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
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ISBN:9781337798686
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Chapter44: Semiconductors
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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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