思考题solution

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1、Zhiming YU,Dept.of MSE,CSU思考题solution Still waters run deep.流静水深流静水深,人静心深人静心深 Where there is life,there is hope。有生命必有希望。有生命必有希望Zhiming YU,Dept.of MSE,CSU7/8/20232Nature of the surface2D crystallographyideal surface5 Bravais lattices,10 point groups,17 space groupsSurface Geometry of Several Structur

2、es PPFclean surfacesurface reconstruction-Rhklp x q-Asurface relaxationreal surface：Surface ScienceSFE:defination,anisotropy,Wulff plotEquilibrium form and growth formsl Surface thermodynamicsl Fundament of solid surfaceZhiming YU,Dept.of MSE,CSU7/8/20233l Fundament of vacuumThin Solid Filmsl Chemic

3、al vapor depositionl Physical vapor deposition Zhiming YU,Dept.of MSE,CSU7/8/202341.What are dangling bond,surface reconstruction and relaxation?3.Prove that for cubic lattices the condition for lying on hkl is thathkl=0.4.Compare the values of the PFs of(111)plane of the sc,bcc,fcc and cth crystals

4、.5.Find the distances between the first three neighboring atoms in the fcc and cth crystals,respectively.6.Compare relative values of the SFEs of(100),(110)and(111)surfaces for sc and cth crystals,respectively.(consider nearest neighboring atoms only)2.What is the equilibrium form(EF)of a crystal?Ho

5、w is the EF determined from Wulff-plot?Zhiming YU,Dept.of MSE,CSU7/8/202357.Calculate the g111(erg/cm2)of diamond.It is known that EC-C=83 kcal/mol,d0=1.54(1 cal=4.184 J).8.=2/n is used to describe rotation symmetry,where n is called a rotation degrees.Show that the rotation degree n=1,2,3,4,6 for c

6、rystal lattices.9.An fcc film is deposited on the(100)surface of a single-crystal fcc substrate.It is determined that the angle between the 100 in the film and the substrate is 63.4,what are the Miller indices of the plane lying on the film surface?10.Estimate approximate Al-Cu melt composition requ

7、ired to evaporate films containing 2 wt%Cu from a single crucible heated to 1350 K.(assuming gCu=gAl，and it is known that at 1350 K PAl(0)=1x10-3 torr,and PCu(0)=2x10-4 torr,atomic weight MAl=27,MCu=63.7).Zhiming YU,Dept.of MSE,CSU7/8/202363.Prove that for cubic lattices the condition for lying on h

8、kl is that hkl=0.Plane abcabcxyzoUnit vector normal to the plane Definition of the Miller index of the plane=(h k l)Zhiming YU,Dept.of MSE,CSU7/8/20237-1100011100101-4、Show structures of the(100),(110)and(111)surfaces of cth crystal with respect to C-C bond length d0.aa=(4/3)d0-110Zhiming YU,Dept.of

9、 MSE,CSU7/8/20238(12)a=(8/3)d0-110(100)(100)Zhiming YU,Dept.of MSE,CSU7/8/20239(12)a=(8/3)d0-110001-110111d100=(1/3)d0d110=(2/3)d0d111=d0/3,d0(110)aZhiming YU,Dept.of MSE,CSU7/8/202310(12)a=(8/3)d0-110a11-2(111)Zhiming YU,Dept.of MSE,CSU7/8/202311sc:d0=abcc:d0=(3/2)a fcc:d0=(2/2)a cth:d0=(3/4)a Surf

10、aces(100)(110)(111)A0 a22a2 (3/2)a2 surface packing fraction:A=nd02/4Area occupied by a atom in latticesc1,(/4)a21,(/4)a2 1/2,(/8)a2 Abcc1,(3/16)a22,(3/8)a2 1/2,(3/32)a2fcc2,(/4)a22,(/4)a2 2,(/4)a2 cth2,(3/32)a24,(3/16)a2 2,(3/32)a2 Zhiming YU,Dept.of MSE,CSU7/8/202312Surfaces Atomic area density:sc

11、rAbccfcccth1001101110.5890.8300.340Zhiming YU,Dept.of MSE,CSU7/8/202313121110010010afcc 1st:D110=(2/2)a 2nd:D100=a3rd:D211=(6/2)aacth1st:D111=(3/4)a 2nd:D110=(2/2)a3rd:D100=a0,0,01/2,1/2,01,1/2,1/2-110111Zhiming YU,Dept.of MSE,CSU7/8/2023141-7、Volume contribution of atom in crystal sc:d0=a,1 x v0=(/

12、6)a3 bcc:d0=(3/2)a,2 x v0=(3/8)a3 fcc:d0=(2/2)a,4 x v0=(2/6)a3 cth:d0=(3/4)a,8 x v0=(3/16)a3 Volume of atom itself:v0=(/6)d03Volume given by one atom for crystal:V=a3/nh/63/82/63/16Zhiming YU,Dept.of MSE,CSU7/8/202315sc:h+k+l(h2+k2+l2)1/2E2d02ghkl=fcc：2h+k(h2+k2+l2)1/23E2d02ghkl=g100:g110:g111=1:2:3

13、 g100:g110:g111=2:3/2:3/3Zhiming YU,Dept.of MSE,CSU7/8/202316h(h2+k2+l2)1/23E8d02ghkl=5261(erg/cm2)83 x 103 x 4.184 x 107 6.02 x 1023 11.542 x 10-16=(3/8)g111=(3/8)E/d02 Zhiming YU,Dept.of MSE,CSU7/8/202317qBqAABBA/AB,both are belong to a same CR,thereforeBA=iAB,in which i is integral.On the other h

14、and,it is easily found that geometricallyBA=AB(1 2cosq)i.e.i=1 2cosqq()06090120180i-10123n16432Zhiming YU,Dept.of MSE,CSU7/8/202318Substrate Film cos 63.4=h/(h2+k2+l2)(hkl)f100s(100)s100f63.4(hkl)=(120)or(102)(hkl)fZhiming YU,Dept.of MSE,CSU7/8/202319FA cA XA PA(0)MB FB cB XB PB(0)MAXB=1/(1+x)XA=x/(

15、1+x)=x 2 1010-4 27 98 210-4 63.7x=0.0664,XAl=1/1.066=0.9377,XCu=0.0623,WAl=0.9377 x 27=25.32WCu=0.0623 x 63.7=3.97Cu(wt%)=3.97/(25.32+3.97)=13.6%WA/MAWB/MBA=Cu,B=Al:MA=63.7,PA(0)=2104 torr;MB=27,PB(0)=1103 torrSolution:XA WA PB(0)MB XB WB PA(0)MA=XA PA(0)MBXB PB(0)MAZhiming YU,Dept.of MSE,CSU7/8/202

16、3202.The Equilibrium Form from Wulff plot1、Wulff-plot;2、inner polygon.giRi=constant SgiAi=min.oACDBRARDRBRCZhiming YU,Dept.of MSE,CSU7/8/202321001110111oecd010001011oex.2 cth crystalghkl=h(h2+k2+l2)1/2E2a2g011=1/2(E/2a2)g111=1/3(E/2a2)g001=1(E/2a2)=oa=ob=oc=oe =odWulff plotEquilibriumpolyhedron(1-10

17、)110100010001Octahedron010001011110100oecdababcZhiming YU,Dept.of MSE,CSU7/8/202322Zhiming YU,Dept.of MSE,CSU7/8/202323Q1.Ns is atomic number of the surface of a cubic particle(fcc)and Nb the bulk.Try to find the radios of Ns/Nb when the particle is consisted of 1000 and 1024 gold atoms,respectively

18、.Q2.Suppose melting heat of iron per unit volume is 3 kJ/cm3,and the value of surface tension at MP is 1.88 N/m,calculate decrease of its melt point when the diameter of its particle is d=20 nm;Drawing the relation of DT and r.EXERCISESZhiming YU,Dept.of MSE,CSU7/8/202324Ns is atomic number of the s

19、urface and Nb the bulk.Try to find the radios of Ns/Nb when the particles are consisted of 1000 and 1024 gold atoms,respectively.Solution:One finds that the volume occupied by one atom(V0)is:as well as the area occupied by one atom(A0100)For fcc crystal,it is known that Zhiming YU,Dept.of MSE,CSU7/8

20、/202325Crystalline volume when Nb=1000 is that:Therefore,a total area(A)of the surface of the cubic particle isThe number of surface atoms is then that Zhiming YU,Dept.of MSE,CSU7/8/202326Generally,for fcc crystal consisting of Nb atoms Zhiming YU,Dept.of MSE,CSU7/8/202327(3/4)c2(100):c2cV=NbV0=(2/2

21、)d03)Nbc=(3V/52)1/3c=(3/52V)1/3 =(3/50)(2/2)d03)Nb1/3=3/10Nb1/3d0 A111=(23)c2=(96/5Nb)2/3d02/3A100=6c2=96/5Nb2/3d02 Ns(100=A100/A0100=(96/5)2/3Nb2/3Ns(111)=A111/A0111=(12/5)2/3Nb2/3Ns=Ns(100)+Ns(111)=(96/5)2/3+(12/5)2/3Nb2/3 4.5Nb2/3(111)Each atom in fcc lattice offersZhiming YU,Dept.of MSE,CSU7/8/2

22、02328xax=(1/2)aA111=23(a2-3x2)A100=6x2V=(2/3)(a3-3x3)A111=(3/2)a2A100=(3/2)a2V=(52/24)a3V=(52/24)a3=(52/24)(2c)3=(52/3)c3 Zhiming YU,Dept.of MSE,CSU7/8/202329(3/4)c2c2cc=(52/3)V1/3Zhiming YU,Dept.of MSE,CSU7/8/202330Q2.Suppose melting heat of iron per unit volume is 3 kJ/cm3,and the value of surface

23、 tension at MP is 1.88 N/m,calculate decrease of its melt point when the diameter of its particle is d=20 nm.Drawing the relation of DT and r.V/L=3 kJ/cm3 g=1.88 N/M=1.88 J/m2=1.88104 J/cm2 Tm=1808 K Zhiming YU,Dept.of MSE,CSU7/8/202331Zhiming YU,Dept.of MSE,CSU7/8/2023324、光学薄膜的种类、功能、制备及应用领域7、固体薄膜的热

24、学性质与应用领域5、纳米多层膜制备方法、性能及其应用6、超硬膜的种类、制备技术及应用8、薄膜技术在IC工业的应用9、薄膜技术在机械工业的应用10、磁控溅射镀膜的技术历史、现状与发展1、表面科学与技术的现状、发展趋势2、透明导电薄膜的制备技术与应用3、磁性薄膜的应用领域及其制备技术SSTSF课程论文Zhiming YU,Dept.of MSE,CSU7/8/2023331、课题的意义（基本原理、关键技术、应用前景）、课题的意义（基本原理、关键技术、应用前景）3、项目的研究内容、研究目标、拟解决的关键科学问题、项目的研究内容、研究目标、拟解决的关键科学问题 4、项目研究特色与创新之处、项目研究特色与创新之处5、拟采取的研究方案（包括有关方法、技术路线、实验手段、拟采取的研究方案（包括有关方法、技术路线、实验手段、关键技术等说明）及可行性分析。关键技术等说明）及可行性分析。6、年度研究计划及预期研究结果（或成果）、年度研究计划及预期研究结果（或成果）“表面科学与薄膜材料”课程论文撰写提纲 2、国内外研究现状及分析、国内外研究现状及分析 研究项目立项申请报告研究项目立项申请报告 Zhiming YU,Dept.of MSE,CSU7/8/202334（封面）项目名称：申 请 人：学 号：硕士导师：联系方式：申请日期：SSTSF课程论文