Ba (AW = 137.327 g/cm3) FCC Atomic radius = 0.115 nm Qv = 1.23ev/atom T = 1230 OC Determine: a) Number of lattice sites, N (atoms/m3) b) Number of vacancies, Nv (atoms/m3)
Ba (AW = 137.327 g/cm3) FCC Atomic radius = 0.115 nm Qv = 1.23ev/atom T = 1230 OC Determine: a) Number of lattice sites, N (atoms/m3) b) Number of vacancies, Nv (atoms/m3)
Understanding Motor Controls
4th Edition
ISBN:9781337798686
Author:Stephen L. Herman
Publisher:Stephen L. Herman
Chapter44: Semiconductors
Section: Chapter Questions
Problem 4RQ
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Ba (AW = 137.327 g/cm3)
FCC
Atomic radius = 0.115 nm
Qv = 1.23ev/atom
T = 1230 OC
Determine: a) Number of lattice sites, N (atoms/m3)
b) Number of vacancies, Nv (atoms/m3)
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