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Calculate the fraction of atom sites that are vacant for lead at its melting temperature of 327°C (600 K). Assume an energy for vacancy formation of 0.55 eV/atom.

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anniwuanyanwu1

Answer:

Fraction of atom sites,Nv/N= 2.41×10^-5

Explanation:

Nv/N = exp(-Qv/kT)

Nv/N= exp[- (0.55ev/atom)/(8.62×10^-5ev/atom)(600k)]

Nv/N= 2.41×10^-5

Fraction of atom sites is [tex]Nv\div N= 2.41\times 10^{-5}[/tex]

Given that,

  • Melting temperature of 327°C (600 K).
  • Assume an energy for vacancy formation of 0.55 eV/atom.

Based on the above information, the calculation is as follows:

Therefore we can conclude that the Fraction of atom sites is [tex]Nv\div N= 2.41\times 10^{-5}[/tex]

Learn more: brainly.com/question/16911495

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