SOLUTION SET Study questions for MSE230 Midterm I 1 Compute the percentage ionic character and percentage covalent character of the interatomic bonds for each of the following compounds MgO GaP CsF CdS and FeO From Callister Chapter 2 What is the primary bonding type Solution The percent ionic character is a function of the electron negativities of the ions XA and XB according to Equation 2 10 The electronegativities of the elements are found in Figure 2 7 For MgO XMg 1 5 and XO 3 5 and therefore IC 1 e 0 25 3 5 1 2 2 100 73 4 Similar calculation results for other materials as tabulated below Material MgO GaP CsF CdS FeO EN 2 3 0 5 3 3 0 8 1 7 Ionic character 73 4 6 1 93 4 14 8 51 4 Bonding type Ionic Covalent Ionic Covalent Ionic covalent 2 Specify the primary type of bonding for the given elements Na C Ar Ge and Ni Solution Na Metallic C Covalent Ar Van der Waals Ge Covalent Ni Metallic 3 The boiling temperatures for some elements and molecules are plotted below How would you explain a the increasing trend for Ne Ar Kr Xe in terms of bonding b the increasing trend for CH4 SiH4 GeH4 SnH4 in terms of bonding c the anomaly of water H2O Hint Think of the factors that affect the van der Waals type of bonding 1 Solution a As the atomic number gets larger the number of electrons in the atom increases and the radius of the atom increases The induced dipole moment which is caused by the movement of electrons is the product of the charge and the distance between the charges If the atomic radius goes up then the vdW bonding increases Therefore the boiling point goes up as the size and atomic number goes up b Similar to a larger molecules are easier to polarize and thus have larger van der Waals forces resulting in higher boiling temperature c The permanent dipole within the water molecule results in large attractive forces between the hydrogen and oxygen in different water molecules This special type of bond is called hydrogen bond The effect of hydrogen bonds in water outweighs the size dependent effect of van der Waals forces and causes the anomaly of water 2 4 How are individual directions planes and families of directions planes denoted Fill in the table xxx xxx xxx xxx Direction Individual xxx Family xxx Plane xxx xxx 5 a What is the family of planes for 111 111 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 Note that the negatives of planes are identical to each other i e 1 1 1 1 1 1 b What is the family of direction for 110 110 1 1 0 1 1 0 1 1 0 1 1 0 1 0 1 1 0 1 1 0 1 1 0 1 0 1 1 0 1 1 0 1 1 0 1 1 c Which directions in the 110 family are located on 1 1 1 plane In a cubic system the vector that is normal to the plane has the same indices as the plane Accordingly 1 1 1 is the normal of 1 1 1 A direction a b c that is located on a plane is perpendicular to the planes normal Therefore it follows a b c 1 1 1 i e the angle between the two is 90 The angle between two vectors u1 v1 w1 and u2 v2 w2 is given as Since cos 90 1 the members of the 110 that are located on 1 1 1 are 110 011 1 1 0 and 0 1 1 6 Below is the geometry for the 110 for BCC a Express the length of the edges in terms of the lattice parameter a and equilibrium atomic radius r b Draw the positions of the atoms and indicate where the atoms touch c Draw the 0 0 1 1 1 0 and 1 1 1 3 Solution 7 Below is the 111 for FCC a Express the lengths of the edges in terms of the lattice parameter a and atomic radius r b Draw the positions of the atoms in the plane and where they touch c Draw 1 0 1 1 0 1 and 0 1 1 on the plane Solution 8 For a cubic crystal put the given dhkl values d200 d101 d220 d311 d110 d100 d111 in order from largest to smallest Solution d100 d110 d101 d111 d200 d220 d311 4 9 Figure below shows the first five peaks of the x ray diffraction pattern for tungsten which has a BCC crystal structure monochromatic x radiation having a wavelength of 0 1542 nm was used a Index i e give h k and l indices for each of these peaks b Determine the interplanar spacing for each of the peaks c For each peak determine the atomic radius for W From Callister Chapter 3 Solution a Since W has an BCC crystal structure only those peaks for which h k l odd will appear Therefore the peaks result by diffraction from 110 200 211 220 and 310 planes at 2 40 59 72 86 and 101 respectively b For each peak in order to calculate the interplanar spacing we must employ Equation 3 13 For the first peak which occurs at 40 d111 n 1 0 1542 nm 0 225 nm 40 2 sin 2 sin 2 The results for all peaks are tabulated below c Employment of Equations 3 14 and 3 1 is necessary for the computation of R for W as R d 3 h 2 k 2 l 2 a 3 hkl 4 4 0 225 nm 3 1 2 1 2 0 2 4 0 138 nm Similar computations are made for the other peaks which results are tabulated below Peak 1 2 3 4 5 h 1 2 2 2 3 k 1 0 1 2 1 l 0 0 1 0 0 2 degrees 40 59 72 86 101 5 d nm 0 225 0 157 0 131 0 113 0 100 a nm 0 318 0 314 0 321 0 320 0 316 r nm 0 138 0 136 0 139 0 138 0 137 10 Identify the microstructural imperfections at A B and C Solution A Interstitial impurity B Edge dislocation C Substitutional impurity 6 11 A metal specimen 300 mm long and having an elastic modulus of 200 GPa is deformed under a stress of 400 MPa with the resulting strain of 0 005 Calculate the final length of the specimen when the stress is released Solution Material is deformed under 400MPa and the strain under this tension is i 0 005 Once the stress is released the material will shrink and a permanent strain will remain f 0 The difference in the strain values can be calculated as f i 400MPa 0 002 E 200GPa Accordingly the final strain is f i 0 005 0 002 0 …
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