Question: Directions: Additional information to complete these problems can be found in your Examination Booklet. --Given Values-- Atomic Radius (nm) = 0.120 FCC Metal = Lead

Directions: Additional information to complete these problems can be found in your Examination Booklet. --Given Values-- Atomic Radius (nm) = 0.120 FCC Metal = Lead BCC Metal: = Iron Temperature (C)= 1454 ( Metal A= Sulfur Equilibrium Number of Vacancies (m-3) = 4.81E+23 Temperature for Metal A(C) = 369 Metal B = Iron 1) If the atomic radius of a metal is the value shown above and it has the face-centered cubic crystal structure, calculate the volume of its unit cell in nm?? Write your answers in Engineering Notation. Your Answer = 1.72E-3 2) What is the atomic packing factor for the BCC crystal structure? Your Answer=0.68 3) Find the theoretical density for the FCC Element shown above in g/cm. Write your answer to ten thousandths place (0.0000): Your Answer=0.113304 4) Calculate the atomic radius, in nm, of the BCC Metal above utilizing the density and the atomic weight provided by examination booklet - Write your answer with 4 significant figures: Your Answer = 5) Calculate the fraction of atom sites that are vacant for copper (Cu) at the temperature provided above. Assume an energy for vacancy formation of 0.90 eV/atom: Your Answer 6) Repeat the calculation in question 5 at room temperature (25 C): Your
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