Respuesta :
Answer : The number of vacancies (per meter cube) = 2.889 × [tex]10^{24}m^{-3}[/tex]
Solution : Given,
Atomic weight of silver (Ag), [tex]A_{Ag}[/tex] = 107.87 g/mol
Density of silver (Ag), [tex]\rho_{Ag}[/tex] = 10.35 [tex]g/cm^{3}[/tex] = 10.35 [tex]g/cm^{3}[/tex] × [tex]10^{6} cm^{3}/m^{3}[/tex] = 10.35 × [tex]10^{6} g/m^{3}[/tex]
Avogadro's number, [tex]N_{A}[/tex] = 6.022 × [tex]10^{23}[/tex] atoms/mol
Fraction of lattice sites that are vacant in silver = 0.5 × [tex]10^{-6}[/tex]
Formula used :
[tex]N_{Ag}=\frac{N_{Ag\times\rho_{Ag} } }{A_{Ag} }[/tex]
Where,
[tex]N_{Ag}[/tex] = Total number of lattice sites in Ag
[tex]N_{A}[/tex] = Avogadro's number
[tex]\rho_{Ag}[/tex] = Density of silver
[tex]A_{Ag}[/tex] = Atomic weight of silver
This problem is solved by two steps:
step 1 : First we calculate the total number of lattice sites in silver.
Now put all the values in above formula, we get
[tex]N_{Ag}=\frac{(6.022\times10^{23}atoms/mol)\times\ (10.35\times10^{6} g/m^{3})}{ 107.87 g/mol}[/tex]
= 5.778 × [tex]10^{30}[/tex] atoms/[tex]m^{3}[/tex]
Step 2 : Now multiply the value of [tex]N_{Ag}[/tex] by the fraction of lattice sites that are vacant in silver, 0.5 × [tex]10^{-6}[/tex], we get the number of vacancies (per meter cube)
The number of vacancies (per meter cube) = 5.778 × [tex]10^{30}[/tex] atoms/[tex]m^{3}[/tex] × 0.5 × [tex]10^{-6}[/tex] = Answer : The number of vacancies (per meter cube) = 2.889 × [tex]10^{24}m^{-3}[/tex]
