Foundations of Materials Science and Engineering
6th Edition
ISBN: 9781259696558
Author: SMITH
Publisher: MCG
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Chapter 2.9, Problem 71SEP
To determine
Write the reason for the formation of ionic compounds when the energy is lowered.
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Chapter 2 Solutions
Foundations of Materials Science and Engineering
Ch. 2.9 - Prob. 1KCPCh. 2.9 - Prob. 2KCPCh. 2.9 - How was the existence of electrons first verified?...Ch. 2.9 - How was the existence of protons first verified?...Ch. 2.9 - What are the similarities and differences among...Ch. 2.9 - Prob. 6KCPCh. 2.9 - Prob. 7KCPCh. 2.9 - Prob. 8KCPCh. 2.9 - Explain the law of chemical periodicity.Ch. 2.9 - Prob. 10KCP
Ch. 2.9 - Prob. 11KCPCh. 2.9 - Prob. 12KCPCh. 2.9 - Prob. 13KCPCh. 2.9 - Prob. 14KCPCh. 2.9 - Prob. 15KCPCh. 2.9 - Prob. 16KCPCh. 2.9 - Prob. 17KCPCh. 2.9 - Describe the terms (a) metallic radius. (b)...Ch. 2.9 - Prob. 19KCPCh. 2.9 - Prob. 20KCPCh. 2.9 - Prob. 21KCPCh. 2.9 - Prob. 22KCPCh. 2.9 - Prob. 23KCPCh. 2.9 - Prob. 24KCPCh. 2.9 - Describe the properties (electrical, mechanical,...Ch. 2.9 - Prob. 26KCPCh. 2.9 - Prob. 27KCPCh. 2.9 - Prob. 28KCPCh. 2.9 - The diameter of a soccer ball is approximately...Ch. 2.9 - Each quarter produced by the U.S. mint is made up...Ch. 2.9 - Sterling silver contains 92.5 wt% silver and 7.5...Ch. 2.9 - Prob. 32AAPCh. 2.9 - Prob. 33AAPCh. 2.9 - Prob. 34AAPCh. 2.9 - Prob. 35AAPCh. 2.9 - Prob. 36AAPCh. 2.9 - Prob. 37AAPCh. 2.9 - Prob. 38AAPCh. 2.9 - Prob. 39AAPCh. 2.9 - Prob. 40AAPCh. 2.9 - Prob. 41AAPCh. 2.9 - Prob. 42AAPCh. 2.9 - Prob. 43AAPCh. 2.9 - Prob. 44AAPCh. 2.9 - Prob. 45AAPCh. 2.9 - Prob. 46AAPCh. 2.9 - Prob. 47AAPCh. 2.9 - Prob. 48AAPCh. 2.9 - Prob. 49AAPCh. 2.9 - Prob. 50AAPCh. 2.9 - Write the electron configurations of the following...Ch. 2.9 - Prob. 52AAPCh. 2.9 - Prob. 53AAPCh. 2.9 - Prob. 54AAPCh. 2.9 - Prob. 55AAPCh. 2.9 - Prob. 56AAPCh. 2.9 - Prob. 57AAPCh. 2.9 - Prob. 58AAPCh. 2.9 - Prob. 59AAPCh. 2.9 - Prob. 60AAPCh. 2.9 - Prob. 61AAPCh. 2.9 - Prob. 62AAPCh. 2.9 - Prob. 63AAPCh. 2.9 - For each bond in the following series of bonds,...Ch. 2.9 - Prob. 65AAPCh. 2.9 - Prob. 66AAPCh. 2.9 - Prob. 67AAPCh. 2.9 - Prob. 68AAPCh. 2.9 - Prob. 69SEPCh. 2.9 - Most modern scanning electron microscopes (SEMs)...Ch. 2.9 - Prob. 71SEPCh. 2.9 - Of the noble gases Ne, Ar, Kr, and Xe, which...Ch. 2.9 - Prob. 73SEPCh. 2.9 - Prob. 74SEPCh. 2.9 - Prob. 75SEPCh. 2.9 - Prob. 76SEPCh. 2.9 - Prob. 77SEPCh. 2.9 - Prob. 78SEPCh. 2.9 - Prob. 79SEPCh. 2.9 - Prob. 80SEPCh. 2.9 - Silicon is extensively used in the manufacture of...Ch. 2.9 - Prob. 82SEPCh. 2.9 - Prob. 83SEPCh. 2.9 - Prob. 84SEPCh. 2.9 - Prob. 85SEPCh. 2.9 - Prob. 86SEPCh. 2.9 - Prob. 87SEPCh. 2.9 - Prob. 88SEPCh. 2.9 - Prob. 89SEPCh. 2.9 - Prob. 90SEPCh. 2.9 - Prob. 91SEPCh. 2.9 - Prob. 92SEP
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- The net potential energy EN between two adjacent ions is sometimes represented by the expression C ; - - - + Dexp(-6) (2.18) r EN in which r is the interionic separation and C, D, and p are constants whose values depend on the specific material. (a) Derive an expression for the bonding energy Eo in terms of the equilibrium interionic separation ro and the constants D and p using the following procedure: 1. Differentiate EN with respect to r and set the resulting expression equal to zero. 2. Solve for C in terms of D, p, and ro.arrow_forwardShow that Ni and Cu are totally soluble in one another using Hume-Rothery rules. Atomic radii, electronegativities and crystal structures of Ni and Cu are given below. Ni Cu Crystal Structure FCC FCC Electronegativities 1.9 1.8 r (nm) 0.1246 0.1278arrow_forwardFor APF, Which one of the following is the assumption a. The atoms overlap b. The atoms are rigid cubes c. The atoms do not overlap d. The atoms are not rigid spheresarrow_forward
- The plot below shows bonding energy vs. interatomic separation for two elements. АВ a A. Which element would you expect to have a lower melting point? A B same can't tell provide a brief justification of your choice. B. Which element would you expect to have a smaller lattice constant? A В same can't tell provide a brief justification of your choice. C. Which element would you expect to have a larger elastic modulus? A justification of your choice. В same can't tell provide a brief can't tell provide a brief D. Which element would you expect to have the highest yield stress? justification of your choice. A В same Bonding energyarrow_forwardCopper is a typical face-centered-cubic metal. In its pure state, it has extremely high electrical conductivity. Oxygen in copper causes a major reduction of conductivity. To investigate this issue further, I want to you to tell me whether oxygen atoms substitute for copper atoms, or if they enter interstitial sites.arrow_forwardGold forms a substitutional solid solution with silver. Compute the number of gold atoms per cubic centimetre for a silver-gold alloy that contains 10 wt% Au and 90wt% Ag. The densities of pure gold and silver are 19.32 and 10.49 g/cm3 respectively.arrow_forward
- Calculate the number of vacancies per m3 for gold at 900 C. the energy for vacancy formation is 0.86 eV/atom.arrow_forwardCopper at 500 °C has 1.10×1015 vacancies/cm³. The atomic weight of Cu is 63.55 g/mol and the density at this temperature is 8.96 g/cm³. Avogadro constant NA ~ 6.022×1023 atoms/mol. Boltzmann constant (K) - 8.62×10-³ ev/atom • K. (a) What is the activation energy required to create a vacancy in Cu? (b) Cu is then cooled to 400 °C. What will be the number of vacancies per cubic centimeter? Assume that the change in the density in the cooling is negligible.arrow_forwardFor a hypothetical material, experiments show that a sample containing 1 billion atoms has 40 vacancies under equilibrium conditions at 600 degrees Celsius. Determine the energy for vacancy formation in the material at 600 degrees Celsius.arrow_forward
- Calculate the fraction of atom sites that are vacant for copper at its melting temperature of 1084degC. assume an energy vacancy formation of 0.90eVlatom. * 6.56 x 10^-5 O 0.9992 O 4.56 x 10^-4 O 2.19 X 10^-4arrow_forwardCalculate the number of vacancies per cubic meter at 1000°C for a metal that has an energy for vacancy formation of 1.22 eV/atom, a density of 6.25 g/cm3, and an atomic weight of 37.4 g/mol.</o:p> (A) 1.49 ×1018 m−3</o:p> (B) 7.18 ×1022 m−3</o:p> (C) 1.49 ×1024 m−3</o:p> (D) 2.57 ×1024 m−3</o:p>arrow_forwardO h. 0.74 Silver (Ag) has the FCC crystal structure. The atomic radius is R = 0.144 nm and the atomic mass is M = 107.87 g/mole. What is the density of Silver in g/m? Given: Avogadro's Number NA = 0.6023 x 1024 (atoms/mole) Select one: a. 10.6x106 O b. 10.6×105 O c. 10.6×10 O d. 10.60 O e. 10.6x10 Chromium (Cr) has the Boody-Centered Cubic (BCC) crystal structure. The edge length is a= 0.288 nm. What is the linear density in atoms/nm along direction (111P Select one: 9:12 PM O a. 7.48 O O e 4) A ENG 15-Apr-2021arrow_forward
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