Session #15: Homework Solutions

Session #15: Homework Solutions Problem #1 Iron ( ρ  7.86 g/cm3 ) crystallizes in a BCC unit cell at room temperature. Calculate the radius of an iron atom in this crystal. At temperatures above 910ºC iron prefers to be FCC. If we neglect the temperature dependence of the radius of the iron atom on the grounds that it is negligible, we can calculate the density of FCC iron. Use this to determine whether iron expands or contracts when it undergoes transformation from the BCC to the FCC structure. Solution In BCC there are 2 atoms per unit cell, so 2 3 a = NA , where Vmolar = A/ρ ; A is Vmolar the atomic mass of iron. 2 3 a = NA  ρ A 1  2A 3 4 r  a=   = 3  NA  ρ   r = 1.24  10-8 cm If we assume that change of phase does not change the radius of the iron atom, then we can repeat the calculation in the context of an FCC crystal structure, i.e., 4 atoms per unit cell and a = 2 2r . ρ= 4A NA (2 2r)3 = 8.60 g/cm3 FCC iron is more closely packed than BCC suggesting that iron contracts upon changing from BCC to FCC. This is consistent with the packing density calculations reported in lecture that give FCC as being 74% dense and BCC 68% dense. The ratio of the densities calculated here is precisely the same: 7.86 0.68 = 8.60 0.74 Problem #2 Determine the total void volume (cm3/mole) for gold (Au) at 27oC; make the hard-sphere approximation in your calculation, and use data provided in the periodic table. Solution First determine the packing density for Au, which is FCC; then relate it to the molar volume given in the periodic table. 16r3 volume of atoms/unit cell 16r3 3 packing density = = = volume of unit cell a3 3a3 packing density  16r3 3  16 2r 3   3 2  0.74  74% void volume = 1 – packing density = 26% From the packing density (74%) we recognize the void volume to be 26%. Given the molar volume as 10.3 cm3/mole, the void volume is: 0.26  10.3 cm3 / mole = 2.68 cm3 / mole Problem #3 Determine the atomic (metallic) radius of Mo. Do not give the value listed in the periodic table; calculate it from other data given. Solution Mo: atomic weight = 95.94 g/mole  = 10.2 g/cm3 BCC, so n = 2 atoms/unit cell a3 = (95.94 g/mole)(2 atoms/unit cell) (10.2 g / cm3 )(6.023  1023 atoms/mole)  10-6 m3 cm3  3.12  10-29 m3 a = 3.22  10-10 m For BCC, a 3 = 4r , so r = 1.39  10-10 m Problem #4 A metal is found to have BCC structure, a lattice constant of 3.31 Å, and a density of 16.6 g/cm3. Determine the atomic weight of this element. Solution BCC structure, so n = 2 a = 3.31 Å = 3.31 x 10–10 m  = 16.6 g/cm3 N atomic weight  106 = A  a3  n atomic weight = (6.023  1023 atoms/mole) (3.31  10-10 m)3 -6 (2 atoms/unit cell)(10 3 3 m /cm )  16.6 g/cm3  181.3 g/mole Problem #5 At 100°C copper (Cu) has a lattice constant of 3.655 Å. What is its density at this temperature? Solution Cu is FCC, so n = 4 a = 3.655 Å = 3.655 x 10–10 m atomic weight = 63.55 g/mole N atomic weight  10-6 = A  a3  n = (63.55 g/mole)(4 atoms/unit cell) 23 (6.023  10 -10 atoms/mole)(3.655  10 3 m ) = 8.64 g/cm3 Problem #6 Determine the second-nearest neighbor distance for nickel (Ni) (in pm) at 100° C if its density at that temperature is 8.83 g/cm3. Solution Ni: n=4 atomic weight = 58.70 g/mole  = 8.83 g/cm3 For a face–centered cubic structure, the second nearest neighbor distance equals “a” (see LN4–11). N atomic weight  10-6 = A  a3  n a3 = (58.70 g/mole)(10-6 m3 /cm3 )(4 atoms /unit cell) (6.023  1023 atoms / mole)(8.83 g / cm3 ) = 4.41  10-29 m3 a  3.61  10-10 m  1012 pm = 3.61  102 pm m Problem #7 Determine the highest linear density of atoms (atoms/m) encountered in vanadium (V). Solution V: atomic weight = 50.94 g/mole  = 5.8 g/cm3 BCC, so n = 2 The highest density would be found in the [111] direction. To find “a”: N atomic weight = a3 A  n  a3 = 50.94  2 5.8  6.023  1023 a = 3.08  10-8 cm = 3.08  10-10 m The length in the [111] direction is a 3 , so there are: 2 atoms / a 3 = 2 atoms/(3.08  10-10 m  3) = 3.75 x 109 atoms/m