-dimensional nanosheets, which can be formed into a range of structures. (A ) MoS 2 powder. (B ) WS 2 dispersed in surfactant solution. (C ) An e xfoliate d MoS 2 nanosheet. (D ) A hybrid material consisting of WS 2 nanosheets embedded in a network of carbon nanotubes.Published by AAASo n J u l y 3, 2013w w w .s c i e n c e m a g .o r g D o w n l o a d e d f r o mLiquid Exfoliation of Layered Materials Valeria Nicolosi,1,2Manish Chhowalla,3Mercouri G.Kanatzidis,4Michael S.Strano,5Jonathan N.Coleman1*Not all crystals form atomic bonds in three yered crystals,for instance,are those that form strong chemical bonds in-plane but display weak out-of-plane bonding.This allows them to be exfoliated into so-called nanosheets,which can be micrometers wide but less than a nanometer thick.Such exfoliation leads to materials with extraordinary values of crystal surface area,in excess of1000square meters per gram.This can result in dramatically enhanced surface activity,leading to important applications,such as electrodes in supercapacitors or batteries. Another result of exfoliation is quantum confinement of electrons in two dimensions,transforming the electron band structure to yield new types of electronic and magnetic materials.Exfoliated materials also have a range of applications in composites as molecularly thin barriers or as reinforcing or conductive fillers.Here,we review exfoliation—especially in the liquid phase—as a transformative process in material science,yielding new and exotic materials,which are radically different from their bulk,layered counterparts.I n1824,Thomas H.W ebb heated a mineral sim-ilar to mica and,by means of thermal ex-foliation,transformed it into what is today a valuable commodity,with applications as an ion exchange resin,an insulating material,and a structural binder in cement.He named the mineral“vermiculite”for its wormlike appear-ance upon exfoliation(Fig.1),from the Latin vermiculare meaning“to breed worms.”Almost 200years later,in2004,Geim and Novosolov showed that thin transparent adhesive tape could be used to exfoliate graphite into single atomic layers of graphene and demonstrated atomical-ly thin devices(1).As a process,exfoliation of layered solids has had a transformative effect on materials science and technology by opening up properties found in the two-dimensional(2D) exfoliated forms,not necessarily seen in their bulk counterparts.Layered materials are defined as solids with strong in-plane chemical bonds but weak out-of-plane,van der Waals bonds.Such materials can be sheared parallel or expanded normal to the in-plane direction.In the extreme limit,these processes yield nanometer-thin—even atomically thin—sheets that are not at all characteristic of the bulk precursor.This production of extremely thin sheets from layered precursors is known as exfoliation or delamination,although in this work we will use the former term.The sheets produced are generally referred to as nanosheets, where“nano”refers to the magnitude of the thick-ness.Although in the ideal case such nanosheets consist of single monolayers,they are often man-ifested as incompletely exfoliated flakes compris-ing a small number(<10)of stacked monolayers.There are many types of layered materials,whichcan be grouped into diverse families(Fig.1).The simplest are the atomically thin,hexagonalsheets of graphene(1–3)and hexagonal boronnitride(h-BN)(4).Transition metal dichalco-genides(TMDs)(such as MoS2and WSe2)(5,6)and metal halides(such as PbI2and MgBr2)(7)have near-identical structures and consist ofa plane of metal atoms sandwiched betweenplanes of halide/chalcogen yered me-tal oxides(such as MnO2,MoO3,and LaNb2O7)(8–11)and layered double hydroxides(LDHs)[such as Mg6Al2(OH)16](8,12)represent a di-verse class of materials with a large variety ofstructures.Similarly,layered silicates,or clays,are minerals and exist as many different types,with well-known examples being montmorilloniteor the micas(13,14).Generally,oxides,LDH,and clay nanosheets are charged and are accom-panied by charge-balancing ions(8,14).Otherinteresting families are the layered III-VIs(suchas InSe and GaS)(15),the layered V-VIs(such asBi2Te3and Sb2Se3)(16),the metal trichalcogenides,and metal trihalides.Although many other lay-ered materials exist(Table1),all share a planar,anisotropic bonding and therefore the potentialto be exfoliated into nanosheets.One substantial advantage of layered materi-als is their diversity.Even before exfoliation,themany families of layered materials display a verybroad spectrum of properties.For example,TMDs(5,6)occur as more than40different types de-pending on the combination of chalcogen(S,Se,or Te)and transition metal(5,6).Depending onthe coordination and oxidation state of the metalatoms,or doping of the lattice,TMDs can be me-tallic,semimetallic,or semiconducting(6).In ad-dition,these materials display interesting electronicbehavior,such as superconductivity or charge-density wave effects(6).Similarly,the many dif-ferent types of layered metal oxides have interestingelectronic,electrochemical,and photonic proper-ties(8).These materials have been fabricated intotransistors,battery electrodes,and magneto-opticdevices(8–10).Thus,even as bulk crystals,lay-ered materials are an interesting and potentiallyuseful material class.This makes them an excitingstarting material for exfoliation into nanosheets.As we will see below,exfoliation dramaticallyenhances the range of properties displayed by analready diverse material type.Why Exfoliate?The simplest effect of exfoliation is to dramati-cally increase the accessible surface area of amaterial.For surface-active or catalytic materials,this can radically enhance their chemical andphysical reactivity.The ion exchange ability ofminerals such as vermiculite to purify water at1000meq/kg depends on its near106-fold increasein surface area after expansion(13).In structuralmechanics,the strength and stiffness of compositesincrease as the thickness of planar fillers,such asclay or graphite,decreases(17).When heat causesexfoliation,a layered material can be used as anintumescent(or thermally expansive)material.Hence,vermiculite and graphite are used for fireretardation in paints and firestop pillows be-cause they reduce their density upon heating andproduce an ash of low thermal conductivity.As interest in nanotechnology has intensi-fied in recent decades,another important advan-tage of exfoliation has emerged.In a layeredcrystal,the electronic wave function extendsin three dimensions.However,after exfoliationelectrons are constrained to adopt a2D wavefunction,thus modifying the electronic bandstructure.Graphite can be transformed into agraphene monolayer after exfoliation,with elec-tronic properties that differ greatly from any othermaterial(1).These include an enormously highcarrier mobility and other exciting properties,such as Klein tunnelling and the half-integer quan-tum Hall effect(1,3).Likewise,the propertiesof MoS2depend strongly on exfoliation state.The bandgap of MoS2changes on exfoliationfrom1.3eV for the bulk crystal to1.9eV foran exfoliated nanosheet.Because the bandgapchanges monotonically with number of mono-layers per nanosheet,this allows the electronicresponse to be chosen at will(18).In addition,although multilayer MoS2is not photolumines-cent,exfoliation-induced changes in its electron-ic structure lead to photoluminescent behaviorin exfoliated monolayers(19).Similar behavioris expected in other layered semiconductors(5).Large-Scale Exfoliation in Liquids?The exfoliation of graphite demonstrated by Geimand Novolosov was achieved essentially by rubbinggraphite on a surface(1).Such mechanical ex-foliation remains the source of the highest-qualitygraphene samples available and has resultedin some major advances(1).However,it suffersfrom low yield and a production rate that is not1School of Physics and Centre for Research on Adaptive Nano-structures and Nanodevices,Trinity College,Dublin,D2Dublin,Ireland.2School of Chemistry,Trinity College,Dublin,D2Dublin,Ireland.3Materials Science and Engineering,Rutgers University,Piscataway,NJ08901,USA.4Department of Chemistry,North-western University,Evanston,IL60208,USA.5Department ofChemical Engineering,Massachusetts Institute of Technology,Cambridge,MA02139,USA.*Corresponding author.E-mail:colemaj@tcd.ie SCIENCE VOL34021JUNE20131226419-1o n J u l y 3 , 2 0 1 3 w w w . s c i e n c e m a g . o r g D o w n l o a d e d f r o mtechnologically scalable in its current form.One possible solution is the exfoliation of lay-ered compounds in liquids to give large quan-tities of dispersed nanosheets.This should allow for methods to obtain sizable quantities of 2D materials that can be processed by using exist-ing industrial techniques,such as reel-to-reel manufacturing.Here,we briefly outline the four main liquid exfoliation techniques for layered materials (schematics are provided in Fig.2,and examples of exfoliated nanosheets are pro-vided in Fig.3).One of the oldest methods of exfoliating lay-ered crystals with low reductive potential is oxi-dation and subsequent dispersion into suitable solvents.The best example is that of graphite (20),in which treatment with oxidizers such as sulphuric acid and potassium permanganate results in ad-dition of hydroxyl and epoxide groups to the basal plane.The resulting hydrophillicity allows water intercalation and large-scale exfoliation to yield graphene oxide upon ultrasonication.The dispersed flakes are predominantly monolayers,typically hundreds of nanometers across,and sta-bilized against reaggregation by a negative sur-face charge at concentrations of up to 1mg/ml.Dispersed graphene oxide can be chemically re-duced in the liquid phase but will then aggregate unless surfactant or polymer stabilizers are present.Although reduction removes most of the oxides,structural defects remain,rendering the properties of oxidatively produced graphene substantially different from pristine graphene.Layered materials can also strongly adsorb guest molecules into the spacing between lay-ers,creating what are called inclusion complexes.This forms the basis of another exfoliation meth-od that is widely applied to layered materials,including graphite (21)and TMDs (22,23).Intercalation,often of ionic species,increases the layer spacing,weakening the interlayer ad-hesion and reducing the energy barrier to exfolia-tion.Intercalants such as n -butyllithium (22,23)or IBr (21)can transfer charge to the layers,re-sulting in a further reduction of interlayer bind-ing.Subsequent treatment such as thermal shock (21)or ultrasonication (22,23)in a liquid com-pletes the exfoliation process.The exfoliated nanosheets can be stabilized electrostatically by a surface charge (23)or by surfactant addition (21).In the case of MoS 2,this method tends to give highly exfoliated nanosheets (22).However,ion intercalation –based methods have drawbacks associated with their sensitivity to ambient condi-tions (22–24).Ion exchange methods take advantage of the fact that LDHs,clays,and some metal oxidesFig.1.Crystalstructures,natural-ly occurring forms,and exfoliated products for four example layered materials.(A )Graphite consists of alternating stacks of hexagonally ar-ranged carbon atoms (black spheres),(B )is a naturally occurring mineral,and (C )exfoliates to single atomic layers of carbon called graphene.(D )Vermicu-lite is a layered silicate hydrate (typ-ically Mg 1.8Fe 0.9Al 4.3SiO 10(OH)2•4(H 2O)that (E )is found naturally as a min-eral and (F )can be exfoliated,for example,upon heating.Silicon atoms are in blue,oxygen atoms are in red,Al/Mg/Fe atoms are in yellow,and interlayer counterions are in black (H and H 2O not shown).(G )MoS 2is a layered arrangement of S and Mo atoms (chalcogen atoms are in yellow,and transition metal are in green)that (H )is found naturally as the mineral molybdenite and (I )can be exfoliated to MoS 2monolayers.(J )Layered manganese dioxide (man-ganese atoms are in yellow,oxygen is in red,and interlayer counterions are in black)occurs naturally (K )as birnessite and (L )can be exfoliated to give MnO 2nanosheets.(C),(I),and (L)are adapted from (48),(87),and (58),respectively.The layer spac-ings for each material are graphite,0.35nm;vermiculite,1.5nm;MoS 2,0.6nm;and MnO 2,0.45nm.21JUNE 2013VOL 340SCIENCE1226419-2REVIEWo n J u l y 3, 2013w w w .s c i e n c e m a g .o r g D o w n l o a d e d f r o mTable 1.Referenced table of families of layered compounds,including structures and information on exfoliation methods,potential applications,and availablility.This table is not exhaustive.Crystal structures were obtained from the CrystalMaker Library (/library/index.html).Graphi t eTop viewSide viewF amily of layered compound StructureExfoliation methodApplicationsCommercial availabilitySonication in surfactantsolution (30, 50–53)Sonication in solvents (27, 45–48)Sonication in polymersolutions (54, 55)Graphene oxide (20,88)Many (1, 89)Widely availableMoS 2top viewMoS 2 side viewSingle-layer transistor (92)Batteries (63, 64)Top-gatephototransistors (93)Thermo-electrics (29, 58)Superconducting composites (94)Raw materials mostly available (purity issues)h-BNTop viewNitrogen BoronSide viewSonication in surfactant solution (58)Sonication in solvents (29, 56)Sonication in polymer solutions (54)Sonication in surfactant solution (58)Sonication in solvents (29, 59, 60)Sonication in polymer solutions (54)Ion intercalation (91)Composites (57)Device substrates (90)YesTransition metal ChalcogenTransition metal dichalcogenides (TMDs) SCIENCE VOL 34021JUNE 20131226419-3REVIEWo n J u l y 3, 2013w w w .s c i e n c e m a g .o r g D o w n l o a d e d f r o mTiTe 3top viewMnPS 3top viewTiTe 3side viewMnPS 3 side viewBatteries (96)No, only by synthesisIon intercalation (95)Wide band-gap semiconductors (97)Magnetic properties (98)No, only by synthesisIntercalation (75)Transition metal ChalcogenTransition metal Chalcogen PhosphorusTransition metal trichalcogenides (TMDs)AMo 3X 3, NbX 3, TiX 3, and TaX 3 (X = S, Se, or Te)Metal phosphorous trichalcogenides (MPX 3), such as MnPS 3, CdPS 3, NiPS 3, ZnPS 3, and Mn 0.5Fe 0.5PS 321JUNE 2013VOL 340SCIENCE 1226419-4REVIEWo n J u l y 3, 2013w w w .s c i e n c e m a g .o r g D o w n l o a d e d f r o mtop viewCrCl PbCl top viewMoCl CrCl 3 side viewPbCl 4 side viewNo (synthesis required)No (synthesis required)No (synthesis required)Ion intercalation (7)Polymer intercalation(99)Ion intercalation (100)Polymer intercalation (99)Ion intercalation (101)Transition metal HalideTransition metal Heavy metal Halide AmmoniumMetal halidesTransition-metal dihalides*Metal MX 3 halides, such as αRuCl 3, CrCl 3, and BiI 3†Layer-type halides with composition MX 4, MX 5, MX 6‡MoCl 2 top 2 side view3top viewHalide SCIENCE VOL 34021JUNE 20131226419-5REVIEWo n J u l y 3, 2013w w w .s c i e n c e m a g .o r g D o w n l o a d e d f r o mNa x (Mn 4+,Mn 3+)2O 4(birnessite) top viewNa x (Mn 4+,Mn 3+)2O 4 (birnessite) side viewVanadium oxide (V 2O 5) top viewV 2O 5 side viewSome raw materials available (purity issues)Most compounds are not availableSupercapacitors (106)Batteries (107)Catalysts (108)Dielectrics (109)Ferroelectrics (109)Ti oxidesIon intercalation (102)Mn oxidesSonication in surfactantsolution (58); Ion intercalation (103)Nb oxidesIon intercalation (104)Va oxidesPolymer intercalation (105)Transition metal Oxygen CatonVanadium OxygenOxidesTransition metal oxides : Ti oxides, Ti 0.91O 2, Ti 0.87O 2, Ti 3O 7, Ti 4O 9, Ti 5O 11; Nb oxides,Nb 3O 8, Nb 6O 17, HNb 3O 8;§ Mn oxides, MnO 2, Ti 3O 7, Na x (Mn 4+,Mn 3+)2O 421JUNE 2013VOL 340SCIENCE 1226419-6REVIEWo n J u l y 3, 2013w w w .s c i e n c e m a g .o r g D o w n l o a d e d f r o mMoO 3top viewMoO 3 side viewSr 2RuO 4 top viewSr 2RuO 4 side viewYesNo (only by synthesis)No (only by synthesis)Ion intercalation (110)Polymer intercalation(111)Intercalation with liquid crystals (115)Ion intercalation (protonation and ion exchange) (116)Amine surfactant (TBA +) under sonication (117)Ion intercalation (protonation and ion exchange) (114)Electrochromics (112)Light emitting diodes (113)Ferroelectrics (118)Photochromic (119)Photoluminescent (120)Layered trirutile phases HMWO (M = Nb, Ta), such as (HNbWO 6 and HTaWO 6)Hydrogen OxygenTransition metal StrontiumRutheniumOxygenOxidesTrioxides, such as MoO 3, TaO 3, and hydrated WO 3Perovskites and niobates, such as Sr 2RuO 4KCa 2Nb 3O 10, H 2W 2O 7, LaNb 2O 7, La 0.90Eu 0.05Nb 2O 7, Eu 0.56Ta 2O 7, Sr 2RuO 4, Sr 3Ru 2O 7, SrTa 2O 7, Bi 2SrTa 2O 9, Ca 2Nb 3O 10, Sr 2Nb 3O 10, NaCaTa 3O 10, CaLaNb 2TiO 10, La 2Ti 2NbO 10, and Ba 5Ta 4O15 SCIENCE VOL 34021JUNE 20131226419-7REVIEWo n J u l y 3, 2013w w w .s c i e n c e m a g .o r g D o w n l o a d e d f r o mTi 2Sb 2O top viewTi 2Sb 2O side viewFeOCl top viewFeOCl side viewNo (only by synthesis)No (only by synthesis)To our knowledge, these have never been exfoliatedIon intercalation (122)Superconductivity (121)Magnetic properties (121)Catalyst (redox properties) (121)Batteries (121)Batteries (123)Transition metal Pnictide CationTransition metal Halide OxygenOxidesOxychalcogenides and oxypnictides: Oxychalcogenides, LaOCuCh(Ch, chalcogenide) and derivatives, Sr 2MO 2Cu 2-δS 2 (M = Mn, Co, Ni),Sr 2MnO 2Cu 2m -0.5S m +1 (m = 1-3), Sr 4Mn 3O 7.5Cu 2Ch 2 (Ch=S, Se); oxypnictides, LaOFeAs II Oxyhalides of transition metals, such as VOCl, CrOCl, FeOCl, NbO 2F, WO 2Cl 2, and FeMoO 4Cl 21JUNE 2013VOL 340SCIENCE 1226419-8REVIEWo n J u l y 3, 2013w w w .s c i e n c e m a g .o r g D o w n l o a d e d f r o mLayered ␣ and ␥ zirconiumphosphates and phosphonatesGaSe top viewGaSe side viewCaHPO 4 top viewCaHPO 4 side viewSome available (maybe purity issues), mostly synthesizedNo (only by synthesis)Ion intercalation (124)Surfactant (125)Nonlinear optical properties, poor thermal conductivity (126)Intercalation (127)Exfoliation in water/acetone mixtures (128)Drug delivery (127)Semiconductor for dye sensitized solar cells (129)S/Se/Te Ga/InOxygen Phosphorus Cation WaterIII–VI layered semiconductorGaX (X = S, Se, Te); InX (X = S, Se Te)¶α-M IV phosphates, α-M IV (O 3P–OH)2·H 2O; and α-Metal IV phosphonates, M IV (O 3P–R)2·n H 2O# o n J u l y 3, 2013w w w .s c i e n c e m a g .o r g D o w n l o a d e d f r o mVermiculite top viewVermiculiteside viewKaolite top viewKaolite side viewNatural minerals,available on themarket Dispersion in water(13)Intercalation (130)Polymer intercalation(131)Catalysis (130)Composites (131)Lightweightnanocomposites forstructural applications(132)Clay-dye complexesand photoactivematerials (131)Organoclays as ionicand electronicconductors (compositeswith conductivepolymers) (131)Thermal and barrierpropertiesnanocomposites (133)Aluminum Oxygen Silicon OH-Oxygen Silicon Cation Intercalates/-C/-OHClays(layered silicates)2:1 Layered silicates**:Smectites, M n+x/n.yH2O[Al4-xMgx](Si8)O20(OH)4;talc, [(Mg3)(Si2O5)2(OH)2];vermiculite, [Mg6.Si6Al2/O20(OH)4] [M n+1/n] (Mg6);biotite [(MgFe)3(Si3Al)O10(OH)2]K;phlogopite, [(Mg6)(Si6Al2)O20(OH)4]Ba;fluorphlogopite, [Mg11/4(Si6Al2)O20F4][(M2+)3/2;margarite, [(Al2)(Si2Al2)O10(OH)2]Ca;and muscovite, [(Al2)(Si3Al)O10(OH)2]K1:1 Layered Silicates ††: Kaolite,[Al4Si4O10](OH)8; halloysite,Al4Si4O10(OH)8.4H2OonJuly3,213www.sciencemag.orgDownloadedfromBrucite (Mg 2+x , Mg 3+x (OH)2 (A n–)x/n . yH 2O)top viewBrucite (Mg 2+x , Mg 3+x (OH)2 (A n–)x/n . yH 2O) side viewTi3C2No, only bysynthesisYesIntercalation (134)Surfactant-assistedexfoliation and intercalation ofmolecules (135)Surfactant exfoliation (136)Solvent exfoliation in DMF (137)Functionalizationfollowed by exfoliation in solvents (138)Biocompatible–bio-hybrids/drug delivery (12, 134)Bionanocomposites with functional and structural properties (139)Extra oxygen and hydrogen at layers surface are present as a consequence of the exfoliation treatment with HF (66, 67).Batteries andsupercapacitors (140)Carbon TitaniumOxygenHydrogenCationLayered double hydroxides (LDHs)Ternary transition metal carbides and nitridesGeneral formula: M(II)1–x M(III)x (OH)2(A n –)x /n . yH 2O, where M(II) = divalent cation; M(III) = trivalent cation; A = interlayer anion; and n – = charge on interlayer anion‡‡Derivatives from MAX phases, where M = transition metal; A = Al or Si;and X=C or N§§Oxygen Hydrogen*These are iso-structural with TMDs.†These are defect CdI2structure types.‡These are heavy metal halides (perovskite type)structurally similar to transition metal dihalides withorganic ammonium interlayers.§Protons emplaced between 2D of Nb 3O 8–anion nanosheets composed of NbO 6octahedra.‖They contain oxide layers separated by distinct layers,which contain the softer chalcogenide (S,Se,and Te)or pnictide (P,As,Sb,and Bi).¶The building block has a trigonal structure,consisting of a pair of (M 3X 3)rings linked by M –M yers interact through van der Waals forces between the X outermost planes.#R is an organic radical,and n is the number of water molecules that can be intercalated in the interlayer region.**The 2:1notation means that the layers consist of two tetrahedral silicate sheets sandwiching one octahedral sheet.††Layer consists of one tetrahedral silicate sheets and one octahedral sheet.‡‡The structure of LDHs can be described by considering Mg(OH)2,which consists of Mg 2+ions coordinated octahedrally by hydroxyl groups.The octahedral units share edges to form infinite,charge neutral layers.In an LDH,isomorphous replacement of a fraction of the Mg 2+ions with a trivalent cation,such as Al 3+,occurs and generates a positive charge on the layers that necessitates the presence of interlayer,charge-balancing,anions.The remaining free space of the interlayer is occupied by water of crystallization.§§Layered M 2X,M 3X 2,M 4X 3,where M =transition metal and X =C or N,can be obtained after removal of the A layer with hydrofluoric acid (HF).o n J u l y 3, 2013w w w .s c i e n c e m a g .o r g D o w n l o a d e d f r o m。