A new approach for calculating the electronic structure
九天国际教育GCSE真题June 2013 (v3) QP - Paper 3 CIE Chemistry IGCSE
READ THESE INSTRUCTIONS FIRSTWrite your Centre number, candidate number and name on all the work you hand in.Write in dark blue or black pen.You may use a pencil for any diagrams, graphs or rough working.Do not use staples, paper clips, highlighters, glue or correction fl uid.DO NOT WRITE IN ANY BARCODES.Answer all questions.Electronic calculators may be used.A copy of the Periodic Table is printed on page 12.You may lose marks if you do not show your working or if you do not use appropriate units.At the end of the examination, fasten all your work securely together.The number of marks is given in brackets [ ] at the end of each question or part question.CHEMISTRY0620/33Paper 3 (Extended)May/June 20131 hour 15 minutesCandidates answer on the Question Paper.No Additional Materials are required.UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS International General Certifi cate of Secondary EducationThis document consists of 11 printed pages and 1blank page.[Turn overIB13 06_0620_33/2RP © UCLES 2013*6258911381*For1S ubstances can be classifi ed as:Examiner’sUseelements mixtures compoundsE lements can be divided into:metals non-metals(a)D efi ne each of the following terms.(i)e lement.................................................................................................................................... (2)(ii)c ompound.................................................................................................................................... (2)(iii)m ixture.................................................................................................................................... (1)(b)C lassify each of the following as either an element, compound or mixture.(i)b rass (1)(ii)c arbon dioxide (1)(iii)c opper (1)(c)W hich physical property is used to distinguish between metals and non-metals?I t is possessed by all metals but by only one non-metal. (1)[Total: 9]© UCLES 20130620/33/M/J/13For2O ne of the factors which determine the reaction rate of solids is particle size.Examiner’sUse(a)A mixture of fi nely powdered aluminium and air may explode when ignited.A n explosion is a very fast exothermic reaction. This causes a large and sudden increasein temperature.E xplain each of the following in terms of collisions between reacting particles.(i)W hy is the reaction between fi nely powdered aluminium and air very fast?.................................................................................................................................... (2)(ii)E xplain why for most reactions the rate of reaction decreases with time..................................................................................................................................... (2)(iii)S uggest an explanation why the rate of reaction in an explosion could increase rather than decrease with time......................................................................................................................................................................................................................................................................... (3)(b)(i)G ive another example of a substance other than a metal which, when fi nelypowdered, might explode when ignited in air. (1)(ii)D escribe a simple test-tube reaction which shows the effect of particle size on the rate at which a solid reacts with a solution......................................................................................................................................................................................................................................................................... (3)[Total: 11]© UCLES 2013[Turn over0620/33/M/J/130620/33/M/J/13© UCLES 2013ForExaminer’sUse3I ron from the blast furnace is impure. It contains 5% of impurities, mainly carbon, sulfur, silicon and phosphorus. Almost all of this impure iron is converted into the alloy, mild steel.(a) (i) S tate a use of mild steel. (1)(ii) N ame and give a use of another iron-containing alloy.name ..........................................................................................................................use (2)(b) T he oxides of carbon and sulfur are gases. The oxides of silicon and phosphorus are not.Explain how these impurities are removed from the impure iron when it is converted into mild steel............................................................................................................................................ ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... (5)[Total: 8]4G ermanium is an element in Group IV. The electron distribution of a germanium atom is 2 + 8 + 18 + 4. It has oxidation states of +2 and +4.(a) G ermanium forms a series of saturated hydrides similar to the alkanes.(i) D raw the structural formula of the hydride which contains three germanium atomsper molecule.[1](ii) P redict the general formula of the germanium hydrides. (1)0620/33/M/J/13© UCLES 2013[Turn overFor Examiner’sUse(b) D raw a diagram showing the arrangement of the valency electrons in one molecule ofthe covalent compound germanium(IV ) chloride, GeC l 4.Use o to represent an electron from a chlorine atom.Use x to represent an electron from a germanium atom.[2](c) D escribe the structure of the giant covalent compound germanium(IV ) oxide, GeO 2.I t has a similar structure to that of silicon(IV ) oxide............................................................................................................................................ ........................................................................................................................................... (3)(d) I s the change GeC l 2 to GeC l 4 reduction, oxidation or neither? Give a reason for your choice............................................................................................................................................ (2)[Total: 9]5A ll metal nitrates decompose when heated. A few form a nitrite and oxygen. Most form the metal oxide, oxygen and a brown gas called nitrogen dioxide.(a) (i) N ame a metal whose nitrate decomposes to form the metal nitrite and oxygen. (1)(ii) C omplete the equation for the action of heat on lead(II ) nitrate.......Pb(NO 3)2 → ......... + ......NO 2 + O 2 [2](iii) S uggest why the nitrate of the metal, named in (a)(i), decomposes less readily thanlead(II ) nitrate..................................................................................................................................... (2)0620/33/M/J/13© UCLES 2013For Examiner’sUse(b) A lmost all samples of nitrogen dioxide are an equilibrium mixture of nitrogen dioxide,NO 2, and dinitrogen tetroxide, N 2O 4.forward reactionreverse reactionN 2O 4(g)colourless2NO 2(g)dark brownI n the forward reaction, a bond forms between the two nitrogen dioxide molecules.NO 2 + NO 2 → O 2N – NO 2(i) E xplain the term equilibrium mixture ..................................................................................................................................... .............................................................................................................................. [1] (ii) T he syringe contains a sample of the equilibrium mixture. The plunger was pulledback reducing the pressure.H ow would the colour of the gas inside the syringe change? Give an explanation foryour answer..................................................................................................................................... .................................................................................................................................... .............................................................................................................................. [3] (iii) A sealed tube containing an equilibrium mixture of nitrogen dioxide and dinitrogentetroxide was placed in a beaker of ice cold water. T he colour of the mixture changed from brown to pale yellow.I s the forward reaction exothermic or endothermic? Give an explanation for your choice..................................................................................................................................... (2)(iv) W hat other piece of information given in the equation supports your answer to (iii)?NO 2 + NO 2 → O 2N–NO 2 (1)[Total: 12]0620/33/M/J/13© UCLES 2013[Turn overForExaminer’sUse6S ulfuric acid and malonic acid are both dibasic acids. One mole of a dibasic acid can form two moles of hydrogen ions.H 2SO 4 → 2H + + SO 42–D ibasic acids can form salts of the type Na 2X and CaX.(a) M alonic acid is a white crystalline solid which is soluble in water. It melts at 135 °C.T he structural formula of malonic acid is given below. It forms salts called malonates.CH 2(COOH)2 or HOOC – CH 2 – COOH(i) H ow could you determine if a sample of malonic acid is pure?technique used .......................................................................................................... result if pure (2)(ii) W hat is the molecular formula of malonic acid? (1)(iii) W hen malonic acid is heated there are two products, carbon dioxide and a simplercarboxylic acid. Deduce the name and molecular formula of this acid..................................................................................................................................... .............................................................................................................................. [2] (iv) M alonic acid reacts with ethanol to form a colourless liquid which has a ‘fruity’ smell.I ts structural formula is given below.CC OOO OCH2CH 2CH 3CH 2CH 3W hat type of compound contains the group which is circled? (1)0620/33/M/J/13© UCLES 2013For Examiner’sUse(b) (i) S uggest why a solution of malonic acid, concentration 0.2 mol / dm 3, has a higher pHthan one of sulfuric acid of the same concentration. (1)(ii) D escribe a test, other than measuring pH, which can be carried out on both acidsolutions to con fi rm the explanation given in (b)(i) for the different pH values of the two acids..................................................................................................................................... (2)(c) C omplete the following equations for reactions of these two acids.(i) sodium hydroxide + malonic acid → .................... + ....................[1].................... (ii) CuO + H 2SO 4 → .................... + .................... [2] (iii) Mg + CH 2(COOH)2 → .................... + ....................[2](iv) K 2CO 3 + H 2SO 4 → .................... + .................... + ....................[2][Total: 16]7 A lkanes and alkenes are both series of hydrocarbons.(a) (i) E xplain the term hydrocarbon ..................................................................................................................................... (1)(ii) W hat is the difference between these two series of hydrocarbons?.................................................................................................................................... (2)(b) A lkenes and simpler alkanes are made from long-chain alkanes by cracking.C omplete the following equation for the cracking of the alkane C 20H 42.C 20H 42 → 2C 4H 8 + 2C 2H 4 + ...............[1]0620/33/M/J/13© UCLES 2013[Turn overForExaminer’sUse(c) Alkenes such as butene and ethene are more reactive than alkanes.A lkenes are used in the petrochemical industry to make a range of products, which includes polymers and alcohols.(i) D ibromoethane is used as a pesticide. Complete the equation for its preparationfrom ethene.C C + Br 2 →HHH H[1](ii) T he structural formula of a poly(alkene) is given below.nC CH 3C CH 3HHD educe the structural formula of its monomer.[2](iii) H ow is butanol made from butene, CH 3 – CH 2 – CH = CH 2? Include an equation in your answer..................................................................................................................................... (2)(iv)C racking changes alkanes into alkenes. How could an alkene be converted into analkane? Include an equation in your answer..................................................................................................................................... (2)0620/33/M/J/13© UCLES 2013ForExaminer’sUse(d) 20 cm 3 of a hydrocarbon was burnt in 175 cm 3 of oxygen. After cooling, the volume ofthe remaining gases was 125 cm 3. The addition of aqueous sodium hydroxide removed carbon dioxide leaving 25 cm 3 of unreacted oxygen.(i) v olume of oxygen used = .......... cm 3 [1] (ii) v olume of carbon dioxide formed = .......... cm 3 [1](iii) D educe the formula of the hydrocarbon and the balanced equation for the reaction..................................................................................................................................... .................................................................................................................................... .................................................................................................................................... (2)[Total: 15]11BLANK PAGEPMT12Permission to reproduce items where third-party owned material protected by copyright is included has been sought and cleared where possible. Every reasonable effort has been made by the publisher (UCLES) to trace copyright holders, but if any items requiring clearance have unwittingly been included the publisher will be pleased to make amends at the earliest possible opportunity.University of Cambridge International Examinations is part of the Cambridge Assessment Group. Cambridge Assessment is the brand name of University of Cambridge Local Examinations Syndicate (UCLES), which is itself a department of the University of Cambridge.G ro u p140C eC e r i u m 58141P rP r a s e o d y m i u m 59144N d N e o d y m i u m 60P mP r o m e t h i u m61150S mS a m a r i u m62152E uE u r o p i u m63157G dG a d o l i n i u m64159T bT e r b i u m65162D yD y s p r o s i u m66165H oH o l m i u m67167E rE r b i u m68169T mT h u l i u m69173Y bY t t e r b i u m70175L uL u t e t i u m71232T hT h o r i u m 90P aP r o t a c t i n i u m 91238UU r a n i u m92N pN e p t u n i u m93P uP l u t o n i u m94A mA m e r i c i u m95C mC u r i u m96B kB e r k e l i u m97C fC a l i f o r n i u m98E sE i n s t e i n i u m99F mF e r m i u m100M dM e n d e l e v i u m101N oN o b e l i u m102L rL a w r e n c i u m1031HH y d r o g e n17L iL i t h i u m 323N aS o d i u m 1124M gM a g n e s i u m1240C aC a l c i u m 2045S c S c a n d i u m 2148T i T i t a n i u m2251V V a n a d i u m 2352C r C h r o m i u m 2455M n M a n g a n e s e 2556F e I r o n 2659C o C o b a l t 2759N i N i c k e l 2864C u C o p p e r 2965Z nZ i n c3070G aG a l l i u m3127A lA l u m i n i u m1311B B o r o n 512C C a r b o n614NN i t r o g e n716OO x y g e n819FF l u o r i n e928S iS i l i c o n1431PP h o s p h o r u s1532SS u l f u r1635.5C lC h l o r i n e1740A rA r g o n1820N eN e o n104H eH e l i u m273G eG e r m a n i u m3275A sA r s e n i c3379S eS e l e n i u m3480B rB r o m i n e3584K rK r y p t o n3639KP o t a s s i u m 1988S rS t r o n t i u m 3889Y Y t t r i u m 3991Z r Z i r c o n i u m4093N b N i o b i u m 4196M o M o l y b d e n u m 42T c T e c h n e t i u m 43101R u R u t h e n i u m 44103R h R h o d i u m 45106P d P a l l a d i u m 46108A gS i l v e r47112C dC a d m i u m48115I nI n d i u m49119S nT i n50122S bA n t i m o n y51128T eT e l l u r i u m52127II o d i n e53131X eX e n o n54137B aB a r i u m 56139L a L a n t h a n u m 57*178H fH a f n i u m72181T a T a n t a l u m 73184W T u n g s t e n 74186R e R h e n i u m 75190O s O s m i u m 76192I rI r i d i u m 77195P t P l a t i n u m78197A uG o l d79201H gM e r c u r y80204T lT h a l l i u m81207P bL e a d82209B iB i s m u t h83P oP o l o n i u m84A tA s t a t i n e85R nR a d o n86F rF r a n c i u m 87227A cA c t i n i u m899B eB e r y l l i u m4II I I I II V V V I V I I 085R bR u b i d i u m 37133C sC a e s i u m 55226R a R a d i u m 88T h e v o l u m e o f o n e m o l e o f a n y g a s i s 24d m 3a t r o o m t e m p e r a t u r e a n d p r e s s u r e (r .t .p .).a Xb a = r e l a t i v e a t o m ic m a s sX = a t o m i c s y m b o lb = p r o t o n (a t o m ic ) n u m b e rK e y *58-71 L a n t h a n o i d s e r i e s 90-103 A c t i n o i d s e r i e s D A T A S H E E T T h e P e r i o d i c T a b l e o f t h e E l e m e n t sPMT。
新视野大学英语第五册单词(注释+例句)~
Unit 1 新视野大学英语第五册第一课A Technological Revolution in Education 课文生词讲解( new words study) prevalenta. (fml.)existing commonly, generally, or widely (in some place or at some time ); predominant(正式)(在某地或某时)流行的,盛行的;普遍的The habit of traveling by aircraft is becoming more prevalent each year.坐飞机旅行一年比一年普遍了。
One simple injection can help to protect you right through the cold months when flu is most prevalent.只要打上一针预防针就能帮助你度过流感盛行的寒冷月份。
continuityn. [U]uninterrupted connection or union (through time or space)连续性,连贯性There is no continuity of subject in a dictionary.词典的主题没有连续性。
This paper lacks continuity.这篇论文缺乏连贯性。
inherenta. existing as an essential constituent or characteristic; intrinsic固有的,内在的,天生的The communication skills that belong to each species of animal, including people, are not inherent.包括人在内的动物所具有的交际技能并不是天生的。
汉译英_高级职称
The mistake escaped my notice.
9.只有在显微镜下才能看见红血球。
A microscope is needed (for us) to see red cells.
10.2000年奥运会将在悉尼举行。
Sydneywill host (be the host city for) the 2000 Olympic Games.
After a careful check of the blueprint, they made comments on his design.
6.据说在质量方面这种仪器是优越的。
It is said that this instrument is high in quality.
2.不同国家的社会习俗往往大相径庭。
Social customs are usually different from country to country.
3.我们应该尽我们最大的努力防止这类事故再次发生。
We should do our best to prevent this kind of accident from happening again.
At the end of this century, the world population will be over six billion.
2.去年,这个国家的工农业总产值为两千亿美元。
Last year, the general output value of industry and agriculture of this country was two hundred billion dollars.
K-vec A New Approach for Aligning Parallel Texts
K-vec starts by estimating the lexicon. Consider the example: fisheries → pêches. The K-vec algorithm will discover this fact by noting that the distribution of fisheries in the English text is similar to the distribution of pêches in the French. The concordances for fisheries and pêches are shown in Tables 1 and 2 (at the end of this paper).1
Abstract
Various methods have been proposed for aligning texts in two or more languages such as the Canadian Parliamentary Debates (Hansards). Some of these methods generate a bilingual lexicon as a by-product. We present an alternative alignment strategy which we call K-vec, that starts by estimating the lexicon. For example, it discovers that the English word fisheries is similar to the Frendistribution of fisheries in the English text is similar to the distribution of pêches in the French. K-vec does not depend on sentence boundaries.
Edexcel GCE
Paper Reference (complete below)CentreNo.CandidateNo.Examiner’s use only Team Leader’s use only Question Number Leave Blank 12345Paper Reference(s)6663Edexcel GCEPure Mathematics C1Advanced SubsidiarySpecimen PaperTime: 1 hour 30 minutes 6Materials required for examinationItems included with question papers 78Answer Book (AB16)Mathematical Formulae (Lilac)Graph Paper (ASG2)Nil 9Calculators may NOT be used in this examination.10Total Instructions to Candidates Tour candidate details are printed next to the bar code above. Check that these are correct and sign your name in the signature box above.If your candidate details are incorrect, or missing, then complete ALL the boxes above.When a calculator is used, the answer should be given to an appropriate degree of accuracy.You must write your answer for each question in the space following the question.If you need more space to complete your answer to any question, use additional answer sheets.Information for CandidatesA booklet ‘mathematical Formulae and Statistical Tables’ is provided.Full marks may be obtained for answers to ALL questions.This paper has 10 questions.Advice to Candidates You must ensure that your answers to parts of questions are clearly labelled.You must show sufficient working to make your methods clear to the examiner. Answers without working may gain no credit.Turn overSurname Initial(s)SignatureLeave blank1.Calculate å=+201)25(r r 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ó(5x + 3Öx) d x.blank 2.Find ôõ(4) .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... 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.................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... ....................................................................................................................................................Leave 3.(a)Express Ö80 in the form aÖ5, where a is an integer.blank(1)(b)Express (4 -Ö5)2 in the form b + cÖ5, where b and c are integers.(3) 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.................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... ....................................................................................................................................................Leaveblank 4.The points A and B have coordinates (3, 4) and (7, -6) respectively. The straight line lpasses through A and is perpendicular to AB.Find an equation for l, giving your answer in the form ax + by + c = 0, where a, b andc are integers.(5) .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... 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.................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... ....................................................................................................................................................6.(a)Solve the simultaneous equationsy + 2x = 5,2x2- 3x-y = 16.(6)(b)Hence, or otherwise, find the set of values of x for which2x2- 3x- 16 > 5 - 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.................................................................................................................................................... .................................................................................................................................................... ....................................................................................................................................................Leaveblank 7.Ahmed plans to save £250 in the year 2001, £300 in 2002, £350 in 2003, and so on untilthe year 2020. His planned savings form an arithmetic sequence with commondifference £50.(a)Find the amount he plans to save in the year 2011.(2)(b)Calculate his total planned savings over the 20 year period from 2001 to 2020.(3)Ben also plans to save money over the same 20 year period. He saves £A in the year2001 and his planned yearly savings form an arithmetic sequence with commondifference £60.Given that Ben’s total planned savings over the 20 year period are equal to Ahmed’stotal planned savings over the same period,(c)calculate the value of A.(4) .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... 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相对论性电子的Compton散射
相对论性电子的Compton散射2010年9月第25卷第3期山东师范大学(自然科学版)JournMofShandongNormMUniversity(NaturMScience)Sep.2010V o1.25No.3相对论性电子的Compton散射姜孟瑞梁大光孙和丽(山东师范大学物理与电子科学学院,250014,济南)摘要用单粒子理论研究了相对论性电子的Compton散射.当散射过程不改变光子运动方向时,存在n倍频效应,也存在光子频率及运动方向都不改变的散射过程.指出了一定条件下可以实现激光场对电子的有效加速.关键词Compton散射;自由电子激光;n倍频效应;激光加速中图分类号O442文献标识码Adoi:10.3969/j.i明n.1001—4748.2010.03.011激光的出现,直接推动了非线性光学的发展,从倍频,混频现象到光参量放大和振荡,多光子吸收,自聚焦,受激散射等,都是非线性效应的具体表现.随着人们对大功率高亮度激光的应用研究,发现电子在超强激光场中表现出一系列新特性…,并由此产生了许多新的研究课题和研究方向.如超短波激光的高次谐波,快速点火"惯性约束聚变"【2等.另一方面,随着加速器加速能力及性能的不断提高,相对论性电子束的辐射新特性逐渐被人们发现,其应用研究的热点之一就是自由电子激光研究[3,43.既然高速电子和超强激光已经存在,那么相对论电子束与超强激光的相互作用问题自然成为科学界的重要课题.在以前对真空中自由电子与赝光子相互作用的研究中J,发现电子在真空中能与激光场发生净能量交换,而且当激光场足够强时,电子在激光束上的小角度非弹性散射现象开始出现,其能量的交换机制主要是非弹性Compton散射].但这些研究大多采用数值模拟方法,这就使得每一次实验数据的处理或每一种设想的验证都要从头计算,对实验的预见度受到很大限制.笔者从另一角度对相对论性电子束与强激光的相互作用进行探讨.结果表明,沿激光束入射方向上的散射光存在严格的n倍频效应.当满足一定条件时,还可以实现激光场对电子的有效加速.这一研究方法对低速电子或非强场情况同样适用.1物理模型以下讨论中,我们对电子采用单粒子模型,并考虑其相对论效应,当其相对论因子一1时,即过渡到低速情况.对于激光场,我们不考虑电磁场的量子化,即在经典电动力学范围内处理电子与激光场的相互作用,这就要求激光场的场强E<<mC/萌1.3×10"V/m.当相对论性电子束射人强激光场中时,由于强激光场的光子数密度很大,所以,电子束中每个电子可能与单个光子碰撞,也可能与几个相同状态的光子组成的光子集团碰撞.无论哪种情况,我们统称为电子与n光子集团碰撞,n=1即为单光子.假设电子与n光子集团发生Compton散射前的四动量分别为P'和np(o,散射后电子和光子的四动量分别为和P.若不考虑光场的极化,则根据碰撞系统的四动量守恒P+,=P+附,(1)即可研究这一散射过程,这就是本文采用的相对论性电子Compton散射的基本方程.另外,根据四动量可以得到以下不变量(,')==0,(2)(P')==一mc,(3)式中m为电子的静止质量,c为真空中光速.2散射光子频率公式为了计算散射光子的频率,将(1)式中的P移到等式左边,再将等式两边平方.考虑到(2)和(3)式后,得到如下方程np(o一一,':0,(4)式中三个标积可按四动量的各分量展开.设激光光子散射前,后的频率分别为‰和,波矢量分别为‰和j},以散射前电子的运动方向为轴正向建立坐标系(∑系),则.'P2.'∽一.'∞/c=印∞启kocos0o一荐Wo/C,'=一.'se/c=p町序kcosO.一8∞荐w/c,=∞一∞/c=n序kokcosO-nh0w/c,收稿日期:2009—11—18通讯作者,男,副教授,硕士生导师(5)(6)(7)37第25卷山东师范大学(自然科学版)第3期式中.和,.'分别为电子和n光子集团碰撞前的普通动量,为散射后光子的动量,.和;.'分别为电子和n光子集团的初始能量,为散射光子能量,Oo和01分别为光子散射前,后运动方向与轴的夹角.0=10一Ool为散射角.将(5)~(7)代入(4)式,并注意到=O~o/C和=,解得一青,㈣'此即为多光子Compton散射过程中光子频率的计算公式.若散射前电子和光子状态已知,则可出该式计算各散射方向的散射光子频率.例如,当0:0o(即散射光与入射光同向,散射角0=0)时,由(8)式可得散射光的频率为60=//OJ0,(9)这是严格的n倍频效应,是n光子集团借助于散射过程将能量合并造成的.要在实验中观察到这种沿入射方向的倍频效应可能有困难,但产生这种困难的原因不是因为同等情况下在该方向上散射的概率比其它方向小,而是因为在这个方向上存在很强的背景光.即沿原方向一直运动的光子大多属于没有与电子发生碰撞的情况,这时n光子集团仍保持原状态,每个光子频率仍为O)0.所以我们应注意,运用(8)式计算的前提条件是有电子参与的,真正发生了的碰撞过程.如果光场不是很强,电子每次只能与单个光子碰撞.此时取n=1,(8)式变为∞=,''Ⅲ,由此,我们可以在实际意义上定义"强激光场"的临界值,凡能够发生多光子碰撞的为强激光场;只能发生单光子散射的为非强场激光.我们还可以根据这一临界值,由激光光子的模体积定义电子的大小,这在理论上和实验上都具有重要意义.当0,=0o(即散射角0=O)时,由(10)式可得散射光的频率为∞=nJ.,(11)可见,单光子散射时,在入射方向上的散射光频率与背景光没有差别,这在以前被认为光子与电子没有发生相互作用,而现在我们认为:虽然频率与不发生相互作用时没什么不同,简单测量中也无法区分,但本质上是不同的.光子沿某一特定路径传播时,与电子发生碰撞必定影响其相位或传播时间.如果实验设计上达到一定的精确度,应该能够测出碰撞与不碰撞的区别.3散射光子行为探讨由(8)式可知,散射结果不仅与电子的初始能量有关,还与初始动量有关,所以,碰撞前电子的运动增加了散射结果的复杂性.若碰撞前电子是静止的,则钟=0,8=,由(8)式可得瑚110)0…,『=百丽'¨对于n=1的单光子散射,有∞=o)otit'二=c=万,(13)这就是人们熟知的Compton散射公式.对于(8)式所描述的运动电子的多光子Compton散射,可以先建立一个惯性系∑,令∑系的运动与电子的初始运动情况相同,并使其坐标轴指向电子的运动方向(与轴同向),则在∑系中,散射光子频率为,,l∞0….,∞'J式中各带撇号的量表示∑系中对应物理量在∑系中测量的结果.同理,对于n:I的单光子散射,∑系中应为,.∞0….,∞丽'¨J若再将∑系变换到∑系(实验室坐标系),就可得到散射光子的行为.由于这一变换涉及到频率和散射角的变换,所以需要应用Doppler效应公式/.fl=(1一osO)或03=037(1+#cosO),(14)通过变换可得南1案11cos0,w+凡序∞o(一JBcos)(一)/(眦)','式中,=v/c,是电子的初始速度,y=(1一)是由相对论性电子确定的Lorentz因子.(15)式即为相对论性电子的多光子Compton散射的完整表达式,该式与(8)式等价.对照Compton散射公式可知,(13)式就是(15)式在卢=0(=1),n=1情况下的特例,可见(15)式具有普遍意义.另外,(15)式中含有∑系测量的散射光方向角0及散射角0,当电子速度很大时,与∑系中的对应量差别很大,不能混淆.这种情况下可根据光行差公式tan0=sin0,/1一/(cosO一卢),(16)计算两系中各角度的对应关系,也可以由(14)式联立解得的eosO=(cos0一g)/(1一/~cos0),(17)来计算.(16),(17)二式等价.假如电子的运动情况一定,改变激光的入射方向,即可以在0~1T之间变化.在电子静止坐标系∑看来,入射方向角满足cos0.=(cosSo一卢)/(1一~cosSo).对于极端相对论情形,以下分两种情况讨论.3.1Oo=0的情形当=0时,由cos0.=(COSO0一卢)/(1一os)=1得到0.=0,这时两惯性系都说是"光子追上电子而发生碰撞".在∑系中,按碰撞结果又可分为两种情况:①00,即碰后光子与电子仍同向运动,这就是(9)式所描述的情形,不再赘述.②0.=盯,即光子被反弹,电子被加速而方向不变.将各角度代入(15)式,得38第3期姜孟瑞,等:相对论性电子的Compton散射第25卷no)0(1一卢)rUOo/(4T),1o,一—1+n~too—/(7mc2)'¨其中用到卢一1以及1—3=(1一)/(1+卢)~1/(27).对于相对论性电子而言,7mc>>rdtoo,所以有to~ntoo/(472)<<tOo,(19)散射光子能量很小,n光子集团将大部分能量交给电子,从而电子被有效加速,这在激光加速方面有重要意义.3.2≠O的情形当≠O时,cos0o=(cos0o-3)/(1-3cos0o)=一1,这时无论入射角如何,对于相对论性电子而言,都将得出0.=1T,即"迎面相撞"的结论.在∑系中,按碰撞结果也可分为两种情况:①0.=竹,此时0=0,将各角度代人(15)式,得∞=///.O0(1-3co~0o)/2,(20)若0o=1T,则碰后光子与电子仍同向运动,这也是(9)式所描述的情形.若Oo=,则n,:rUao/2,(21)这一结果是很有意义的,正好是n倍频的一半,可以称其为"半n倍频".②0=O,此时0=1r,将各角度代入(15)式,得72,0(1-~OS0o)(1+)27,(1—3cos0o)一了二丽T=丽'(22)若0o=,则=27n~O0,(23)若…则∞=署=雨472rtto0,考虑到7mc.>>∞0,有∞=47no)0.(24)由(23),(24)可知,在2一叮r之间变化时,散射光子频率将在272no,t0-47?Mr)0取值这就是说,散射光子频率可以是It倍频的27一472倍,我们不妨称其为"超n倍频效应".因此,可以通过相对论性电子与n光子集团的Compton散射过程,使激光频率成几万倍甚至几十万倍地增加.此时,散射光子的能量来源除n光子集团外,绝大部分来自于相对论性电子,这对于射线激光或超短波自由电子激光研究无疑是一个令人振奋的结果.4结语.本文运用单粒子模型研究了相对论性电子的Compton散射,得到了散射公式.通过对散射公式的讨论,至少可以得出以下结论:1)当0.=(即散射光与入射光同向,散射角0=0)时,将发生严格的n倍频效应,即=砌..2)散射角0=0的单光子散射(n=1)也是一种散射过程,虽然散射频率与不发生相互作用时没什么不同,简单测量中也无法区分,但本质上是不同的.3)当电子束与激光束沿相同方向入射时(即0o=0),将发生散射光的n倍频效应或散射电子的加速效应.4)当电子束与激光束入射方向互相垂直(即=Ir/2)时,根据散射角的不同将发生"半n倍频效应"或"超lrt倍频效应".5)由超r/,倍频效应,经过单次Compton散射,光子频率可以达到∞=47nto..本文描述了n光子集团与相对论性电子的Compton散射,但是由于没有考虑电磁场的量子化,无法得到散射光场的强度等其他信息,所以n与激光强度有何关系以及n是否与电子运动速度有关,有待进一步研究.5参考文献[1]KongQing,ZhuLij.n,WangJiaxiang,eta1.Electrondynamicsintheextra—intensestationarylaserfield[J].ActaPhysicaSinica,1999,48(4):93—103.[2]ZhangJiatai,HeBin,HeXiantu,eta1.Studyonthemechanismofthefastignitioninlaserfusion[J].ActaPhysicaSinica,2001,50(5):126—130.[3]JiangMengrui,ZhuJiaqing.Non—longitudinalelectricfieldion—ripplewigglerforfree —electron—laser[J].JOptoelectronicsLaser,2001,12(4):385—387.[4]JiangMengrui,ZhangLei.Smallsig,~lgainofnon—longitudinalelectricfieldiontipplelasers[J].JOptoelectronicsLaser,2003,14(6):653—654.[5]ZhuJiaqing.Energyconversionoffreeelectronlaser[J].ActaPhysicaSinica,1996,45(1): 52—57.[6]WangJx,HoYK,ScheidW,HoraH.NonlinearComptoneffectandelectroninelasticscatte ringbyanintensestationarylaserbeam[J].PhysLeaA,1997,31(3-4):139—143.[7]衡耀付,郝东山.Compton散射对强双折射光纤矢量调制不稳定性的影晌[J].河南大学:自然科学版,2008,38(3):246-250.[8】潘国华,高雁军.推迟势积分中的相对论不变量[J】.广西师范大学:自然科学版,1992,10(4):72—74.(下转第42页)39第25卷山东师范大学(自然科学版)第3期ERRORDISCUSSIONINCALCULATIoNOFTHEDIFFRACTION AMPLITUDEWITHZoNECoNSTRUCTIoNoFFRESNEL WangZhenjiuLliHongshenSunPing(SchoolofPhysicsandElectronics,ShandongNormalUniversity,250014,Jinan,China) AbstractWhencalculatingtheamplitudeofFresneldiffractionwithzoneconstructionofFres nel,wetradi- tionallvdonotconsiderthephasechangeswhichareintroducedbyopticalpathchangesbetwe enareaelementsin'sideazonec0nstmctionofFresnelandthepointatwhichthelightdisturbanceisdetermined.Theamplitudeof Fresneldiffractionisanalyzedwhenthephasechangeisconsideredinthispaper.Theamplitu deexpressionisgivenwhenaD0intsourceisused.Theexpressionisdifferentfromthetraditionalsolution.Thereare errorsbetweenthemwhenthevareusedtosoluteexamples.Thustheeffectofphasechangeshouldbeconsideredw henzoneconstruc'tionofFresnelisusedtoanalyzesomeproblems.Keywordsfresneldiffraction;zoneconstructionofFresne1;phasechange;error(上接第39页)NONLINEARCOMPTONSCA TTERINGBYRELATIVISTICELECTRON JiangMengruiLiangDaguangSunHell(SchoolofPhysicsandElectronics,ShandongNormalUniversity,250014,Jinan,China) AbstractThesingle.particlemodelisusedtostudythenonlinearComptonscatteringbyrelati visticelectron?Then—times—frequencyeffectwillhappenwhilethedirectionofphotondoesnotchangebyscattering.Itisin exist- encethatboththefrequencyandthedirectionofphotondonotchange.Theconceptionsofhalf-n-times—frequencyandofuhra..times-frequencvareputforward.Underthecertaincondition,acceleratingelectr onbeameffectivelybylaserfieldcanberealized.KeywordsComptonscattering;free—electron—laser;n—times—frequencyeffect;laseracceleration42。
小学下册Y卷英语第5单元综合卷(有答案)
小学下册英语第5单元综合卷(有答案)英语试题一、综合题(本题有100小题,每小题1分,共100分.每小题不选、错误,均不给分)1.My dad is a ______. He works at an office.2.The plants are _______ (growing) fast.3.The _______ (The Age of Exploration) led to European colonization of the Americas.4. A ______ is a geographical feature that influences local climates.5._____ (环境) plays a big role in plant health.6.I want to _______ a picture for my mom.7.The ______ is wearing a blue dress.8.The chemical symbol for silver is ________.9. A thermometer measures ______.10.Gardening can offer a sense of community and ______ among participants. (园艺可以在参与者之间提供社区感和团结感。
)11.The wind is _______ (howling) outside.12.We will have a ________ (家庭聚会) soon.13.The nurse gives _____ (疫苗) to children.14.The bison roams freely across the ____.15.The Sun is a medium-sized star in the ______ galaxy.16.The __________ is a region known for its deserts.17.What is the term for a young pelican?A. ChickB. HatchlingC. CalfD. Kit答案:a18.geothermal) energy comes from the Earth's heat. The ____19.The _______ can attract various pollinators.20.The main component of cells is ______.21.The _____ (海豚) is known for its playful behavior.22.My ________ (玩具名称) is made of plastic and metal.23.The sun is ___ in the morning. (rising)24.My favorite thing about school is ________ (学习).25.The _______ of an object can be calculated by measuring its volume.26.The ancient Egyptians wrote on ______ (草纸).27.The chemical formula for calcium hydroxide is __________.28.The _______ (狮子) is king of the jungle.29. A _______ can help to illustrate the concept of gravity.30.What is the primary ingredient in hummus?A. LentilsB. ChickpeasC. PeasD. Beans答案:B31. A __________ (实验观察) can lead to new discoveries in science.32.__________ reactions involve the release of energy.33.I have a dream of traveling to ________.34.The _______ (Hellenistic period) followed the conquests of Alexander the Great.35.ts can reproduce through ______ rather than seeds. (某些植物可以通过扦插而不是种子繁殖。
高三英语询问技术创新单选题50题
高三英语询问技术创新单选题50题1. Many tech companies are investing heavily in ______ to improve data security.A. artificial intelligenceB. blockchainC. virtual realityD. augmented reality答案:B。
解析:本题考查新兴科技词汇的理解。
A选项人工智能主要用于模拟人类智能,如语音识别、图像识别等,与数据安全关联不大。
B选项区块链是一种分布式账本技术,以其安全性和不可篡改的特性被广泛用于数据安全领域,符合题意。
C选项虚拟现实主要是创建虚拟环境,与数据安全不是直接相关。
D选项增强现实是将虚拟信息叠加到现实世界,和数据安全关系不紧密。
2. The ______ technology has made it possible for self - driving cars to navigate complex roads.A. 5GB. cloud computingC. big dataD. Internet of Things答案:A。
解析:5G技术具有低延迟、高带宽等特性,这些特性使得自动驾驶汽车能够在复杂的道路上进行导航,因为它能快速传输数据。
B选项云计算主要是提供计算资源的网络服务,与自动驾驶汽车导航关系不直接。
C选项大数据侧重于数据的收集、存储和分析,不是直接助力自动驾驶导航的关键。
D选项物联网强调设备之间的连接,并非自动驾驶汽车导航的最主要技术支持。
3. Tech startups are exploring the potential of ______ in the field of medical diagnosis.A. quantum computingB. gene editingC. nanotechnologyD. all of the above答案:D。
科技英语考试翻译
考试日期:05 年12 月31 日1.为此,必须对电路(circuit)的基本内容有一个很好的了解。
For this purpose, it is necessary to have a good understanding of the fundamentals of the circuit.2.这个系数(coefficient)有待确定。
This coefficient remains to be determined.3.现有的教科书均没有提这一点。
None of the textbook available mentions this point.4.由于反馈(feedback)在电子线路(electronic circuit)中起着重要作用,所以对它的研究极为重要。
Since feedback plays an important role in the electronic circuit, its study is very important.5.有迹象表明,计算机的价格将进一步下跌。
There is an indication that the price of computers will fall further.6.这个方法的确管用,不过到底该在什么情况下使用它尚不清楚。
This method does work, but it is not clear in what condition it is that it should be used.7.十年前该厂的产量比现在高五倍。
The output of this factory 10 years ago is 6 times what it is now.8.激光(laser)是二十世纪六十年代引入的一项新技术,它能穿透(pierce)特别硬的物质。
A new technology introduced in the 1960s, laser can pierce very hard substances.9.为使晶体管(transistor)正常工作,必须给其电极(electrode)加(apply)上合适的电压(voltage)。
Ray tracing (physics)
Ray tracing (physics)From Wikipedia, the free encyclopediaJump to: navigation, searchThis article is about the use of ray tracing in physics. For computer graphics, see Ray tracing (graphics).In physics, ray tracing is a method for calculating the path of waves or particles through a system with regions of varying propagation velocity, absorption characteristics, and reflecting surfaces. Under these circumstances, wavefronts may bend, change direction, or reflect off surfaces, complicating analysis. Ray tracing solves the problem by repeatedly advancing idealized narrow beams called rays through the medium by discrete amounts. Simple problems can be analyzed by propagating a few rays using simple mathematics. More detailed analyses can be performed by using a computer to propagate many rays.When applied to problems of electromagnetic radiation, ray tracing often relies on approximate solutions to Maxwell's equations that are valid as long as the light waves propagate through and around objects whose dimensions are much greater than the light's wavelength. Ray theory does not describe phenomena such as interference and diffraction, which require wave theory (involving the phase of the wave).Co TeRay is ad Ray as a dis The loc thi a co the adj pro suc witsys ontents∙1 Techni ∙2 Uses o o o o o ∙3 See als ∙4 Refere ∙5 Extern chnique tracing of a b dvanced by a y tracing w a large nu tance, pos ray trac al derivats locatio omplete pa ray may ustments perties o h as inten h as manytem.que 2.1 Radio sig 2.2 Ocean ac 2.3 Optical d 2.4 Seismolo 2.5 Plasma P so nces al linksebeam of light small amoun works by a umber of ve ssibly ver er will a tive of th n, a new r ath is gen be tested to the ra f the ray nsity , wavy rays as gnals coustics esign ogy Physics t passing thro nt, and then t assuming t ery narrow ry small, o dvance the he medium t ray is sen nerated. I d for inte ay's direc may be altvelength ,are neces ough a mediu the direction that the p w beams (r over which e ray over to calcula nt out and If the sim ersection ction if a tered as t or polarissary to u um with chan is re ‐calculat particle o ays ), and h such a ra r this dis ate the ray d the proc ulation in with them a collisio the simula ization . T understand nging refracti ed.or wave ca that ther ay is loca stance, an y's new dir cess is re ncludes so m at each on is foun ation adva he proces d the beha ive index . Th an be mode re exists s lly straig nd then us rection. F epeated un olid objec step, mak nd. Otherances as we s is repea avior of t e ray eled some ght. se aFrom ntil cts, king ell, ated theU R SeeRadibase One tra are invpro medray to h thr The opt concom ion ray ele the sep com(gr Usesadio sign also: Radio p io signals trac e of the 3D gr particul aces radio refracted volves the pagation o dia such a y tracing help deter rough the image at t ical ray nstant ref mplexities nospheric e y trajecto vation an magnetic arately r mponent folreen) comp nalspropagationced from the rid).ar form o signals, d and/or r e integrat of electro s the iono is shown rmine the p ionospher the right tracing w fractive i of a spa electron d ories. Two gles. Whe field spl ray traced llows a paponent.e transmitter of ray tra modeled a eflected b ion of di omagnetic osphere. A to the rig precise be re.illustrat where the ndex , sig tially var densities o sets of n the main lits the s d through thcompletat the left to acing is r as rays, th back to th ifferentia waves thr An example ght. Radio ehavior of esthe com medium be gnal ray t rying refr influence signals a n signal p signal int the ionos telyindep o the receive radio sign hrough the e Earth. T al equatio rough disp e of physi o communic f radio sig mplexity of etween obj tracing mu ractive in e the refra are broadc penetrates to two com sphere. Th pendent of r at the right nal ray tr e ionosphe his form o ons that d ersive and ics-based cators use gnals as th f the situa jects typi ust deal w ndex, wher active ind cast at tw s into the ponent wav he ordinar the extrao t (triangles on racing, wh ere where t of ray trac describe t d anisotro radio sig e ray trac hey propag ation. Unl ically has with the re changes dex and hen wo differe e ionosphe ves which ry wave (r ordinary w n the hich they cing the opic gnal cing gate like s a s in nce, ent ere, are red) waveOc See Sou and Thi tenof s eff the int acoA ra path O See Ray as app is uor systra ean acou also: Underw und veloci d temperatu s local mi nds to bend sound thro fects of th ocean sur ensity ma ustics , un y tracing of a h can be seen Optical de also: Optical y tracing m in camera lication used to de optical i tem to be acer in a ∙Dispersio ∙Polarizat o o ∙ Laserlig usticswater acoustic ty in the ure , reach nimum, cal d towards ough the o he SOFAR c rface and b ay be comp nderwater acoustic wav n to oscillate a esignlens designmay be use as , micros in this fi escribe th nstrument modeled. straightf on leads to ch tion Crystal optics Fresnel equa ht effectscsocean var hing a loca lled the SO it. Ray t ocean up to hannel, a bottom. Fr puted, whi acoustic vefronts prop about the SO d in the d copes , te ield dates he propaga , allowin The follo forward fa hromatic abe s ationsries with al minimum OFAR chann tracing ma o very lar s well as rom this, ich are us communica pagating thro OFAR channel.design of l elescopes ,s back to t ation of l ng the ima wing effe ashion:errationdepth due m near a dep nel , acts a ay be used rge distan reflectio locations seful in t ation , and ugh the vary .lenses and and bino the 1900s.light rays age-formincts can be to change pth of 800as a wavegu to calcul ces, incor ons and ref of high a the fields acousticying density o d optical s oculars , a Geometri through a ng properte integrat es in dens –1000 met uide , as so late the p rporating fractions and low sig s of ocean thermomet of the ocean systems , s and itsic ray trac a lens sys ties of th ted into a sity ters. ound path the off gnal n try .. The such cing stem he rayThin film interference (optical coating, soap bubble) can be used to calculate the reflectivity of a surface.For the application of lens design, two special cases of wave interference are important to account for. In a focal point, rays from a point light source meet again and may constructively or destructively interfere with each other. Within a very small region near this point, incoming light may be approximated by plane waves which inherit their direction from the rays. The optical path length from the light source is used to compute the phase. The derivative of the position of the ray in the focal region on the source position is used to obtain the width of the ray, and from that the amplitude of the plane wave. The result is the point spread function, whose Fourier transform is the optical transfer function. From this, the Strehl ratio can also be calculated.The other special case to consider is that of the interference of wavefronts, which, as stated before, are approximated as planes. When the rays come close together or even cross, however, the wavefront approximation breaks down. Interference of spherical waves is usually not combined with ray tracing, thus diffraction at an aperture cannot be calculated.These techniques are used to optimize the design of the instrument by minimizing aberrations, for photography, and for longer wavelength applications such as designing microwave or even radio systems, and for shorter wavelengths, such as ultraviolet and X-ray optics.Before the advent of the computer, ray tracing calculations were performed by hand using trigonometry and logarithmic tables. The optical formulas of many classic photographic lenses were optimized by roomfuls of people, each of whom handled a small part of the large calculation. Now they are worked out in optical design software such as Code-V, Zemax, OSLO or TracePro from Lambda Research. A simple version of ray tracing known as ray transfer matrix analysis is often used in the design of optical resonators used in lasers. The basic principles of the mostly used algorithm could be found in Spencer and Murty's fundamental paper: "General ray tracing Procedure".[1]SeThis com In s and vel ben geo ear In p rig Pl Ene the wav sol pro Stu fouSe eismolog ray tracing o mplicated, and seismology d tomograph ocity var nd and ref physical rthquake, particula ght) allowe lasma Ph rgy transp wave heat ves through utions of pagation o udies of w und in [5].ee also∙Ocean ac ∙Ray tran ∙Gradient ∙ Raytraci gyofseismic wa d reveals tellin y , geophys hic recons ies within flect. Ray model, fo or deduci r, the dis ed scienti hysicsport and t ting of pl h a spatia f Maxwell’of waves i ave propa coustic tomo sfer matrix a t index optics ing(graphics)ves through ng informatio sicists use struction n and bene y tracing ollowing t ng the pr scovery of ists to de the propag lasmas. Po ally nonun s equati n the plas gation in graphy nalysis s )the interior o on about the e ray trac of the Ea eath Earth'may be us them back roperties f the seis educe the p gation of w ower-flow niform pla ions. Anot sma medium plasmas u of the Earth s structure of ing to aid arth's int 's crust , sed to com to their of the in smic shado presence o waves play trajector sma can be ther way o m is by usi using ray shows that p our planetin earthqu erior .[2][3] causing t mpute path source, s tervening w zone (il of Earth's ys an impo ies of ele e computed of computi ing Ray tra tracing m paths can be uake locat Seismic w hese wave hs through such as an g material llustrated molten co ortant rol ectromagne d using dir ing theacing meth method can quite tion wave s to h a n [4]. d at ore. e in etic rect hod. n beReferences1.^ G. H. Spencer and M. V. R.K. Murty (1962). "General ray tracing Procedure" (PDF). J.Opt. Soc. Am. 52 (6): 672–678. doi:10.1364/JOSA.52.000672./archive/nasa//19620005046_1962005046.pdf.2.^ Rawlinson, N., Hauser, J. and Sambridge, M., 2007. Seismic ray tracing and wavefronttracking in laterally heterogeneous media. Advances in Geophysics, 49. 203‐267.3.^ Cerveny, V. (2001). Seismic Ray Theory. Cambridge University Press.4.^ Purdue University5.^ Bhaskar Chaudhury and Shashank Chaturvedi (2006). "Comparison of wavepropagation studies in plasmas using three‐dimensional finite‐difference time‐domain and ray‐tracing methods". Physics of Plasmas 13: 123302. doi:10.1063/1.2397582./resource/1/phpaen/v13/i12/p123302_s1.。
基于作战网络可靠度的体系贡献率评估
0 引 言
随着网络技术和 装 备 种 类 的 不 断 升 级,现 代 战 争 正 逐 渐转变为体系与 体 系 之 间 的 对 抗。 武 器 装 备 的 发 展,也 正 由以平台为中心,向 以 体 系 为 中 心 转 变[12],在 装 备 论 证 和 方案评估阶段都需要从体系角度来对所研装备进行评估。
收稿日期:2020 09 30;修回日期:2021 01 03;网络优先出版日期:2021 03 24。 网 络 优 先 出 版 地 址 :http:∥kns.cnki.net/kcms/detail/11.2422.TN.20210324.1015.007.html 基 金 项 目 :装 备 预 研 领 域 基 金 (61400040503)资 助 课 题 通讯作者. 引 用 格 式 :周 琛 ,宋 笔 锋 ,尚 柏 林 ,等 .基 于 作 战 网 络 可 靠 度 的 体 系 贡 献 率 评 估 [J].系 统 工 程 与 电 子 技 术 ,2021,43(7):18751883. 犚犲犳犲狉犲狀犮犲犳狅狉犿犪狋:ZHOU C,SONGBF,SHANGBL,etal.Systemofsystemscontributionrateevaluationbasedonoperationalnetwork reliability[J].SystemsEngineeringandElectronics,2021,43(7):18751883.
些装备往往具有多种功能,如战斗机具有 目 标 侦 察、火 力 打 击、指挥控制等功能;预警机具有侦察、指 控 等 功 能;还 有 众 多空、天、地、海的综合型装备。因此,为 了 使 作 战 环 方 法 更 具适用性,就需要将 综 合 型 装 备 恰 当 地 引 入 到 建 模 和 求 解 过程中来。针对这个问题,目前的处理 思 路 有 两 种:一 是 遵 循“合”的思想,将综合型装备抽象为 单 独 的 第 5 类 节 点,这 种思路使得该类节点 在 作 战 环 中 的 功 能 定 位 较 为 模 糊,可 能形成节点间的双向连接,继而涌现出复 杂 的 连 接 关 系;二 是遵循“分”的思想,将 综 合 型 装 备 分 成 多 个 单 一 功 能 的 节 点,这些节点均属于侦察、决策、攻击 3 类 之 一,这 种 处 理 思 路较为清晰,且由于不改变作战环的组成 要 素,可 以 在 现 有
lead calculation method
Lead Calculation MethodLead refers to the time it takes for a potential customer to express interest in a product or service, to the point of actually making a purchase. It is an important metric for businesses to understand, as it can help them gauge the effectiveness of their marketing and sales strategies. Calculating lead time accurately can provide valuable insights into consumer behavior and help businesses make informed decisions. In this article, we will explore different methods for calculating lead time and discuss their advantages and disadvantages.1. Simple Lead Time CalculationThe simplest method for calculating lead time is to measure the time it takes for a lead to convert into a customer. This can be done by tracking the date when a potential customer first interacts with the business, such as by visiting the website or requesting more information, andparing it to the date of their first purchase. The lead time is then calculated as the difference in days between these two dates.Advantages:- Easy to implement and understand- Provides a basic understanding of lead timeDisadvantages:- Does not take into account the various stages of the customer journey- May not accurately reflect the true lead time for different types of leads2. Funnel-based Lead Time CalculationA moreprehensive approach to lead time calculation involves using a funnel-based model to track leads through each stage of the sales process. This method breaks down the customer journey into multiple steps, such as initial contact, lead nurturing, product demonstration, and purchase, and measures the time it takes for leads to move through each stage. By analyzing lead time at each stage, businesses can identify potential bottlenecks in the sales process and make improvements to increase conversion rates.Advantages:- Provides a more det本人led and nuanced understanding oflead time- Helps identify areas for improvement in the sales processDisadvantages:- Requires moreplex data tracking and analysis- Can be time-consuming to implement and m本人nt本人n 3. Cohort Analysis for Lead TimeCohort analysis involves grouping leads based on shared characteristics, such as the date of initial interaction or the source of the lead, and tracking their behaviors and conversion rates over time. This method allows businesses topare the lead time and conversion rates of different cohorts, g本人ning insights into the impact of different marketing channels or customer segments on lead time. By understanding how lead time varies across different cohorts, businesses can t本人lor their marketing and sales strategies to better target high-converting leads.Advantages:- Offers det本人led insights into the impact of different variables on lead time- Helps identify the most effective marketing channels and customer segmentsDisadvantages:- Requires robust data collection and analytics capabilities- May be challenging to identify meaningful cohorts for analysis4. Predictive Modeling for Lead TimePredictive modeling uses statistical techniques to forecast lead time based on historical data and other relevant variables, such as demographic information, online behavior, and purchase history. By building predictive models, businesses can estimate the likelihood and timing of lead conversion for different types of leads, allowing for more targeted and efficient marketing and sales efforts. This method is especially valuable for businesses with large volumes of lead data andplex customer profiles.Advantages:- Enables proactive and targeted lead nurturing- Provides valuable insights for resource allocation and budget planningDisadvantages:- Requires advanced statistical and data science expertise- Relies on the av本人lability of high-quality and relevant dataIn conclusion, lead time calculation is a crucial aspect of understanding customer behavior and improving sales and marketing performance. By using abination of different methods, businesses can g本人n aprehensive understanding of lead time and make data-driven decisions to optimize their customer acquisition strategies. Whether using simple lead time calculation, funnel-based analysis, cohort analysis, or predictive modeling, businesses can leverage lead time insights to drive growth and achieve their goals.。
FESTO EXCM型号轨迹式梯架产品说明书
Planar surface gantries EXCM2d Internet: /catalogue/...Subject to change – 2023/04Planar surface gantries EXCMKey featuresAt a glance GeneralApplication examples• A gantry that is characterised by ex-cellent functionality in compact in-stallation spaces• The drive concept has a low moving mass • The kinematics are driven by 2 step-per motors with built-in optical en-coders (closed loop)• Flexible motor mounting possible• Feeding, pressing, joining compo-nents• Dispensing liquid media• Mounting electronic componentsEXCM-30EXCM-40Functional principleA slide is moved in a two-dimensional space (XY-axis) via a toothed belt. The system is powered by 2 fixed motors in posi -tion-controlled operation (closed loop). The motors are coupled to the toothed belt. The belt is guided via guide pulleys so that the slide can move to any position in a working space when the motors are actuated accordingly.Motor 1M o t o r 2EXCM-30EXCM-40Motor 1Motor 2Motor 2Motor 1H- -NoteAdditional multi-axis controller re-quired for interpolation (e.g. CPX-E-CEC-M1-...).Planar surface gantries EXCM Key features1) Rated load = tool load (attachment components) + payload2) Vertical/horizontal installation position. For vertical installation, we recommend consulting a sales engineer from Festo.3 2023/04 – Subject to change d Internet: /catalogue/...Planar surface gantries EXCMKey featuresEXCM-30 – Motor mounting variants Additional technical data a page 8 BelowEXCM-30-...-B2 – Cable outlet to the rearEXCM-30-...-B1 – Cable outlet to the frontEXCM-30-...-B3 – Cable outlet on the insideEXCM-30-...-B4 – Cable outlet on the outside4d Internet: /catalogue/...Subject to change – 2023/04Planar surface gantries EXCM Key featuresEXCM-30 – Motor mounting variantsOn topEXCM-30-...-T2 – Cable outlet to the rearEXCM-30-...-T1 – Cable outlet to the frontEXCM-30-...-T3 – Cable outlet on the inside5 2023/04 – Subject to change d Internet: /catalogue/...Planar surface gantries EXCMKey featuresEXCM-40 – Motor mounting variants Additional technical data a page 22 EXCM-40-...-B – Motor underneathEXCM-40-...-T – Motor on top6d Internet: /catalogue/...Subject to change – 2023/04Planar surface gantries EXCMType codes72023/04 – Subject to change d Internet: /catalogue/...Planar surface gantries EXCM-30Peripherals overviewVariants and accessories[1] With protection against particles EXCM-...-P8[2] With adjusting kit EADC-E11The cover protects the guide of the Y-axis against contamination.With the adjusting kit, the gantry can be aligned after installation.8d Internet: /catalogue/...Subject to change – 2023/04Planar surface gantries EXCM-30 Peripherals overview-NoteH-Homing is always carried out using the mechanical stop in combination with thedrive package from Festo; the sensor mounting and proximity switch are not re-quired in this case.9 2023/04 – Subject to change d Internet: /catalogue/...Planar surface gantries EXCM-30Datasheet1) Rated load = tool load (attachment components) + payload2) Vertical/horizontal installation position.3) Perpendicular to working plane, at standstill4) In case of a load supply of 48 V/24 VMotors with brake must be used in the case of vertical installation5)10d Internet: /catalogue/...Subject to change – 2023/04MaterialsStandard/with protection against particles P81)H--NoteEngineering software Handling Guide Online/handling-guideAcceleration a as a function of the rated load m and stroke of the Y-axisThe following data applies to a horizontal installation position and refers to the service life of the mechanical system of 3500 km. For vertical installation positions, please get in touch with your local contact at Festo.The centre of gravity of the slide is at the height of the slide in the Z-direction and in the centre of the slide in the X-/Y-directions.Stroke, Y-axis = 110/160/210 mmStroke, Y-axis = 260 mmStroke, Y-axis = 310 mmStroke, Y-axis = 360 mmStroke, Y-axis = 410 mmStroke, Y-axis = 460 mmStroke, Y-axis = 510 mmNo-load torque M as a function of rotational speed nLoad valuesThe centre of gravity of the slide is at the height of the slide in the Z-direc-tion and in the centre of the slide in the X-/Y-directions.The system is subject to the greatest load in the case of 45° travel.The following data apply in this case:Formula for calculating the required torque M and the required rotational speed n M45° = a x (4.28 x m L + 2.14 x m Ay + 23.38 x J m + 0.56) x 10–3 + M Rn45° = 60000 / feed constant(mm) x sqrt(2)a = acceleration [m/s2]v = speed [m/s]m Ay = product weight of the Y-axis [kg] a page 11m L = attachment component (Z-axis) [kg] with payloadJ m = moment of inertia of the motor [kgcm2] a table belowM R = no-load torque [Nm] a page 12n45° =rotational speed at 45° travel [rpm] Sample calculationAssuming:Planar surface gantry EXCM-30-700-410-KF-ST a max = 10 m/s2v max = 2 m/s Payload = 0.5 kg H--NoteThe following data applies to a horizontal installation position. For a vertical in-stallation position, please get in touch with your local contact at Festo.The centre of gravity of the slide is at the height of the slide in the Z-directionand in the centre of the slide in the X-/Y-directions.X-/Y-axisZ-axisCalculation:1. What is the max. acceleration permitted by the mechanical system?Moving mass m L on the Y-axis:m L= 2 kgStroke of the Y-axis:410 mmResults:With a moving mass m L of 2 kg, the maximum permissible acceleration is13 m/s2.The required acceleration of 10 m/s2is thus permissible.Stroke, Y-axis = 110/160/210 mmStroke, Y-axis = 260 mmStroke, Y-axis = 310 mmStroke, Y-axis = 360 mmStroke, Y-axis = 410 mmStroke, Y-axis = 460 mmStroke, Y-axis = 510 mmSample calculation2. Is the attached motor sufficient for this load?Assuming:a max = 10 m/s 2v max = 0.35 m/s m Ay = 1.32 kg m L = 2 kg J m = 0.082 kgcm 2M 45° = a x (4.28 x m L + 2.14 x m Ay + 23.38 x J m + 0.56) x 10–3 + M R n 45° = 60000 / feed constant(mm) x sqrt(2)a = acceleration [m/s 2]v = speed [m/s]m Ay = product weight of the Y-axis [kg] a page 11m L = attachment component (Z-axis) [kg] with payloadJ m = moment of inertia of the motor [kgcm 2] a table below M R = no-load torque [Nm] a page 12n 45° = nominal rotational speed at 45° travel [rpm]H- -NoteThese requirements for the dynamic response apply to 45° travel.The dynamic values may be higher for travel only in the X- or Y-direction.Determining M45°n 45°= 60000 / feed constant(mm) x sqrt(2)No-load torque:EXCM-30M R = 0.12 NmM 45° = a x (4.28 x m L + 2.14 x m Ay + 23.38 x J m + 0.56) x 10–3 + M RM 45° = 10 m/s 2 x (4.28 x 2 kg + 2.14 x 1.32 kg + 23.38 x 0.082 kgcm 2 + 0.56) x 10–3 + 0.12 Nm = 0.26 NmResults:The torque value lies below the motor characteristic curve.The design is thus acceptable.Minimum number of profile mountingsDepending on the installation position and the stroke of the X-axis, a different number of profile mountings is required.Stroke reduction in combination with mounting kit EAHT-E9The reduction is influenced by the following factors:• [1] The mounting kit EAHT-E9 is wider than the slide of the Y-axis• [2] By adjusting kits EADC-E11 or profile mountings MUE that are mounted on the inside of the X-axis• [3] When using an additional mounting surface for the cover in combinationwith EXCM-...-P8 (with protection against particles)22Planar surface gantries EXCM-30 Ordering data – Modular product system[1] B, T Not in combination with stepper motors ST and SB. Option if third-party motors are mountedPlanar surface gantries EXCM-40 Peripherals overviewProximity switch for sensing the position of the slide on the Y-axis10Planar surface gantries EXCM-40Peripherals overviewSelection of attachment componentsThe gantry is delivered as standard in the configuration without attachment ele -ments.The "Handling Guide Online" engineering software can be used to configure the planar surface gantry with other attachment components, such as a pneumatic or electric Z-axis.EXCM-... (without attachment component)The following are pre-installed:• 2 supply ports for e.g. Z-axis• Multi-pin plug distributor for bundling signals:– e.g. proximity switchPlanar surface gantries EXCM-40 Datasheet1) Rated load = tool load (attachment component (Z-axis) + e.g. gripper) + payload2) These values must also be complied with when installing third-party motors3) At v=0.2 m/s and 45° travel.4) This data applies only under ideal conditions.For a precise configuration, please consult a sales engineer from Festo.Additional information a page 271) Note operating range of proximity switches and motorsPlanar surface gantries EXCM-40 DatasheetMaterials5Planar surface gantries EXCM-40 Datasheet1) Weight per component1) Feed constant at 45° travel-NoteH-Engineering softwareHandling Guide Online/handling-guidePlanar surface gantries EXCM-40DatasheetNo-load torque M as a function of rotational speed nLoad valuesThe centre of gravity of the slide is at the height of the slide in the Z-direc-tion and in the centre of the slide in the X-/Y-directions.The system is subject to the greatest load in the case of 45° travel.The following data apply in this case:Formula for calculating the required torque M and the required rotational speed n M 45° = a x (9.79 x m L + 4.89 x m Ay + 10.21 x J m + 19.58) x 10–3 + M R n 45° = 60000 / feed constant(mm) x sqrt(2)a = acceleration [m/s 2]v = speed [m/s]m Ay = product weight of the Y-axis [kg] a page 26m L = attachment component (Z-axis) [kg] with payloadJ m = moment of inertia of the motor [kgcm 2] a table below M R = no-load torque [Nm] a page 27n 45° =nominal rotational speed at 45° travel [rpm]Planar surface gantries EXCM-40DatasheetSample calculationAssuming:Planar surface gantryEXCM-40-1000-500-KF-SB-B-PF7-HE1-…with attached motor EMMS-ST-57-M-SEB-G2a max = 2 m/s 2v max = 0.5 m/s Payload = 0.5 kgAttachment component on Z-axis: EGSL-BS-45-100-10PX-/Y-axis Z-axisSample calculation2. Is the attached motor sufficient for this load?Assuming:a max = 2 m/s 2v max = 0.5 m/s m Ay = 10.65 kg m L = 3.8 kg J m = 0.5 kgcm 2M 45° = a x (9.79 x m L + 4.89 x m Ay + 10.21 x J m + 19.58) x 10–3 + M R n 45° = 60000 / feed constant(mm) x sqrt(2)a = acceleration [m/s 2]v = speed [m/s]m Ay = product weight of the Y-axis [kg] a page 26m L = attachment component (Z-axis) [kg] with payloadJ m = moment of inertia of the motor [kgcm 2] a table below M R = no-load torque [Nm] a page 27n 45° = nominal rotational speed at 45° travel [rpm]H- -NoteThese requirements for the dynamic responseapply to 45° travel.The dynamic values may be higher fortravel only in the X- or Y-direction.DatasheetSample calculationDetermining M45°n45° = 60000 / feed constant(mm) x sqrt(2)No-load torque:EXCM-40 M R = 0.4 NmM45° = a x (9.79 x m L + 4.89 x m Ay + 10.21 x J m + 19.58) x 10–3 + M RM45° = 2 m/s2 x (9.79 x 3.8 kg + 4.89 x 10.65 kg + 10.21 x 0.5kg cm2 + 19.58) x 10–3 + 0.4 Nm = 0.63 Nm Results:The torque value lies below the motor characteristic curve.The design is thus acceptable.Planar surface gantries EXCM-40DatasheetMinimum number of profile mountingsIrrespective of the installation position, a different number of profile mountings needs to be used depending on the stroke of the X-axis.The required number is mounted on delivery.Distances between the profile mountingsThe profile mountings must be evenly spaced by distance l.ll1=ll+141nn−1l1 = distancel = stroken = number of profile mountings per axisInstallation position of attachment components Array Due to manufacturing tolerances andthe backlash in the guides, the anglebetween the mounting plane and theattachment component, e.g. Z-axis,may not be exactly 90° in certain cir-cumstances.Max. deviation:á = ±1.1°Pin allocationsMotors on the X-/Y-axisMotor EncoderSelection of cable lengths2 cable lengths (5 m or 10 m) can beselected using the modular productsystem a page 34. This specifica-tion relates to the output of the energychain at the X-axis (dimension L) anddescribes the minimum length bywhich the cables and tubing protrude.The selected length applies to the fol-lowing components:• Tubing• Plug sockets with cableSample product image1) With brakeH--NoteDepending on the stroke of the X-ax-is, a different number of profile mountings is required. The distance between the profile mountings must always be the same (a page 30).The tension of the toothed belt must be set before commissioning. The tools required to do this (e.g. fre-quency meter) are not included in the scope of delivery.Ordering data – Modular product systemH--NoteIn combination with key feature W (without motor), the planar surface gantry EXCM is provided without a coupling housing and without a cou-pling.H--NoteThe planar surface gantry can only be operated with a load voltage of 48 V.H- -NoteDepending on the combination of motor and drive, it may not be possible to reach the maximum feed force of the drive.Third-party motors that have an overly high driving torque may damage the linear gantry. When selecting the motors, please observe thelimits specified in the technical data.1) The input torque must not exceed the max. permissible transferable torque of the axial kit.Profile mounting MUEFor size 30Material:Anodised aluminiumRoHS-compliantFor mounting the planar surface gantry (scope of delivery: 1 pair)Included in the scope of delivery of the planar surface gantry:X-stroke < 500 mm: 2 pairsX-stroke ›500 mm: 3 pairsAdjusting kit EADC-E11 For size 30Material:Anodised aluminiumRoHS-compliantFor mounting and aligning the planarsurface gantry. The kit is height adjust-able.Mounting kit EAHT-E9 For size 30Material:Anodised aluminiumRoHS-compliantPrepared hole patterns for:• Mini slide EGSL-35• Mini slide DGSL-8/-10/-12• Electric slide EGSK-20/-26• Electric cylinder EPCO-16•Mini slide EGSC-BS-25/-32Sensor mounting EAPR For size 30(incl. switch lug)Material:Retaining bracket: Wrought aluminium alloySwitch lug: SteelEnergy chain and connection set for size 30 Ordering data – Energy chainEADH-U-30-30EADH-U-30-40Adjusting kit EADC-E12For size 40Material:Anodised aluminiumRoHS-compliantFor mounting and aligning the planar surface gantry. The kit is height adjust-able.Mounting kit EAHM-E12For size 40Material:Anodised aluminiumRoHS-compliantFor mounting the planar surface gan-try. The kit is not height adjustable.Planar surface gantries EXCM AccessoriesFor size 40Material:Switch lug: SteelSensor bracket: Wrought aluminiumalloyRoHS-compliantFor proximity switchesSIES-V3B and SIES-Q8B(for sensing the position of the slide onthe X-axis)41 2023/04 – Subject to change d Internet: /catalogue/...Planar surface gantries EXCMAccessories-NoteH-For homing in combination withthird-party motors.42d Internet: /catalogue/...Subject to change – 2023/04Planar surface gantries EXCM Accessories1) Cables especially suitable for the motor controller and motor.Degree of protection to IP65 (in assembled state)43 2023/04 – Subject to change d Internet: /catalogue/...Festo - Your Partner in AutomationConnect with us/socialmedia 1Festo Inc.2Festo Pneumatic 3Festo Corporation 4Regional Service Center 5300 Explorer DriveMississauga, ON L4W 5G4CanadaAv. Ceylán 3,Col. Tequesquináhuac 54020 Tlalnepantla, Estado de México1377 Motor Parkway Suite 310Islandia, NY 117497777 Columbia Road Mason, OH 45040Festo Customer Interaction CenterTel:187****3786Fax:187****3786Email:*****************************Multinational Contact Center 01 800 337 8669***********************Festo Customer Interaction Center180****3786180****3786*****************************S u b j e c t t o c h a n g e。
人教版高中英语选择性必修四 UNIT 1 SCIENCE FICTION Section Ⅱ
被动语态
语法图解
探究发现 1.It was going to be tested out by Larry’s wife,Claire. 2....but Larry persuaded her that the robot wouldn’t harm her or allow her to be harmed. 3.Now she was being looked after by a robot that looked so human,and it was disturbing.
2.【教材原文】When it comes to games,2017 became a year in which AI defeated humans in so many games that AI researchers say that it will not be long before AI can win at almost everything.(page 4) 谈到比赛,2017年成了人工智能在那么多比赛中击败人类的 一年,以至于人工智能的研究者们说,人工智能不久就可能在 几乎所有方面获胜。
语法精析
被动语态 英语中动词有两种语态:主动语态和被动语态。主动语态表 示主语是动作的执行者;被动语态表示主语是动作的承受者。 汉语中常用“被”“给”“由”“受”等词来表示被动,而英语中由 “be+动词过去分词”构成。
小学上册I卷英语第2单元期中试卷
小学上册英语第2单元期中试卷英语试题一、综合题(本题有100小题,每小题1分,共100分.每小题不选、错误,均不给分)1.The baby is sleeping ___. (quietly)2. A parakeet loves to socialize with ________________ (人).3.I like to play with my ________ (玩具名称) after school.4. A ______ (食品生产) relies heavily on plant cultivation.5.我的朋友喜欢 _______ (活动). 她觉得这很 _______ (形容词)6.The process of fermentation produces __________ from sugars.7. A _____ is an area of land that rises sharply.8.I enjoy attending ______ (音乐会) to see my favorite artists perform.9.What is the weather like when it snows?A. HotB. ColdC. RainyD. Windy10.What is 5 + 7?A. 11B. 12C. 13D. 14C11.What do we call the top of a tree?A. TrunkB. LeafC. BranchD. CanopyD12.The _____ (balloon) is colorful.13.What is the capital city of Indonesia?A. JakartaB. BaliC. BandungD. SurabayaA14.What do you call a large body of salt water?A. RiverB. LakeC. OceanD. PondC15.Which insect can produce honey?A. AntB. FlyC. BeeD. MosquitoC16. A chemical reaction can produce _____ and light.17.The ancient Greeks are known for their _____ and philosophy.18.The __________ (历史的视角变化) enrich insights.19.The boiling point of water is _____ degrees Celsius.20.The _____ (小鸟) is learning to fly. It flaps its wings excitedly. 小鸟正在学习飞翔。
小学上册第14次英语第四单元测验试卷
小学上册英语第四单元测验试卷英语试题一、综合题(本题有100小题,每小题1分,共100分.每小题不选、错误,均不给分)1.The ________ jumps high in the air.2.What do you call a person who studies fossils?A. PaleontologistB. ArchaeologistC. HistorianD. Geologist答案:A3.What is the term for a person who creates art?A. SculptorB. ArtistC. PainterD. Designer答案:B4.I want to learn how to ________.5.What is the name of the famous clock tower in London?A. Big BenB. Eiffel TowerC. Leaning TowerD. Tower of Pisa答案:A6.I can ___ (run) very fast.7.I love _______ (吃水果).8.Many plants have a ______ (生命周期) that lasts for years.9.What is the primary color of a stop sign?A. GreenB. YellowC. RedD. Blue答案:C10. A chemical reaction can change the ______ of a material.11.What is 12 ÷ 4?A. 1B. 2C. 3D. 4答案:C12.What is the capital of Portugal?A. LisbonB. PortoC. FaroD. Coimbra答案:A13.I love the sound of ______ (雨) falling on the roof. It makes me feel cozy inside.14.The __________ is a large area of grassland in Africa. (塞伦盖蒂)15.Where do fish live?A. TreesB. WaterC. GroundD. Air16.The raccoon has a black mask around its ______ (眼睛).17.The __________ is a large area of wetlands in the United States. (大沼泽地)18.Butterflies go through _______ (变形) stages.19.The __________ (历史的经历) enriches our lives.20. (99) is a popular sport in many countries. The ____21.The main product of cellular respiration is _____.22.What is the name of the northernmost point on Earth?A. South PoleB. North PoleC. EquatorD. Tropic of Cancer答案:B23.Which animal is known for its stripes?A. LeopardB. ZebraC. CheetahD. Tiger24.I like to make ______ (纸飞机) on windy days.25.The cake is ___. (sweet)26.What is the sound of a cow?A. BaaB. MooC. QuackD. Roar27.What is the sum of 9 + 6?A. 14B. 15C. 16D. 17答案:A28.What do we call a person who studies space?A. GeologistB. BiologistC. AstronomerD. Meteorologist答案:C29. A planet's atmosphere can trap heat through the greenhouse ______.30.The country known for kangaroos is ________ (以袋鼠闻名的国家是________).31.The _____ (树) provides shade.32.My teacher is very __________ (善于沟通).33.The _____ (繁殖) of certain plants can happen through cuttings.34.The chemical symbol for potassium is __________.35.I can ______ (克服) challenges with determination.36.What do you call a baby dog?A. KittenB. PuppyC. CalfD. Foal答案:B37.What do you call a group of musicians?A. BandB. ChoirC. OrchestraD. All of the above答案:D38.What is the process by which plants get energy from sunlight called?A. RespirationB. PhotosynthesisC. DigestionD. Germination39.I see a _____ (cat/dog) in the garden.40. A parakeet enjoys hanging upside down on its ______ (栖木).41.I enjoy _____ (听音乐) at home.42.I enjoy ________ (制作) my own crafts.43.What is the name of the famous wizard school in Harry Potter?A. HogwartsB. BeauxbatonsC. DurmstrangD. Ilvermorny44.She speaks ________ languages.45.My favorite dish is ______ (中国菜).46. A chemical reaction can be affected by the _____ of reactants.47. A ____ buzzes around and collects nectar from flowers.48.What do we use to write?A. BrushB. PenC. RulerD. Scissors答案:B49.What do we use to cut paper?A. GlueB. TapeC. ScissorsD. Ruler答案:C50.The discovery of ________ has had extensive implications for health.51.An enzyme is a biological _____ that speeds up reactions.52.The _______ (蜥蜴) can be green or brown.53.My ________ (玩具) is full of surprises.54.Which is the largest ocean on Earth?A. AtlanticB. IndianC. ArcticD. Pacific答案:D55.The __________ (历史的传承) is vital for future generations.56.What is the capital of Hungary?A. BudapestB. DebrecenC. SzegedD. Pécs57.The __________ (社会变革) can lead to significant political changes.58.The ______ (根茎) helps store energy for the plant.59.The _____ (cabbage) is a leafy vegetable.60.What is the main source of energy for the Earth?A. The MoonB. The SunC. StarsD. Wind61.Which of these is a vegetable?A. AppleB. CarrotC. BananaD. Grape62.The _____ (computer/tablet) is useful.63.The capital of Malaysia is ________ (吉隆坡).64.The _____ (植物保护) is important for biodiversity.65.My grandma grows beautiful ______ (玫瑰) in her garden. They smell very ______ (香).66.What is the opposite of happy?A. SadB. AngryC. ExcitedD. Joyful67.What do you call the events that happen around the world?A. NewsB. StoriesC. MythsD. Legends答案:A68.The law of definite proportions states that a chemical compound contains the same ______.69.The main component of nucleic acids is ______.70.I write in my _____ (笔记本).71.What do we call a device that takes pictures?A. CameraB. ProjectorC. TelevisionD. Computer答案:A72.The chemical formula for vanadium pentoxide is _____.73.The smallest unit of an element is an _____.74.What do you call the large body of salt water that covers most of the Earth?A. LakeB. RiverC. OceanD. Pond75.I enjoy _______ (与朋友一起)外出.76.What is the name of the famous landmark located in Washington D.C.?A. Statue of LibertyB. Golden Gate BridgeC. Lincoln MemorialD. Mount Rushmore答案:C77.The ancient Egyptians created a calendar based on the ______ (太阳).78.The _______ (Great Depression) started in 1929 and affected the whole world.79.What do you call a person who performs in a circus?A. AcrobatB. ClownC. JugglerD. All of the above答案:D80.The chemical formula for lithium carbonate is ______.81.The __________ (历史的反馈) informs future studies.82.Which word means "happy"?A. SadB. JoyfulC. AngryD. Tired83.What do you call a large body of saltwater?A. LakeB. PondC. OceanD. River84.The chemical formula for potassium iodide is _____.85.What do we call the study of the weather?A. MeteorologyB. ClimatologyC. AstronomyD. Geology答案:A Meteorology86.The chemical symbol for yttrium is _______.87.The chocolate is very _______ (rich).88.The ______ helps with the movement of legs.89.______ is the smallest unit of an element.90.What is the shape of a basketball?A. SquareB. OvalC. RoundD. Triangle答案:C91.What is the capital of Uruguay?A. MontevideoB. Colonia del SacramentoC. SaltoD. Paysandú答案:A92.The rabbit is ______ (hopping) through the grass.93.I like to go ______ (钓鱼) with my dad on weekends.94. A garden needs regular ______ (浇水) to keep plants healthy.95. A reactant is a starting ______ in a chemical reaction.96.In organic chemistry, we study compounds that contain _____.97.The chipmunk stores nuts in its ______.98.What is the term for a young sheep?A. CalfB. FoalC. LambD. Kid答案:C99.I enjoy practicing mindfulness to stay present in the __________.100.The _____ (内陆地区) may have unique plant species.。
浩海学校高中英语Unit 1 LIFE STYLES同步练习2 1
涝酒州涉消市浩海学校Unit 1 LIFE STYLES 同步练习2阅读理解AIn a new measure to ensure that an increasing number of Chinese people can speak good English, all universities and colleges across China will have to use English as a teaching language for professional subjects starting this year.In a circular issued last month, the Higher Education Department under the Ministry of Education (MOE) said universities must launch (开始) bilingual(双语的)courses from this year. In three years' time, five to ten percent of universities' total courses must be taught in a bilingual manner.To help accomplish(实现)this, more original editions of English specialized textbooks will be imported and published in China in the coming years."Higher education in situations across the country must come up with implementation plans as soon as possible, "said Zhang Yaoxue, director of the Higher Education Department, MOE. "We need more bilingual profess sionals for China's entry into the WTO."One major problem blocking the development of bilingual education is the lack of qualified bilingual teachers. Many science teachers cannot speak good English, and English teachers may not be able to give science lessons.To solve this problem, the MOE will send more Chinese teachers overseas to receive training. Fifty teachers will be sent abroad at the end of this year.To boost(推动)the trend of bilingual education, Tsinghua University Press imported and published 43 English books on September 25.The books cover six subjects—Management, Economics, Finance, Accounting, Marketing and MBA.They can be used in undergraduate, postgraduate and MBA courses.All the books boast (拥有)the latest information in their specialized fields and most are used as text books by world famous universities, including Cambridge and Havard.This is not the first time Tsinghua University Press has published original editions of books in specialized fields. In 1997, it published the first ever original English books on Economics and Management in China.Then other publishing houses, such as Renmin University of China Press and Machinery Industry Press, also began to import books from English-speaking countries.Since 2000, Tsinghua has been using original English textbooks in its specialized core courses, such as Economics and Information Technology. In some courses, English is the only classroom language: teachers present classes in English and students discuss cases in English.Such classes are proving very popular among students. In the MBA program, among the total of 120 new students, 90 chose such English courses. After the lessons had begun, some students in the Chinese class wanted to change to the English class."The popularity of these courses is that they can promote (促进) students' competitiveness in future working environments, in which English plays an important role. English jargon must be learned through English." said Qian Xiaojun, director of the MBA program in the School of Economics and Management, Tsinghua University.1 Why should all universities in China launch bilingual courses?A.To import and publish more original English .B.To help more people to speak good English.C.To get more qualified bilingual teachers.D.To improve higher education.解析:文章的第一段有足够的信息支持此答案答案:B2 What's the most difficult problem in the development of bilingual education?A.Many Science teachers can't speak good English.B.English teachers can't give science lessons.C.There are not any good original books.D.There are not enough qualified bilingual teachers.解析:答案在倒数第五段答案:D3 At least ________ publishing houses have published original English books.A.1 B.2 C.3 D.4解析:细节题,答案在倒数第四段答案:C4 Which of the following is true according to the text?A.Tsinghua has been using original English text books in all courses since 2000.B.Most of the MBA students like to attend the English class.C.The English courses can't help students' competitiveness in future work.D.The original English books cover many subjects, such as mathematics.解析:答案在倒数第二段答案:BBInterviewing is one of those skills that you can only get better at. You will never again feel so ill at ease as when you try it for the first time, you'll probably never feel entirely comfortable trying to get from another person answers that he or she maybe too shy to reveal. But at least half of the skill is mechanical. The rest is instinct(天赋), which can all be learned with experience.The basic tools for an interview are paper and two or three well-sharpened pencils. But keep your notebook or paper out of sight until you need it. There's nothing less likely to relax a person than the arrival of someone with a note-taking pad.Both of you need time to get to know each other. Take a while just to chat, judging what sort of person you're dealing with, getting him or her to trust you.Never go into an interview without doing whatever homework you can. If you are interviewing a town official, know his voting record.If it's an actor, know what plays he has been in. You will not be liked if you inquire about facts that you could have learned in advance.Many beginning interviewers are afraid that they are forcing the other person to answer questions and have no right to inquire about his personal secrets. The fear is almost 100 percent unnecessary. Unless the person really hates being interviewed, he is delighted that somebody wants to interview him.Most men and women lead lives that are uninteresting, and they grasp any chance to talk to an outsider who seems eager to listen.This doesn't necessarily mean that it will go well. In general you will be talking to people who have never been interviewed before, and they will get used to the process(程序)awkwardly (笨拙地), perhaps not giving you anything that you can use. Come back another day; it will go better. You will both even begin to enjoy it—proof that you aren't forcing your victim to do something he doesn't really want to.1 The word "reveal" could best be replaced by________.A.ask B.question C.give D.seek解析:泄露答案:C2 According to the passage, during an interview notebooks or paper should________.A.never be used B.be used only when necessaryC.be kept at home D.be given to the interviewer解析:细节题,答案在第二段答案:B3 It can be inferred from Paragraph4 that most men and women ________.A.are afraid to meet interviewers B.hate being interviewedC.like being interviewed D.fear to speak to outsiders解析:第四段的意思是说很多人对第一次采访很害怕,其实大可不必如此,因为大多数人是喜欢被采访的,除非他很讨厌被采访。
烧结的优化
NEW APPROACH FOR FIRING OPTIMISATION IN CRYSTALLINE SILICON CELL TECHNOLOGYJaap Hoornstra, Arvid van der Heide, Arthur Weeber, Filip GranekECN Solar Energy, PO Box 1, NL-1755 ZG Petten, the Netherlandshoornstra@ecn.nl, tel: +31 224 564697, fax: +31 224 568214ABSTRACT: A new approach for firing optimization is introduced and demonstrated on 10 different pastes. The approach uses fewer settings and is faster than the methods traditionally used. Therefore the approach will make evaluation of new pastes and processes easier. The method uses a cross belt temperature profile to subject a cell with screen-printed paste to different temperatures while firing in the infrared furnace. J-V data, spatially resolved wafer peak temperatures and spatially resolved contact resistances using the Corescan, are used to determine settings for final optimization. Optimal settings for the ten evaluated pastes were obtained in only two steps and using few cells. With only three cells, a scan over a temperature range of about 180°C could be performed. In this way also the firing window could be assessed. The method is applicable both in industry and R&D.Keywords: metallization, IR-firing, Corescan1 INTRODUCTIONFiring of front and rear contacts in the IR firing furnace dominates the quality of the electrical contact. This is especially important in evaluating new front side metallisation schemes, such as new pastes, AR coatings, or emitters. To find optimum furnace settings is an elaborate effort, since many settings of temperature and belt speed may have to be explored. With new or alternative front side pastes it is even more difficult or tedious to find an optimum, since no data is available in the furnace temperature and belt speed domain. Another draw-back is that firing furnaces of most PV manufacturers are constantly in use for cell production; there is little time for testing, let alone that all kind of firing furnace settings can be explored to determine if a paste "works". This means that in practice, there is only very limited time available for evaluating or testing new pastes. In order to shorten the effort and to use fewer settings to determine if a paste works well and to find near optimum settings, a new approach is invented at ECN, and presented in this paper. In the next section the approach is explained in more detail, and also the method to demonstrate the approach is given.2 EXPERIMENTAL METHODThe approach is based on relating local contact resistance to local cell temperature. Spatially resolved wafer temperatures are measured as a function of relevant furnace settings using a Datapaq temperature profiler. The Corescan instrument is applied to map the contact resistance distribution of the front contact. To reduce the number of trials effectively the approach uses the cross belt temperature gradient to fire a cell. This means that in one run, various locations on the cell are exposed to different temperatures. By relating the local contact resistance to the local wafer temperature, the temperature setting for optimal firing can easily be established.In order to limit the number of Corescan analyses, standard J-V and Suns-Voc measurements are performed first to check if an electrical front contact has been established. On cells with relevant fill factor, the Corescan instrument is used.The approach assumes that the front contact is the limiting factor in achieving optimal cell results. Further the method applies furnace settings in an establishedtemperature and belt speed domain; i.e. rear side contacts and BSF are formed. The approach is elucidated in the next paragraph.2.1 Details of new approach for firing optimization2.1.1 Datapaq measurementsThe cross belt temperature profile, as measured on a wafer with fine wire thermocouples attached, is measured using a Datapaq temperature profiler [1] as depicted in figure 1. The application and procedure are reported in [2].1Figure 1: Data logger placed in opened thermal barrier with TCs attached to a wafer as used in [2].The data logger is the Datapaq 9000 type DP9064, equipped for 6 K-type thermocouples. The thermal barrier used in this study is model TB2015. The TCs are 0.5 mm in diameter, and the high temperature junction is attached onto the silicon wafer surface using ceramic cement. As demonstrated in [2] the use of fine wire thermocouples enables approximation of real wafer temperatures.2.1.2 Furnace settingsThe cross belt temperature distribution in our furnace is influenced by side heating and by the presence of wear strips. Our IR firing furnace (Radiant Technologies Co.) uses wear strips of fused quartz to support the belt. There are 3 wear strips of about 20 mm in diameter, parallel to the belt direction and located at the edges of the belt and in the center. In our furnaces the center wear strip in the high temperature section was removed by RTC to realize a flat cross belt temperature distribution at the center [2]. The furnace is perfectly capable of processing the 16.5% multi-crystalline cells as being presented at this conference [3]. Now the presence of the side wear strip is deliberately used to achieve a large temperature difference over the cell, when placed at the edge of the belt. In addition the side1044heater, normally used to flatten the cross belt profile, was turned off.2.1.3 CharacterizationThe contact resistance (R c ) of the front side metallization is mapped using the Corescan instrument, which has been developed by ECN [4]. Reports on the background of the instrument and its applications are available at [5]. Application of the Corescan to study the contact formation on lowly doped emitters is presented at this conference [6]. From earlier work it is known that the contact resistance is strongly related to the firing temperature and is often very non-uniform [5,7]. Also it was found that there is an excellent correlation between the variation of the measured temperatures over a wafer and the contact resistances found with the Corescan [2].Suns-Voc setup [8] gives the series resistance less fill factor. The method is to expose the cell to a light flash and analyze the response of the Voc. The value of the series resistance less fill factor indicates whether the cell has been over-fired. At the same time the difference between the Suns-Voc FF and the FF from the J-V analysis is a good measure of the cells series resistance. For near optimal screen printed cells a difference of about 2.5 - 3.5% in FF value is realistic. Larger values are mostly attributed to higher contact resistance. See also [6].2.2 Summary procedureThe procedure is summarized here below.1. Realize non-flat cross-belt settings: UseDatapaq profiler to evaluate wafer temperatures and establish cross belt profiles.2. Cell firing and characterization: Fire on edge ofbelt. Characterization using solar simulator, Suns-Voc, and Corescan. Relate R c to wafer temperature to determine best settings.3. Final optimization: Fire with flat profiles atsettings around optimum found in 2.2.3 Experimental set-upFor a series of furnace settings, based on setting for our standard firing, cross belt temperature profiles were made using the Datapaq system. The side heater was switched off in order to realize a maximum temperature difference near the edge of the belt. Various settings of the high temperature zone were explored to cover a wide range of temperature profiles. Five TCs were attached to a 156 cm² SiN coated and texturized mono crystalline wafer. The TCs are mounted as depicted in figure 2.Figure 2: Wafer with 5 K-type thermocouples. Figure 3: Voc map at setting CBM (full scale450 - 620mV)The wafer with thermocouples perpendicular to the belt direction was first placed on one edge of the belt, and in different runs moved towards the center of the belt. In this way a detailed cross belt distribution was achieved. To demonstrate the new method, cells are made of 156 cm² square texturized mono crystalline silicon (CZ) with a 40 - 45 Ohm/square emitter and SiN anti reflection coating. Two cells were processed for each firing furnace setting.The approach is demonstrated in this work by optimizing 10 different pastes. Some of the pastes are commercially available; the others are research pastes. The metal contacts are screen printed on the cells and co-fired in the IR firing furnace using fixed settings for the belt speed. Also on several cells only the rear side contact was printed and fired. These cells were used for Voc mapping with the Corescan instrument.To show how successful the new approach is, additional firing steps were done using standard, i.e. a flat cross belt temperature distributions with different belt speeds.3 RESULTS AND DISCUSSION3.1 Voc mappingTo ensure functional rear side contacts and BSF for the settings used in this work Voc scans were performed using the Corescan instrument on cells with only rear side metallization. In figure 3 a Voc mapping is presented of a cell fired by using the non-flat Cross Belt profile and at Medium peak temperature (CBM). The figure shows a variation of Voc of about 25 mV, with for the right part of the cell a lower value. This is due to the lower local temperature as caused by the wear strip. For all conditions used a proper BSF was realized.3.2 Wafer temperature measurementsThree cross belt profiles with settings CBH, CBM, and CBL are measured on a SiN coated wafer without contacts (figure 4).Figure 4: Cross belt temperature distribution at one belt speed for three peak zone temperatures.The TC data used are the peak temperatures as obtained from the longitudinal temperature profile measured during travel of TCs on wafer on the belt through the furnace. The temperature profiles show a temperature difference over the cell width of about 60°C. For the three settings in total about 180°C is covered this way.3.3 Cross belt firingCells were made using the three set temperatures of the peak temperature zone at a fixed belt speed (v+151045cm/min). In table 1 the averaged J-V fill factor results are presented.Tset CBH CBM CBL Paste FF [%] FF [%]FF [%]1 69.1 47.8 32.92 63.0 46.3 28.0 3 63.1 39.9 32.04 27.5 27.2 27.2 5 28.2 27.8 27.16 32.4 28.9 27.4 735.0 29.3 29.0 8 46.0 33.3 29.2 9 31.2 26.8 24.6 10 65.2 50.7 28.6Table 1: J-V fill factor resultsFrom the results, it can be seen that pastes 1, 2, 3 and 10 give already reasonable cells. For all pastes the FF increases at higher temperature in this range. From the other pastes only paste 8 shows some ability to make a cell, although it seems that an optimum furnace setting is not reached. None of the pastes seems to be (too) aggressive; no firing through the emitter has been observed, as is indicated by fill factors from Suns-Voc analysis that are all in the range of 80 to 82%.Examples of the contact resistance distribution for pastes 1 and 2 are given in figure 5.875°C 790°C 875°C 790°C Paste 1; FF = 70.3% Paste 2; FF = 63.4% Scale 0 - 50mV Scale 0 - 50mVFigure 5: Corescan examples for the cells fired with the cross belt temperature profile CBH.In the Corescan figures the dark color corresponds to "good" contact resistance, while the light color corresponds to "bad" contact resistance. The range of the color scale settings is 50mV. A value of about 5 - 10mV represents a good contact resistance (<10 m Ωcm 2).The area on the left side of the graph is exposed to a high temperature of about 875°C. On the right side the temperature drops to about 790°C due to shadowing by the wear strip. Paste 1 shows a large area of good contact resistance; only the area at the wear strip has too high contact resistance. This paste seems already to be fired at a near optimum setting. By comparing paste 1 and 2 in this temperature range paste 1 shows a wider operating window than paste 2.Figure 6 shows a compilation of Corescan mappings for pastes 8 and 10 as a function of the wafer temperature. The Corescan pictures for the settings CBH, CBM, and CBL have been used. For maximum visibility of scan values, the range of the color scale is increased to 150mV. For both pastes, the contact resistance decreases with increasing temperature. Figure 6: Contact resistance mappings as function of wafer temperature for pastes 8 and 10. Temperature scale is not linear.The local contact resistance data obtained with the Corescan are related to the local wafer temperatures obtained with the TCs for pastes 1, 2, 3, 8, 9, and 10 (figure 7). Some of the pastes reach a minimum R c . Pastes 8 and 9 still need a higher temperature to reach a minimum, as could also be observed for paste 8 in the Corescan mapping in figure 6.Figure 7: Local contact resistance as function of measured wafer temperature for pastes 1, 2, 3, 8, 9, and 10.Based on these results a final firing step is performed with a flat profile to assess optimal furnace settings. In this case it was decided only to vary the belt speed, in effect increasing the cell temperature.3.4 Results after final firingFor the 10 pastes again two cells were fired at the three settings as concluded above: same set temperature, three different belt speeds, but now with the side heaters on and the cells positioned at the center of the belt. So subjected to a flat cross belt temperature profile. In table 2 the best fill factors from the J-V analyses are given per paste and setting.Pastes 1, 2, 3, 8, and 10 improved with the used settings in their contacting behavior, leading to better FFs. Pastes 1, 2, 3, and 10 perform well for the three belt speed settings. The FF values of these pastes change less than 2%, showing that these pastes are not very sensitive1046to temperature profile changes. On the other hand it seems that paste 8 has not reached its maximum yet. The FF for paste 8 differs from 63 - 47% for different belt speeds.Belt speedV-15[cm/min]V[cm/min]V+15[cm/min]Paste FF [%] FF [%] FF [%]1 76.4 76.3 76.32 67.5 66.9 68.23 75.8 75.9 76.14 28.1 28.3 27.35 28.7 28.1 27.26 36.4 33.6 34.57 34.4 31.0 31.58 63.1 52.0 46.79 36.5 38.4 36.710 71.7 71.7 70.8 Table 2: Fill factors from J-V results.Pastes 4, 5, 6, 7, and 9 do not lead to FF results >60%. Different firing settings outside the scope of this work might be successful.To summarize, the best results from the J-V analyses are given in table 3 together with the line resistance R L. J-V data are measured with two Kelvin probes per busbar and not corrected for mismatch.Since the paste rheology is different per paste, and only one screen printer setting has been used, the print definition of the front side pattern might not be the same for the pastes. This is reflected in the line resistance. The line resistance also might show that there is some difference per paste in the silver conduction of the fingers. Calculations [9] show that for paste 10, about 2.4% relative is added to a fill factor loss by the high line resistance value as compared to using the value of paste 1.PasteJ-VFF [%]Suns-Voc FF[%]eta[%]Jsc[mA/cm2]Voc[mV]R L[Ω/cm]1 76.4 81.6 16.7 35.61 613 0.4432 68.2 81.9 14.8 35.41 611 0.5493 76.1 81.4 16.5 35.43 614 0.4024 28.3 82.0 5.1 29.35 618 0.6485 28.7 81.9 5.2 29.48 616 0.6976 36.4 82.6 7.8 35.01 612 0.6397 36.9 82.1 7.9 35.12 613 0.6318 63.1 82.0 14.0 36.30 613 0.7469 38.4 81.9 8.4 35.67 615 0.63110 71.7 81.4 16.1 36.80 612 1.033 Table 3: Best cell results.With the new approach it was found that cell results for paste 1 and 3 are near optimal. Pastes 2, 8, and 10 proveto be viable pastes but would need further statistical optimization on larger number of cells in case of industrial application.4 CONCLUSIONSA new method for firing optimization, using a cross belt temperature distribution over the cell, has been successfully used to assess the capabilities of pastes to make cells with good electrical contacts. With only three cells, a scan over a temperature range of about 180°C has been performed. Also in this way the firing window could be assessed.The new approach, and inherent use of the Datapaq and the Corescan, is applicable in furnaces for production or R&D and can be used as a time saving alternative for existing procedures for firing optimization.5 REFERENCES[1] For details on Datapaq furnace temperature profiler,see: [2] "Simple, detailed & fast firing furnace temperatureprofiling for improved efficiency", Hoornstra, J.;Heide, A.S.H. van der; Bultman, J.H.; Weeber, A.W.,Proceedings PV in Europe Conference and Exhibition, Rome, Italy, 2002.[3] "16% mc-Si Cell Efficiencies using Industrial In-LineProcessing", A.W. Weeber, et al., to be presented atthis conference, Paris, 2004.[4] A.S.H. van der Heide, J.H. Bultman, J. Hoornstra andA. Schönecker, Solar Energy Materials and SolarCells 74, p. 43, 2002.[5] For details on Corescan instrument and applications,see: www.sunlab.nl[6] "Contact Optimisation on Lowly Doped Emittersusing the Corescan On Non-Uniform Emitter Cells",Heide, A.S.H. van der, Goris, M., to be presented atthis conference, Paris, 2004.[7] A.S.H. van der Heide, J.H. Bultman, J. Hoornstra, A.Schönecker, G.P. Wyers and W.C. Sinke, Proceedings 29th IEEE Photovoltaic Specialists Conference, New Orleans, USA, 2002.[8] "Possibilities for Process-control Monitoring ofElectronic Material Properties during Solar Cell Manufacture", Sinton, R.A., Proceedings of the 9th NREL workshop on Crystalline Silicon Materials andProcesses, p.67 - 73, 1999.[9] "Optimizing Metalization Patterns", A.R. Burgers,J.A. Eikelboom, Proceedings 26th IEEE PhotovoltaicSpecialists Conference, Anaheim, USA, 19976 ACKNOWLEDGEMENTSDatapaq Ltd, Cambridge, UK, is gratefully acknowledged for the use of the equipment.This work is supported by the EC in the frame of the FP5 program within the EC2Contact project (#ENK6-CT-2001-00560). Partners Metalor Technologies and RWE Schott Solar are acknowledged for input of pastes and cell material.1047。
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Section 1.Synthesis,structure and properties of glassesA new approach for calculating the electronic structure and vibrational properties of non-crystalline solids:Effective charges for infrared-active normal mode vibrations in oxide and chalcogenide materialsG.Lucovskya,*,L.S.Sremaniak b ,T.Mowrer a ,J.L.WhittenbaDepartments of Physics,Electrical and Computer Engineering,Materials Science and Engineering,North Carolina State University,Raleigh,NC 27695,USAbDepartments of Chemistry,North Carolina State University,Raleigh,NC 27695,USAAbstractThis paper applies ab initio calculations to small clusters with short range bonding order for oxide and chalcogenide dielectrics that form continuous random networks (CRN),including,respectively,SiO 2and GeO 2,and As 2S 3and GeS 2.The calculations have been used to determine the dependence of the total energy on the bond-angle at the twofold coordinated O-and S-atom sites,and the infrared effective charges of the normal mode network vibrations associated with symmetry determined asymmetric bond-stretching,symmetric bond-stretching (bending)and out-of-plane bond rocking displacements,including decomposition of effective charges in equilibrium distribution,and dynamic charge redistribution components.The results of these calculations are in very good to excellent agreement with experiment.The approach is extended to silicates and their chalcogenide analogs in which pseudo-binary alloy formation causes a disruption of the CRN hosts through formation of ionic bonding arrangements of the metal atoms.This is accom-panied by increases in the coordination of O-,S-,Se-and As-atoms of the respective oxide and chalcogenide host CRNs,including SiO 2and As 2Se 3,and leads to a breakdown of 8ÀN bonding coordination.Ó2003Published by Elsevier B.V.1.IntroductionEmpirical force constant analyses of infrared (IR)and Raman scattering spectra have focused on significant differences between non-crystalline ox-ides and chalcogenides in spite of the fact that Si,As and Ge oxides,and As and Ge sulfides and selenides form continuous random networks (CRNs)with equivalent 8ÀN rule bonding coor-dination [1,2].This paper demonstrates through ab initio electronic structure calculations that differ-ences between oxides and chalcogenides are Ôa di-rect result of differences in their electronic structures Õ.In particular differences in total energy as a function of the bond-angle at the twofold co-ordinated O-,S-and Se-atom bonding sites corre-late directly with the Pauling bond ionicity [3].The*Corresponding author.Tel.:+1-9195153468;fax:+1-9195157331.E-mail address:gerry_lucovsky@ (G.Lucovsky).0022-3093/$-see front matter Ó2003Published by Elsevier B.V.doi:10.1016/S0022-3093(03)00369-7Journal of Non-Crystalline Solids 326&327(2003)1–14IR effective charges of O-and S-atoms also reflect quantitative differences between equilibrium charge distributions,and dynamic changes in these dis-tributions during normal mode motions.The nor-malized equilibrium charge distributions obtained from both experiment and the ab initio theory also show scaling trends with bond-ionicity.The ab inito calculations have been extended to non-crystalline oxides in which the atomic coordi-nations do not conform to the8ÀN rule[4].Non-crystalline alloys comprised of group IIIB and IVB transition metal oxides,and rare earth oxides, such as Y(La)2O3and Zr(Hf)O2,and Gd(Dy)2O3, respectively,and SiO2and Al2O3,have been pro-posed as candidate alternative high-k gate dielec-trics for aggressively scaled advanced Si devices with decreased lateral dimensions and increased levels of integration[5].This has lead to the de-velopment in Ref.[4]an empirical classification scheme for non-crystalline oxide dielectrics that includes these oxides as well the CRN oxides identified above.This classification scheme of Ref.[4]separates non-crystalline oxides into three different bonding groups[6],where the average Pauling electroneg-ativity differences,D X,bond-ionicities,f i,provide convenient metrics for this separation.Non-crystalline oxides with D X up to about1.5form covalently bonded CRNs,in which the constituent atoms have a coordination that reflects their pri-mary chemical valence;e.g.,two for O,three for P and As,and four for Si and Ge[6,7].The bonding in these CRN oxides obeys the8ÀN rule independent of any p-bonding contributions that arise from back donation into d-like polarization states of the P,As,Si or Ge atoms.The glass forming character of these materials,as well as their low defect den-sities in thinfilms and bulk glasses,is correlated with the number of bonding constraints per atom and has been discussed in detail in a series of sem-inal papers by Phillips and his coworkers[8–11].The classification scheme has also provided in-sights into the local chemical bonding arrange-ments in the small clusters used in ab initio calculations.These calculations have established correlations between the energies of transition metal atomic states,and the conduction band offset energies between(i)transition metal oxides,and silicate and aluminate alloys and(ii)Si[12].These offset energies determine(i)the barrier for direct tunneling between the Si substrate and the transi-tion metal dielectric,and(ii)the activation energy for injection into localized transition metal anti-bonding states in the silicate and aluminate alloys. In addition the scheme has lead to a new under-standing of the fundamental electronic structure of transition metal and rare earth oxide non-crystalline alloys,which can be transported and applied to non-crystalline chalcogenides as well. For example,it establishes a theoretical approach for electronic structure of amorphous transition metal chalcogenide alloys in which the coordina-tions of constituent atoms fall outside of the con-fines of the8ÀN rule that applies to amorphous Se,and As2S(Se)3,GeS(Se)2.This establishes that atomic coordination is a property of alloy compo-sition,and differences with respect to8ÀN coor-dination can be understood with ab initio quantum chemistry calculations[13].The classification scheme also has provided the stimulus for revisiting the infrared and Raman response of non-crystalline oxides and chalcogenides,and identifying the mi-croscopic origin of differences in the infrared(ir) effective charges of the respective normal mode O-atom,and S-and Se-atom motions.The specific objectives of this paper are twofold: (i)to apply ab initio calculations to relatively small clusters that characterize the bonding of O-and S-and Se-atoms in dielectrics with continuous ran-dom structures in order to determine(a)the total energy as a function of the bond-angle at the twofold coordinated oxygen and chalcogen atoms sites,and(b)the IR effective charges of the normal mode vibrations of the CRNs,and(ii)to extend the concepts developed in(i)to the metal–atom modified/disrupted silicate and metal chalcogenide pseudo-binary alloys with As2S(Se)3and GeS(Se)2.2.Analysis of IR spectra of oxides and chalcoge-nides with CRN morphologiesAnalyses of the IR reflectance and absorption spectra of fused silica glass(SiO2),thermally relaxed thinfilms of SiO2annealed at900°C [14,15],and bulk glasses of As2S3[16],As2Se3[16],2G.Lucovsky et al./Journal of Non-Crystalline Solids326&327(2003)1–14and GeS2[17,18],have yielded the optical fre-quency dielectric functions,e1and e2,from which the vibrational frequencies and IR effective charges in Table1have been determined for the normal mode vibrations displayed in Fig.1.The effective charges for the normal mode vibrations of GeO2 and GeSe2,not included in Table1,are similar to those of SiO2and GeS2,respectively,but the normal mode vibrational frequencies reflect the increase mass of Ge with respect to Si,and Se,with respect to O and S[14,19].Table1specifically in-cludes the experimentally determined frequencies for the asymmetric bond-stretching,symmetric bond-stretching(some times also referred to as bending),and out-of-plane bond-rocking modes. It is important to note that the IR effective charges in Ref.[16]were determined with respect to a particular empirical force constant model that does reflect the correct mode masses associated with the normal mode vibrations of the CRN structure.This has been remedied in Table1.The values of the effective charges for As2S3in Table1 are therefore different from those reported in Ref.[16].The effective charges of Table1were obtained directly from the experimentally determined values of e2by through the usual definition of the ma-croscopic IR effective charge,eÃ,are given by Eq.(1),eÃi¼1:43Â10À6m iðS iÂm i=N iÞ0:5;ð1Þwhere S i is the oscillator strength,m i is the mode mass and m i is the vibrational frequency,all of the i th vibrational mode,and N i is the density of os-cillators per unit volume and is equal to the oxygen or chalcogenide atom density is this normal mode representation[20].The mode masses reflect the respective motions of(i)the Si-,Ge-or As-atoms, and(ii)the O-,S-or Se-atoms in the network normal mode vibrations in Fig.1that satisfy conservation of momentum constraints.The mode masses for all of the Si–O,Ge–S and As–S normal mode vibrations in SiO2,GeS2and As2S3,respec-tively,are the same,and are given in Eqs.(2a)–(2c),mðSiO2Þ¼2mðSiÞmðOÞ=ð2mðSiÞþmðOÞÞ;ð2aÞmðGeS2Þ¼2mðGeÞmðSÞ=ð2mðGeÞþmðSÞÞ;andð2bÞmðAs2S3Þ¼3mðAsÞ2mðSÞ=ð3mðAsÞþ2mðSÞÞ;ð2cÞTable1Experimentally determined normal mode frequencies and IR effective chargesMode SiO2(f i¼0:45)frequency(±4cmÀ1)Effectivecharge(±0.1e)GeS2(f i¼0:08)frequency(±3cmÀ1)Effectivecharge(±0.1e)As2S3(f i¼0:05)frequency(±3cmÀ1)Effectivecharge(±0.1e)Asymmetricstretch1075 2.29367 1.53309 1.30Symmetric stretch8100.933280.523400.61 Out-of-plane rock4600.921490.281620.29 Ratio m AS=m SS 1.33±0.02– 1.12±0.02–0.91±0.02–Ratio eÃAS =eÃSS– 2.45±0.4– 2.94±0.7– 2.14±0.6Ratio eÃAS =eÃR– 2.48±0.4– 5.52±1– 4.48±1G.Lucovsky et al./Journal of Non-Crystalline Solids326&327(2003)1–143where the chemical symbols denote either the designated compound mode mass in atomic units, or the designated atomic mass in the same units.In spite of large Pauling bond-ionicity(f i)dif-ferences between Si–O,and the As–S(Se)and Ge–S bonds,there are relatively small differences in their respective IR effective charges.Equally impor-tantly these charges show significant differences for the different normal modes.This has previously been noted in Ref.[14]where it was proposed that dynamic charge transfer,or effectively dynamic rehybridization of bonding orbitals[21]occurred during normal mode motions and was responsible for the relatively large IR effective charges for the bond-stretching normal modes of As2S3and As2Se3.Table1also includes ratios of the vibra-tional frequencies of the asymmetric and sym-metric stretching modes,m AS=m SS,and ratios of the effective charges for the asymmetric and symmetricstretching modes,eÃAS =eÃSS,and the asymmetricstretching and bond rocking modes,eÃAS =eÃR.Thesignificance of these ratios for the effective charges will become apparent as the electronic structure is determined from the ab initio calculations pre-sented below.The ratio of m AS=m SS for SiO2is smaller than anticipated from central and valence forcefield empirical force constant calculations, but is in better agreement with values that can be determined directly from ab initio calculations like those reported below.This paper an outline of the ab initio quantum chemistry calculations of the electronic structure [22,23]that have been used for the determination of the IR effective charges of the normal mode motions of the twofold coordinated O-,S-and Se-atoms as they occur in these CRNs.The theory has also been applied to disrupted CRNs including pseudo-binary silicates such as SiO2and ZrO2,and the pseudo-binary chalcogenides such as As2Se3–NiSe[24–26]and As2Se3–Cu2Se[27].3.Ab initio calculationsThe ab initio calculations arefirst applied to IR effective charges(also sometimes designated as dynamic dipoles)of SiO2to illustrate the method and new insights that can be obtained from this approach[28].In particular,a decomposition of these IR effective charges into equilibrium charge and charge redistribution contributions is demon-strated.The electronic structure calculations are ab initio in character,and employ variational methods in which an exact Hamiltonian is used. They are qualitatively different from the empirical tight-binding and valence forcefield approaches used for the respective CBL electronic and vibra-tional calculations of Joannopoulos and cowork-ers[29,30],and those of Ref.[19].In addition no core potential,or exchange approximations are assumed[22,23].The Hamiltonian,H,is given in Eq.(3),H¼R iðÀ1=2r2iþR kðÀZ k=r ikÞÞþR i<j1=r ij;ð3Þcontaining kinetic energy(1=2r2i),nuclear-elec-tron attraction(ÀZ k=r ik)and electron–electron re-pulsion(1=r ij)contributions.The calculations are done at two levels,initially through a self-consistentfield(SCF)Hartree–Fock calculation with a single determinant wave func-tion of the general form,W¼C detð/að1Þ/bð2Þ/cð3Þ/dð4Þ.../xðNÞÞ;ð4Þwhere functions/are molecular orbitals contain-ing spatial and spin components.This approach does not include effects due to electron correlation. Following this,there is a configuration interaction (CI)refinement of the bonding orbitals that is based on a multi-determinant expansion wave function,W¼R i c i U i;ð5Þwhere the U i denote determinants of the form specified in Eq.(4)with different choices of occu-pied molecular orbitals.Expansions of this type include electron correlation effects.The expansion coefficients c i are determined by energy minimi-zation.Since the Hamiltonian is exact,through electrostatic interactions,the energy variational theorem is satisfied.The SCF and CI calculations have been applied to the clusters in Figs.2and3.The local cluster in Fig.2is embedded mathematically in a larger net-work structure through a1-electron embedding potential VðrÞ,and basis functions S1and S2, which are represented by SiÃin the diagram.The4G.Lucovsky et al./Journal of Non-Crystalline Solids326&327(2003)1–14functions S1and S2are Gaussian expansions of long and short range functions representing an sp 3hybrid orbital of Si.The mixing parameter,k ,in S1þk S2,is set such that all Si core energy levels are correct.This also assures a correct overall charge distribution in the cluster,so that the clus-ters in Fig.2in 3have zero dipole moments.The approach is conceptually similar to the cluster Bethe lattice (CBL)approach of Ref.[29,30];however,for these ab initio calculations,the Hamiltonian is exact and the calculation is ab initio rather than empirical.The CBL and the embedding potential approach of this paper are each used to emulate the properties of the CRN amorphous morphology of non-crystalline SiO 2[4,6].3.1.Total energyThe total energy for the cluster in Fig.2has been calculated as a function of the Si–O–Si bondangle a (see Fig.4)at three different computational levels:(i)SCF with d-polarization on the Si atoms,(ii)SCF +CI with no d-polarization,and (iii)SCF +CI with d-polarization.Since the Hamilto-nian is exact,any increase in the total energy through a variational process represents an im-proved solution to that calculation,and addition-ally to the IR effective charge calculations that use the equilibrium cluster geometry as a point of reference.This variational approach to binding energy optimization unequivocally establishes that the SCF +CI with d-polarizations is the best ap-proach for calculating the total binding energy,and therefore for other properties of the cluster to be addressed as well.It is significant to note that the calculated minimum in total energy occurs at a Si–O–Si bond angle of approximately 145–150°,very close the average bond angle determined experimentally for vitreous SiO 2,144°[31].In addition,the broad minimum in total energy is consistent with radial distribution functions studies on fused silica that indicate that the Si–O–Si bond angle distribution is broad,extending from approximately 120–180°.It is also consistent with the unusually small value of the empirical Si–O–Si three-bodybond-bendingG.Lucovsky et al./Journal of Non-Crystalline Solids 326&327(2003)1–145force constant[6,8–11].The spread in bond-angle also contributes significantly to the configurational entropy that promotes the high metastability of non-crystalline SiO2.The calculated total energy dependence on the Si–O–Si bond angle a adds additional validity of this local cluster approach for other properties which depend primarily on the local Si–O–Si bonding geometry.It also indicates that the terminal O-atom groups and embedding terminal atom SiÃfunctions emulate the connec-tivity of the Si–O–Si group to SiO2CRN.3.2.IR effective chargesThe IR effective charges,D l j=D q i,q i¼x;y and z,j¼x;y;z have been calculated for the asym-metric bond-stretching(AS)vibration,m y,symmet-ric bond-stretching(SS or B),m z,and out-of-plane bond-rocking(R),m x,infrared active vibrations of non-crystalline SiO2[14].The normal mode mo-tions of O-atoms of these vibrations are defined with respect to the coordinate system displayed in Fig. 1.For the asymmetric bond-stretching vi-bration,m y,the normal coordinate is y,and therelative motion of the O-atom is parallel to a line joining the two Si-atom neighbors.For the sym-metric bond-stretching vibration,m z,the normal coordinate is z,and the relative motion of the O-atom is in the direction of the bisector of the Si–O–Si bond angle,a.Finally,for the bond-rocking vibration,m x,the normal coordinate is x,which is perpendicular to the plane of the Si–O–Si bonding group.The IR effective charges or dynamic dipoles D l y=D y,D l z=D z and D l x=D x have been calculated from SCF and CI calculations,with and without d-polarizations,and as a function of a.Hereafter the subscripts x,y and z will not be used for the normal mode dynamic dipoles defined above.The displacements x,y and z for the calculations were <5%of a nominal Si–O bond length of0.16nm,or $3atomic units(1a.u.¼0.053nm).The dynamic dipoles in Fig.5are in units of the electronic charge(e)to facilitate comparisons with other theoretical and experimental studies.The inclusion of the d-symmetry polarizations increases the magnitude of the dynamic dipole for the asym-metric stretching mode with the normal coordinate displacement,y.The dynamic dipoles for the other two normal modes are each significantly smaller than that of the asymmetric stretching mode,and are decreased in magnitude when the d-polariza-tions are included.For the calculations including the d-symmetry polarizations,D l=D x is essentially independent of a,whilst D l=D y,D l=D z and dis-play opposite dependencies on a.D l=D y decreases (becomes more negative)with increasing a, whereas D l=D z increases,but with a total positive change in D l=D z that is about a factor of7smaller than the total negative change in D l=D y,$0.1e for D l=D z as compared to$0.7e for D l=D y.The influence of next nearest neighbor Si–O–SiÃbond angles on the dynamic dipoles for a¼150°is displayed in Fig.6.Changes in the dynamic di-poles are about5%of their150°values for±30°excursions of the Si–O–SiÃbond angle.This dem-onstrates that the dynamic dipoles are determined primarily by nearest neighbor bonding within the Si–O–Si portion of the cluster,and are effectively insensitive to the details of the next-nearest neighbor O-atom bonding arrangements with embedding SiÃterminations.This establishes that6G.Lucovsky et al./Journal of Non-Crystalline Solids326&327(2003)1–14the IR effective charges for the infrared active normal modes of SiO2are a local property of the CRN amorphous morphology.3.3.Decomposition of IR effective chargesThese calculations compare the IR effective charges for two different electronic structures.The total IR effective charges for the normal mode displacements have been determined by comparingthe equilibrium structure x,y and z dipole moments for a given bond angle a with the dipole moments of dynamic structures that include respectively the x,y and z displacements of the normal modes.This was done at(i)SCF with d-polarizations,(ii) SCF+CI without d-polarizations,and(iii)SCF+ Ci with d-polarization levels.For the determina-tion of the equilibrium charge density contribu-tion,D l=D q ieq,the change in the dipole moment was calculated for normal mode displacements of the O-atom orbitals,with all other orbitalsfixed. The contribution due to the orbital displacements, i.e.,theÔdynamic rehybridization of the bondingorbitalsÕ,D l=D q ioc,is then the difference betweenthe total IR effective charge,D l=D q itotal,and the contribution from the equilibrium charge densityshift,D l=D q ieq,as in Eq.(5),D l=D q itotal¼D l=D q ieqþD l=D q ioc:ð6ÞThese calculations have been performed for two different Si–O–Si bond angles,a¼150°,the opti-mum bond angle,and for a¼180°.The results of these calculations are also included in Table2.The average value of the equilibrium charge density contribution is about35%larger(a¼150°)for the asymmetric stretching mode with the normal co-ordinate y,than the symmetric stretching mode and rocking modes with respective normal mode coordinates,x and z.These equilibrium charge density contributions are negative,consistent with the Pauling electronegativities of1.90for Si,andTable2Decomposition of IR effective charges aa¼150°Effective charges(e)Mode Equilibrium charge Orbital change Totalx,rocking)1.0+0.1)0.9z,symmetric stretch)1.0+0.1)0.9y,asymmetric stretch)1.3)1.2)2.5a¼180°Mode Equilibrium charge Orbital change Totalx,rocking)1.0+0.1)0.9z,symmetric stretch)1.0+0.1)0.9y,asymmetric stretch)1.4)1.3)2.7a The calculations have precision of at least one part in a hundred;however,a realistic assessment of the model is consistent with uncertainty of the order of at least0.1e.G.Lucovsky et al./Journal of Non-Crystalline Solids326&327(2003)1–1473.44for O[32].Finally the orbital change contri-butions for the symmetric stretching and rocking modes are relatively small and positive,and de-crease the magnitude of the total dynamic dipole, whilst the orbital change contribution for the asymmetric mode is more than an order of mag-nitude larger,and is negative,so that it increases the magnitude of the total dynamic dipole.There is also an increase of about10%in the orbital change contribution when the Si–O–Si angle is increased from150°to180°,whilst the equilibrium density contribution increases about7%.Finally,these decompositions demonstrate a significant charge redistribution contribution for asymmetric bond-stretching modes in which the two bonds of the respective bridging-atoms change in different directions,one increasing and the other decreasing(Fig.1).In contrast,the charge redis-tribution contributions of the symmetric stretching and out-of-plane rocking modes,in which both Si–O bond lengths change in the same sense,are sig-nificantly smaller.3.4.Total energy calculationThe total energy ab initio calculations of this paper represent an improvement over the more empirical approaches in previous publications. The calculations of Pantelides and Harrison[33], based on empirical tight-binding theory gave an optimum a of90°,which is a significant depar-ture from the experimentally determined average Si–O–Si bond angle of approximately145–150°[31].However,attempts to improve the tight-binding empirical calculations by Harrison through the addition of an empirical Si–Si repul-sion term resulted in a small improvement,in-creasing a to approximately114°[34].Revesz and Gibbs performed ab initio calcula-tions on a pyrosilicate acid molecule,(HO)3Si-OSi(OH)3or equivalently H6Si2O7,at the SCF level[35].This molecule terminates the Si–O–Si arrangement of interest with terminal OH groups. Their calculation specifically included Si3d-states in the basis set,as opposed to the two sets of Gaussians used in these calculations.They ob-tained results that are qualitatively similar to those presented in this paper in Fig.4;i.e.,their results also display a shallow minimum in binding energy, but at approximately140°rather than145–150°as in our calculations[35].The energy of this mini-mum of about5%lower than the value of the binding energy at180°,whilst in Fig.4,it is lower by<1.5%for the SCF+CI with d-polarization calculations.The calculations of Lucovsky and Yang in Ref.[36]at the SCF+CI level,but without the Si d-polarizations,give essentially the same results as in corresponding curve of Fig.4.These calcula-tions show a continuous improvement in binding energy between120°and180°,with no shallow minimum in the vicinity of150°or elsewhere.As evident in Fig.4,the inclusion of the Si d-polar-ization gives a large,$2.7eV,increase in the total energy,and also produces a shallow minimum in binding energy close to the experimentally deter-mined Si–O–Si value of145–150°.From a comparison between the calculations of Lucovsky and Yang[36],Revesz and Gibbs[35], and those presented in this paper,it can be con-cluded that the inclusion of Si3d states as in Ref.[35],and Si d-symmetry polarizations as in this paper,appear to be the determinant factor in generating the shallow minimum in the binding energy curve.3.5.Calculations of total energy and IR effective charges for chalcogenidesSimilar ab initio calculations have been per-formed for GeS2and As2S3based on clusters that are the analogs of those in Fig.2.The primary differences between these calculations and those for SiO2are quantitative in nature and relate the differences in the basis sets for the oxygen and sulfur atoms.Differences between the atomic basis sets required for energy minimization for O-and S-atoms contribute very little to differences in the Si–O–Si,and the Ge–S–Ge and As–Ge–As bond angles,and the angular dependence of the total effective charges;whilst differences in bond ionicity between the oxides($30–50%)and sul-fides(<10%)the determinant factor.The basis sets for the As-,Ge-,Si-,O-,and S-atoms include the core levels,and respective valence bonding states,4s and4p for As and Ge,3s and3p for Si8G.Lucovsky et al./Journal of Non-Crystalline Solids326&327(2003)1–14and S,and2s and2p for O.In addition,the basissets for Si,S,As and Ge also include compactGaussian functions with d-like symmetries.Thisreflects the increased size of these atoms with re-spect tofirst row atoms such as O,and C and N.The inclusion of these Gaussian d-states differ-entiates S(and also Se)from O and is quantita-tive in nature.Table3compares the IR effective charges ofSiO2,GeS2and As2S3,including the decomposi-tion into equilibrium and charge redistributioncomponents.Fig.7display energy as a function ofbond angle for calculations at the SCF and CIlevels with and without d-state symmetry contri-butions to the basis for S,for Ge–S–Ge andAs–S–As bonding arrangements,respectively,andthese should be compared with Fig.4for Si–O–Si.The marked differences between the total energyversus bond angle for SiO2,and GeS2and As2S3derive from two effects:(i)predominantly fromdifferences in bond-ionicity,and(ii)weakly fromthe inclusion of the Gaussian d-state polarizationsfor the sulfur-atoms.Fig.8is a plot of bond angleversus Pauling bond-ionicity that demonstratesthat the inclusion of d-symmetry contributions tothe basis sets for Si,Ge and As,and S yields acalculated bond angle that agrees with experiment,and furthermore,additional experimentally deter-mined bond angles for GeO2and As2O3combinedwith the calculations indicate a linear scaling be-tween bond angle and Pauling bond ionicity.Similar considerations apply to the variation of IReffective charges,where the results for Ge–S–Geand As–S–As are presented in Fig.9and should becompared with Fig.5for Si–O–Si.Table3Calculated effective charges for O-atoms in SiO2and S-atoms in GeS2and As2S3Mode SiO2GeS2As2S3Total (±0.1e)Equili-brium(±0.1e)Charge re-distribution(±0.1e)Total(±0.1e)Equili-brium(±0.1e)Charge re-distribution(±0.1e)Total(±0.1e)Equili-brium(±0.1e)Charge re-distribution(±0.1e)Asymmetricstretch)2.51)1.24)1.27)1.30)0.80)0.50)1.31)0.59)0.72Symmetricstretch)0.93)0.98+0.05)0.54)0.56+0.02)0.37)0.36)0.01Out-of-plane rock )0.89)0.96+0.07)0.48)0.39)0.09)0.15)0.35+0.20G.Lucovsky et al./Journal of Non-Crystalline Solids326&327(2003)1–149。