第二章材料的結構,中英文對照答案

1、第二章 材料的結構Chapter 2 Structure of materials1. 原子間的結合鍵共有幾種？各自特點如何？How many kinds of binding bonds among atoms? What are their characteristics?原子間的結合鍵共有5種，分別是金屬鍵、離子鍵、共價鍵、氫鍵和范德華鍵。There are five kinds of binding bonds among atoms, namely, metallic bond, ionic bond, covalent bond, hydrogen bonding and van

2、der Waals bond.1）金屬鍵是金屬中的正離子和自由電子之間形成的鍵合。其特點是電子共有化，可以自由流動。金屬鍵無方向性和飽和性。金屬鍵合力較強，鍵能為幾百kJ/mol。Metallic bond is formed between the positive ions and free electrons. Metal bond is characterized by many sharable electrons and free mobile electrons. It is nondirectional and unsaturated. Metal Binding force

3、in metal bond is strong and the bond energy may up to hundreds of kJ / mol.2）離子鍵是正負離子之間由于靜電吸引而形成的鍵合，離子鍵無方向性也無飽和性，配位數(shù)高。離子鍵具有較強的鍵合力，鍵能為幾百到幾千kJ/mol。Ionic bond is formed through electrostatic attraction between oppositely charged ions. It is nondirecitonal and unsaturated. It has high coordination numbe

4、r, and the bond force may up to several hundreds to thousands of kJ/mol.3）共價鍵是原子間共用電子對而形成的鍵合，有方向性和飽和性，配位數(shù)低。共價鍵的鍵強度較高，鍵能通常為幾百kJ/mol。Covalent bond is formed by sharing of pairs of electrons between atoms. It is directional and saturated. It has small coordination number and the bond energy is several

5、hundreds of kJ/mol.4）氫鍵是由氫原子與電負性較大的原子之間形成的XHY的鍵合。具有方向性和飽和性。氫鍵的鍵合較弱，一般為幾十kJ/mol，但是對材料性能的影響較大。Hydrogen bond is formed between hydrogen atom and two electronegative atoms such as NHO. It is directional and saturated. The hydrogen bond is weak, and it is only dozens of kJ/mol，but is has much influence o

6、n the materials properties.5）范德華鍵是分子間形成的一種作用力，其鍵能很弱，為幾到幾十kJ/mol。不具有方向性和飽和性。作用范圍在幾百個皮米之間。它對材料的沸點、熔點、汽化熱、熔化熱、溶解度、表面張力、粘度等物理化學性質有決定性影響。The van der Waals force is the sum of attractive or repulsive forces between molecules. It is weak and the bond energy is several to dozens of kJ/mol. The van der Waals

7、 force is nondirectional and unsaturated, and the range of the force is only several hundreds of picometers. It has much influence on materials boiling point, melting point, heat of vaporization and fusion, solubility, surface tension, viscosity and other physical and chemical properties.2. 為什么可將金屬單

8、質的結構問題歸結為等徑圓球的密堆積問題？Why the structure of metallic single matter can be considered as the close-packed accumulation of balls with equal radius?答：金屬鍵是由金屬陽離子和自由電子形成的結合鍵，它既無方向性，也無飽和性。因此，金屬晶體中的原子不存在受臨近質點的異號電荷限制和化學量比的限制，在一個金屬原子的周圍可以圍繞著盡可能多的又符合幾何圖形的臨近原子。這樣的結構由于充分利用了空間，從而使體系的勢能盡可能降低，使體系穩(wěn)定，故金屬晶體具有較高的配位數(shù)。綜合以上

9、原因，可將金屬單質的結構問題歸結為等徑圓球的密堆積問題。Metallic bond is formed between the positive metallic ions and free electrons. It is nondirectional and unsaturated. Therefore, the atoms in metal are not limited by the opposite charges on neighboring particles or by stoichiometric ratio.  Atoms can be aro

10、und as many as possible around a metal atom. In such a structure, the space is fully utilized, so as to minimize the potential energy of the system, so the system is stable, and the metal crystals always have high coordination number. For these reasons, the structure of metallic

11、single matter can be considered as the close-packed accumulation of balls with equal radius?3. 計算體心立方結構和六方密堆結構的堆積系數(shù)。Please calculate the accumulation factor of body-centered cubic and hexagonal close-packed structure. 1）體心立方結構 2）六方密堆結構1) body-centered cubic 2) hexagonal close-packed structure4. 試確定簡

12、單立方、體心立方和面心立方結構中原子半徑和點陣參數(shù)之間的關系。Please determine the relationship between the atomic radius and lattice parameter in simple cubic, body-centered cubic and face-centered cubic structure.1）簡單立方結構 a=b=c=2R a =b =g = 90°2） 體心立方結構 如題3的圖， 則a=b=c=2.309 a =b =g = 90°3）面心立方結構 點陣參數(shù)為a=b=c=2.829 a =b =

13、g = 90°5. 金屬銣（Rb）為A2型結構，Rb的原子半徑為0.2468 nm，密度為1.53g/cm3，試求晶格參數(shù)a和 Rb的相對原子質量。Structure of rubidium (Rb) belongs to A2-type, the atomic radius of Rb is 0.2468 nm, the density is 1.53g/cm3, please calculated the lattice parameter a and the relative atomic mass of Rb.答：A2型結構為體心立方結構， 則a=(4/1.732)×

14、;R0.57 nm 鎳的晶格參數(shù)為a=b=c=0.57 nm a =b =g = 90°單位晶胞原子數(shù)為2 設其相對原子質量為M g/mol，則 m/V=(2/NA)×M/V NA=6.02×1023個/molV=a3=(0.57 nm)3 已知1.53 g/cm3代入得M=85.3 g/molAnswer: A2 type structure is body-centered cubic, as shown in the figure. From the figure, we can get that a=(4/1.732)×R0.57 nm. The

15、 number of atoms in unit cell is 1+8×(1/8)=2. According to the formulam/V=(2/NA)×M/V, and as is known, NA=6.02×1023個/mol, V=a3=(0.57 nm)3,1.53 g/cm3, so M=85.3 g/mol, that is to say, the relative atomic mass of Rb is 85.3 g/mol.6. FCC結構的鎳原子半徑為0.1243 nm，試求鎳的晶格參數(shù)和密度。Radius of Ni atom in

16、 FCC structure is 0.1243 nm, please calculate the lattice parameter and density of Ni.答：FCC為面心立方結構，其中 已知R0.1243 nm 則a= 0.3516 nm 鎳的晶格參數(shù)為a=b=c=0.3516 nm a =b =g = 90°鎳的原子量為59 g/mol 單位晶胞原子數(shù)為6×1/28×1/84 NA6.02×1023個/mol則密度為m/V=(4/NA)×59 g/mol/a3=9.02 g/cm3Answer：As we known, FC

17、C is face-centered cubic structure, so = 0.3516 nm, so the lattice parameter of Ni are a=b=c=0.3516 nm, a =b =g = 90°. As you known, the atomic mass of Ni is 59 g/mol, and the number of unit cell is 6×1/28×1/84, NA6.02×1023個/mol, som/V=(4/NA)×59 g/mol/a3=9.02 g/cm3.7. 鈾具有斜方結

18、構，其晶格參數(shù)為a=0.2854 nm，b=0.5869 nm，c=0.4955 nm；其原子半徑為0.138 nm，密度為19.05g/cm3。試求每單位晶胞的原子數(shù)目及堆積系數(shù)。答：斜方結構是指正交點陣結構，其特點為abc，a =b =g = 90° 沒有指明是簡單、底心、體心還是面心？所以要根據(jù)計算得到的單位晶胞原子數(shù)來確定。查表得到鈾的原子量M238 g/mol設單位晶胞原子數(shù)為n，則單位晶胞的質量為m=(n/NA)×M 已知晶胞的體積Vabc密度19.05g/cm3 根據(jù)m/V 計算的n=4 為面心斜方結構已知原子半徑R0.138 nm 則堆積系數(shù)nVatom/V

19、cell=4×(4/3)R3/abc=0.5318. 6個O2-環(huán)繞一個Mg2+，將離子看成硬球，其半徑分別是rO2-=0.140 nm，rMg2+=0.066 nm，試求O2-之表面間距離。 待定9. 已知金屬鎳為A1型結構，原子間接觸距離為249.2 pm，請計算：1）Ni立方晶胞的參數(shù)；2）金屬鎳的密度；3）分別計算（100），（110），（111）晶面的間距。答：1）根據(jù)原子間接觸距離為249.2 pm可知，2R=249.2 pm，得R=124.6 pm，晶胞參數(shù)的計算見第6題。2）密度的計算見第6題。3）立方晶系的晶面間距為其中h, k, l 分別為相鄰晶面的米勒指數(shù)，據(jù)此

20、可得各晶面的晶面間距為：d100=a0=0.3524 nm d110= 0.2492 nm d111= 0.2035 nm10. 試計算體心立方鐵受熱而變?yōu)槊嫘牧⒎借F時出現(xiàn)的體積變化。在轉變溫度下，體心立方鐵的晶格參數(shù)是0.2863 nm，而面心立方鐵的點陣參數(shù)是0.3591 nm。Please calculate the volume change of iron when it was heated and change from body-centered cubic to face-centered cubic. As is known, under the transition te

21、mperature, the lattice parameter of BCC is 0.2863 nm, while that of FCC is 0.3591 nm.更正后的正確答案 答：根據(jù)體心立方和面心立方結構中，半徑與晶胞參數(shù)關系的不同，可知，鐵原子的半徑是會變化的，根據(jù)“體心立方鐵的晶格參數(shù)是0.2863 nm，而面心立方鐵的點陣參數(shù)是0.3591 nm。”這個條件，可以計算出體心立方時，鐵原子的半徑為0.124nm，面心立方時為0.127nm。據(jù)此數(shù)據(jù)得到的堆積系數(shù)與我們之前學過的內(nèi)容是相符的。試計算體心立方鐵受熱而變?yōu)槊嫘牧⒎借F時出現(xiàn)的體積變化。體心立方鐵的體積應為2×

22、;（4/3）××（0.124）30.01597nm 3面心立方鐵的體積應為4×（4/3）××（0.127）30.03432nm 3故，體心立方鐵受熱而變?yōu)槊嫘牧⒎借F時，體積膨脹了0.01835 nm 3體心立方鐵的單位晶胞原子數(shù)為2 則其原子堆積系數(shù)為體2×（4/3）×R3/（a體）30.714 則此時晶胞的空隙率為10.714=0.286=28.6%面心立方鐵的單位晶胞原子數(shù)為4 則其原子堆積系數(shù)為面4×（4/3）×R3/（a面）30.724 則此時晶胞的空隙率為10.724=0.276=27.6%所以

23、，體心立方鐵受熱而變?yōu)槊嫘牧⒎借F時體積膨脹了1。Answer: according to the document, the radius of Fe atom is 0.126 nm. In the BCC structure, the number of unit cell is 2, so the atomic accumulation factor is體2×（4/3）×R3/（a體）30.714, here, the porosity of the cell is 10.714=0.286=28.6%. In the FCC structure, the numb

24、er of unit cell is 4, so the atomic accumulation factor is面4×（4/3）×R3/（a面）30.724, here, the porosity of the cell is 10.724=0.276=27.6%. 28.6%27.6%1 Therefore, the volume of iron increase for 1% when it was heated and change from body-centered cubic to face-centered cubic.11. 單質錳有一種同素異形體為立方

25、結構，其晶胞參數(shù)為0.6326 nm，密度7.26 g/cm3,原子半徑r=0.112 nm，計算Mn晶胞中有幾個原子，其堆積系數(shù)是多少？答：設單位晶胞原子數(shù)為n， 則單位晶胞的質量為m=(n/NA)×M 已知晶胞參數(shù)a=0.6326 nm 則晶胞的體積為Va3 已知密度7.26g/cm3 ，Mn的原子量M54.94 根據(jù)m/V 計算的n=20 為復雜立方體結構已知原子半徑R0.112 nm 則堆積系數(shù)nVatom/Vcell=20×(4/3)R3/a3=0.46512. 固溶體與溶液有何異同？固溶體有幾種類型？What are differences between so

26、lid solution and solution? Give the types of solid solution.答：固溶體：是指一種或多種溶質組元溶入晶態(tài)溶劑并保持溶劑的晶格類型所形成的單相晶態(tài)固體。兩組元在液態(tài)下互溶，固態(tài)也相互溶解，且形成均勻一致的物質。有置換型固溶體和間隙型固溶體兩種類型。溶液：是一種或幾種物質分散到另一種物質里，形成的均一的、穩(wěn)定的混合物。溶液各處的組成和性質完全一樣；溫度不變，溶劑量不變時，溶質和溶劑長期不會分離。一般包括氣體溶液、固體溶液（固溶體）、液體溶液。Answer: Solid solution refers to single-phase crys

27、talline solid formed by one or more variety of solute elements integrated into crystalline solvent and remain the structure of solvent. Two groups are miscible both in liquid and solid state, they form homogeneous substance. There are two types of solid solution, they are substitutional solid soluti

28、on and interstitial solid solution.Solution is a homogeneous and stable mixture formed by one or several substances dispersed into another substance at molecule or ionic form. They have the same component and property. When the temperature and quantity of solvent are constant, solute and solvent can

29、 not separate from each other for a long time. Generally, solution includes gas solution, solid solution (solid solution), and liquid solution.13. 試述影響置換型固溶體固溶度的因素。答：1）離子半徑 當溶質和主晶體的原子半徑相對差值超過14%15%時，尺寸因素不利于固溶體的生成，兩固體間的固溶度是很有限的。離子尺寸相近時，生成連續(xù)固溶體。隨著離子尺寸差值的增大，固溶度下降，生成化合物傾向增大。1) Atomic radius. When the re

30、lative radius difference of main crystal atom and solute is more than 14% to 15%, it is not favorable to form solid solution, the solid solubility of the two substance is very limited. When they have similar radius, they will form continuous solid solution. With the increase of the radius difference

31、, the solid solubility decreased, there is a tendency to form compound.2）離子價 離子價對固溶體的生成有明顯的影響。兩種固體只有在離子價相同或同號離子的離子價總和相同時，才可能滿足電中性的要求，生成連續(xù)固溶體。在異價置換的系統(tǒng)里，一般來說，隨著離子價差別的增大，固溶度降低，中間化合物的數(shù)目增多。2) Valence The valence has a significant impact on the formation of solid solution. Only when the two solid has the

32、same valence or the same sum of positive or negative valence, can they meet the requirement of electrically neutral, and will they form a continuous solid solution. In a system of ionic replacement with different valence ion, in general, with the increase of the valence difference, the solid solubil

33、ity decrease, and the number of intermediate compounds increase. 14. 說明下列符號的含義。 Please explain the meaning of the following symbols. Na+格點位置的空位，帶一個負電荷 Vacancy of lattice site of Na +, have one negative charge. Cl- 格點位置的空位，帶一個正電荷。Vacancy of lattice site of Cl-, have one positive charge. Ca2+取代K+的格點位置

34、，帶一個正電荷。Ca2+ replace the lattice site of K+, have one positive charge. Ca2+占據(jù)Ca2+的格點位置，電荷為零。Ca2+ locates on the lattice site of Ca2+, it has no charge. Ca2+進入間隙位置，帶兩個正電荷。Ca2+ locates in the interstitial site, have two positive charges.15. 寫出CaCl2溶解在KCl中的各種可能的缺陷反應式。Write the possible defect interacti

35、on equations when CaCl2 dissolves in KCl.答：CaCl2中有一個Ca2+和2個Cl-，則CaCl2溶解在KCl中可能的形式有1） Ca2+取代K+，2個Cl-都處于間隙位置。 不合理 同種離子，如果有空位的話，優(yōu)先選擇空位，而不是間隙位置。2） Ca2+取代K+，一個Cl-處于間隙位置，另一個Cl-取代KCl中Cl-的位置。 合理3） Ca2+取代K+，2個Cl-都取代KCl中Cl-的位置。 合理4） Ca2+處于間隙位置，2個Cl-都處于間隙位置。 不合理 正負離子處于間隙位置，但是又沒有形成空位，不合理。5） Ca2+處于間隙位置，一個Cl-處于間隙

36、位置，另一個Cl-取代KCl中Cl-的位置。 合理？ 6）Ca2+處于間隙位置，2個Cl-都取代KCl中Cl-的位置。 合理綜上所述，可能出現(xiàn)的只有2）、3）、5）和6）四種情況。16. 說明為什么只有置換型固溶體的兩個組分之間才能相互溶解，而填隙型固溶體則不能。答：因為置換型固溶體中溶劑和溶質原子的半徑差異不大（一般小于15）；離子價相同或同號離子的離子價總和相同，以保持電中性；化學鍵性質相近，并且具有相同的晶體結構類型。而間隙型固溶體的形成無需滿足以上四個條件，此外，晶體中間隙位置是有限的，容納雜質質點能力10%，并且間隙式固溶體的生成，一般都使晶格常數(shù)增大，增加到一定的程度，使晶格變得不

37、穩(wěn)定而離解。所以只有置換型固溶體的兩個組分可形成連續(xù)固溶體。In a substitutional solid solution, there are little difference the radius of solvent and solute atoms (generally less than 15%); they have same valence or the sum of the negative or positive ions is same in order to maintain electrical neutral. The bond property is sim

38、ilar, and they have the same crystal structural type. However, Formation of interstitial solid solution does not need to meet above four conditions. In addition, the interstitial sites in the crystal is limited, the capacity to accommodate impurity particle less than 10%. The formation of interstiti

39、al solid solution makes the lattice constant increases, when it increase to a certain extent, the lattice becomes unstable and dissociation. Therefore, only two components of substitutional solid solution can form a continuous solid solution.17. 銅的空位形成能為1.7×10-19 J，試計算1000時，1 cm3的銅中所包含的空位數(shù)。已知銅的密度為8.9 g/cm3，相對原子量為63.5，波爾茲曼常數(shù)k=1.38×10-23J/K。答：已知1 cm3的質量為m=V=8.9 g/cm3 ×1 cm38.9 g則其中質點的總數(shù)目NNA×（m/M）6.02×1023×8.9/63.5=8.44×1022 個1000時1 cm3的銅中包含的空位數(shù)為NdN exp (Ed/kBT)5.293×101818. 在Fe中形成1 mol空