Ciência e Engenharia dos Materiais (CAP04)

(Parte 1 de 2)

Imperfections in the Atomic and Ionic Arrangements

4–1Calculate the number of vacancies per cm3expected in copper at 1080oC (just below the melting temperature). The activation energy for vacancy formation is 20,0 cal/mol.

 n=(4 atoms/u.c.)

4-2The fraction of lattice points occupied by vacancies in solid aluminum at 660oC is 10−3. What is the activation energy required to create vacancies in aluminum?

4–3The density of a sample of FCC palladium is 1.98 g/cm3and its lattice parameter is 3.8902 Å. Calculate (a) the fraction of the lattice points that contain vacancies and (b) the total number of vacancies in a cubic centimeter of Pd.

 (x)(106.4 g/mol)

4–4The density of a sample of HCPberyllium is 1.844 g/cm3and the lattice parameters are ao =0.22858 nm and co

=0.35842 nm. Calculate (a) the fraction of the lattice points that contain vacancies and (b) the total number of vacancies in a cubic centimeter.

Solution: Vu.c. =(0.22858 nm)2(0.35842 nm)cos30 =0.01622 nm3

(a) From the density equation:

 (x)(9.01 g/mol)

4–5BCC lithium has a lattice parameter of 3.5089× 10−8cm and contains one vacancy per 200 unit cells. Calculate (a) the number of vacancies per cubic centimeter and (b) the density of Li.

 Solution:(a)1 vacancy =1.157× 1020vacancies/cm3

(b) In 200 unit cells, there are 399 Li atoms. The atoms/cell are 399/200:

 r=(399/200)(6.94 g/mol) =0.532 g/cm3

4–6FCC lead has a lattice parameter of 0.4949 nm and contains one vacancy per 500 Pb atoms. Calculate (a) the density and (b) the number of vacancies per gram of Pb.

Solution:(a) The number of atoms/cell =(499/500)(4 sites/cell)

 r=(499/500)(4)(207.19 g/mol)

(b) The 500 Pb atoms occupy 500 / 4 =125 unit cells:

BCC structure; eventually an alloy is produced that has a lattice parameter of 0.32554 nm and a density of 1.95 g/cm3. Calculate the fraction of the atoms in the alloy that are tungsten.

Solution:

 90.94xW=62.366 or xW=0.69 Watoms/cell

vacancy

125cells

36The Science and Engineering of MaterialsInstructor’s Solution Manual

There are 2 atoms per cell in BCC metals. Thus:

Solution:8.772 g/cm3= (xSn)(118.69 g/mol) +(4 −xSn)(63.54 g/mol)

 280.5 =5.15xSn+254.16 or xSn=0.478 Sn atoms/cell

There are 4 atoms per cell in FCC metals; therefore the at% Sn is:

4–9We replace 7.5 atomic percent of the chromium atoms in its BCC crystal with tantalum. X-ray diffraction shows that the lattice parameter is 0.29158 nm. Calculate the density of the alloy.

Solution:

4–10Suppose we introduce one carbon atom for every 100 iron atoms in an interstitial position in BCC iron, giving a lattice parameter of 0.2867 nm. For the Fe-C alloy, find (a) the density and (b) the packing factor.

Solution:There is one carbon atom per 100 iron atoms, or 1 C/50 unit cells, or 1/50 C per unit cell:

 (2)(5.847 g/mol) +(1/50)(12 g/mol)

(b)

4–11The density of BCC iron is 7.882 g/cm3and the lattice parameter is 0.2866 nm when hydrogen atoms are introduced at interstitial positions. Calculate (a) the atomic fraction of hydrogen atoms and (b) the number of unit cells required on average to contain one hydrogen atom.

 2(5.847 g/mol) +x(1.00797 g/mol)

The total atoms per cell include 2 Fe atoms and 0.0081 H atoms. Thus:

(b)Since there is 0.0081 H/cell, then the number of cells containing H atoms is:

cells =1/0.0081 =123.5 or 1 H in 123.5 cells

CHAPTER 4Imperfections in the Atomic and Ionic Arrangements37

4–12Suppose one Schottky defect is present in every tenth unit cell of MgO. MgO has the sodium chloride crystal structure and a lattice parameter of 0.396 nm. Calculate (a) the number of anion vacancies per cm3and (b) the density of the ceramic.

(b)

4–13ZnS has the zinc blende structure. If the density is 3.02 g/cm3and the lattice parameter is 0.59583 nm, determine the number of Schottky defects (a) per unit cell and (b) per cubic centimeter.

Solution:Let xbe the number of each type of ion in the unit cell. There normally are 4 of each type.

 x(65.38 g/mol) +x(32.064 g/mol)

4–14Suppose we introduce the following point defects. What other changes in each structure might be necessary to maintain a charge balance? Explain.

4–22What are the Miller indices of the slip directions (a) on the (1) plane in an FCC unit cell (b) on the (011) plane in a BCC unit cell?

x y x y

38The Science and Engineering of MaterialsInstructor’s Solution Manual

4–23What are the Miller indices of the slip planes in FCC unit cells that include the [101] slip direction?

4–24What are the Miller indices of the {110} slip planes in BCC unit cells that include the [1] slip direction?

4–25Calculate the length of the Burgers vector in the following materials: (a)BCC niobium(b)FCC silver(c)diamond cubic silicon

Solution:(a) The repeat distance, or Burgers vector, is half the body diagonal, or:

(b) The repeat distance, or Burgers vector, is half of the face diagonal, or:

(c) The slip direction is [110], where the repeat distance is half of the face diagonal:

4–26Determine the interplanar spacing and the length of the Burgers vector for slip on the expected slip systems in FCC aluminum. Repeat, assuming that the slip system is a (10) plane and a [1–1] direction. What is the ratio between the shear stresses required for slip for the two systems? Assume that k =2 in Equation 4-2.

x y x y x y x y x y

CHAPTER 4Imperfections in the Atomic and Ionic Arrangements39

(c) If we assume that k=2 in Equation 4-2, then

tbexp(−2(0.408))

(Parte 1 de 2)