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1、优秀精品课件文档资料,8/1/2019,2,Nature of the surface 2D crystallography ideal surface 5 Bravais lattices, 10 point groups, 17 space groups Surface Geometry of Several Structures PPF clean surface surface reconstruction-Rhklp x q-A surface relaxation real surface：,Surface Science,SFE: defination, anisotropy,
2、Wulff plot Equilibrium form and growth forms,Surface thermodynamics,Fundament of solid surface,8/1/2019,3,Fundament of vacuum,Thin Solid Films,Chemical vapor deposition,Physical vapor deposition,8/1/2019,4,1. What are dangling bond, surface reconstruction and relaxation?,3. Prove that for cubic latt
3、ices the condition for lying on hkl is that hkl = 0.,4. Compare the values of the PFs of (111) plane of the sc, bcc, fcc and cth crystals.,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)
4、 and (111) surfaces for sc and cth crystals, respectively. (consider nearest neighboring atoms only),2. What is the equilibrium form (EF) of a crystal? How is the EF determined from Wulff-plot?,8/1/2019,5,7. Calculate the g111(erg/cm2) of diamond. It is known that EC-C = 83 kcal/mol, d0 = 1.54 (1 ca
5、l = 4.184 J).,8. q = 2p/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 crystal 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
6、 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 required 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 P
7、Al(0) = 1x10-3 torr, and PCu(0) = 2x10-4 torr, atomic weight MAl = 27, MCu = 63.7).,8/1/2019,6,3. Prove that for cubic lattices the condition for lying on hkl is that hkl = 0.,Plane abc,a,b,c,Unit vector normal to the plane,Definition of the Miller index of the plane,= ( h k l ),8/1/2019,7,1-4、 Show
8、 structures of the (100), (110) and (111) surfaces of cth crystal with respect to C-C bond length d0.,a = (4/3)d0,8/1/2019,8,(12) a = (8/3) d0,-110,(100),(100),8/1/2019,9,d100 = (1/3)d0 d110 = (2/3)d0 d111 = d0/3, d0,(110),8/1/2019,10,(111),8/1/2019,11,sc: d0 = a bcc: d0 = (3/2)a fcc: d0 = (2/2)a ct
9、h: d0 = (3/4)a,Surfaces (100) (110) (111) A0 a2 2a2 (3/2)a2,surface packing fraction:,A = nd02/4,Area occupied by a atom in lattice,sc 1, (/4)a2 1, (/4)a2 1/2, (/8)a2,A,bcc 1, (3/16)a2 2, (3/8)a2 1/2, (3/32)a2,fcc 2, (/4)a2 2, (/4)a2 2, (/4)a2,cth 2, (3/32)a2 4, (3/16)a2 2, (3/32)a2,8/1/2019,12,Surf
10、aces,Atomic area density:,sc,rA,bcc,fcc,cth,100,110,111,0.589,0.830,0.340,8/1/2019,13,010,fcc 1st: D110 = (2/2)a 2nd: D100 = a 3rd: D211 = (6/2)a,cth 1st: D111 = (3/4)a 2nd: D110 = (2/2)a 3rd: D100 = a,-110,111,8/1/2019,14,1-7、Volume contribution of atom in crystal,sc: d0 = a, 1 x v0 = (/6) a3 bcc:
11、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)d03 Volume given by one atom for crystal: V = a3/n,/6 3/8 2/6 3/16,8/1/2019,15,g100 : g110 : g111 = 1 : 2 : 3,g100 : g110 : g111 = 2 : 3/2 : 3/3,8/1/2019,16,= 5261
12、(erg/cm2),g111 = (3/8) E/d02,8/1/2019,17,BA / AB, both are belong to a same CR, therefore BA = iAB, in which i is integral. On the other hand, it is easily found that geometrically BA = AB(1 2cosq) i.e. i = 1 2cosq,8/1/2019,18,cos 63.4 = h / (h2 + k2 + l2 ),(hkl)f,(hkl) = (120) or (102),(hkl)f,8/1/2
13、019,19,XB = 1/(1 + x) XA = x/(1 + x),= x,XAl = 1/1.066 = 0.9377, XCu = 0.0623,WAl = 0.9377 x 27 = 25.32 WCu = 0.0623 x 63.7 = 3.97,Cu (wt%) = 3.97/(25.32 + 3.97) = 13.6%,A = Cu, B = Al: MA = 63.7, PA(0) = 2104 torr; MB = 27, PB(0) = 1103 torr,Solution:,8/1/2019,20,2. The Equilibrium Form from Wulff
14、plot,1、Wulff-plot; 2、inner polygon.,SgiAi = min.,8/1/2019,21,g011 = 1/2(E/2a2) g111 = 1/3(E/2a2) g001 = 1(E/2a2),= oa = ob = oc,= oe = od,Octahedron,8/1/2019,22,8/1/2019,23,Q1. 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 part
15、icle is consisted of 1000 and 1024 gold atoms, respectively.,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.,EXERCI
16、SES,8/1/2019,24,Ns is atomic number of the surface 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 c
17、rystal, it is known that,8/1/2019,25,Crystalline volume when Nb =1000 is that:,Therefore, a total area (A) of the surface of the cubic particle is,The number of surface atoms is then that,8/1/2019,26,Generally, for fcc crystal consisting of Nb atoms,8/1/2019,27,(3/4)c2,(100): c2,c,V = NbV0 = (2/2)d0
18、3)Nb,c = (3V/52)1/3,c = (3/52V)1/3 = (3/50)(2/2)d03)Nb1/3 = 3/10Nb1/3d0,A111= (23)c2 = (96/5Nb)2/3d02/3,A100 = 6c2 = 96/5Nb2/3d02,Ns(100 =A100/A0100= (96/5)2/3Nb2/3,Ns(111)=A111/A0111 = (12/5)2/3Nb2/3,Ns = Ns(100) + Ns(111) = (96/5)2/3 + (12/5)2/3Nb2/3 4.5Nb2/3,(111),Each atom in fcc lattice offers,
19、8/1/2019,28,x= (1/2)a,A111 = 23(a2-3x2),A100 = 6x2,V = (2/3)(a3-3x3),A111 = (3/2)a2,A100 = (3/2)a2,V = (52/24)a3,V = (52/24)a3 = (52/24)(2c)3 = (52/3)c3,8/1/2019,29,(3/4)c2,c2,c,c = (52/3)V1/3,8/1/2019,30,Q2. Suppose melting heat of iron per unit volume is 3 kJ/cm3, and the value of surface tension
20、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,8/1/2019,31,8/1/2019,32,4、光学薄膜的种类、功能、制备及应用领域,7、固体薄膜的热学性质与应用领域,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
21、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,8/1/2019,33,1、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,4、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、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）,“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, 研究项目立项申请报告,8/1/2019,34,（封面）,