Respuesta :
Explanation:
(a) Number of atoms present in FCC are 4 atoms. Hence, number of atoms present in the given cell are as follows.
No. of atoms/cell = [tex]4 \times \frac{499}{500}[/tex]
= 4
Density will be calculated as follows.
Density = [tex]\frac{ZM}{Na^{3}}[/tex]
= [tex]\frac{4 \times 207}{6.023 \times 10^{23}} \times (0.4949 \times 10^{-7})^{3}[/tex]
= 11.34 [tex]g/cm^{3}[/tex]
Therefore, density of the given substance is 11.34 [tex]g/cm^{3}[/tex].
(b) Now, there is 1 vacancy present for every 500 lead (Pb) atoms. Therefore, 500 Pb atoms occupied by four lattice points is as follows.
[tex]\frac{500}{4}[/tex]
= 125 unit cells
[tex]\frac{\frac{1}{125}}{(0.4949 \times 10^{-7})^{3}} \times \frac{1}{11.34 g/cm^{3}}[/tex]
= [tex]5.82 \times 10^{18}[/tex]
Thus, we can conclude that number of vacancies present per gram are [tex]5.82 \times 10^{18}[/tex].
A) The density of lead is : 11.33 g/cm³
B) The number of vacancies per gram of Pb : 5.8246 * 10¹⁸ vacancies
Given data :
Lattice parameter = 0.4949 nm
contains one ( 1 ) vacancy per 500 Pb atoms
A) Calculate the Density of lead atoms
First step : Determine the number of lead atoms per unit cell
number of lead atoms per unit cell
= ( number of Pb - vacancy ) / number of Pb * 4 sites ---- ( 1 )
= ( 500 - 1 ) / 500 * 4 = 3.992
Next step : Calculate the density of lead atoms
p = ( number of lead atoms per unit * Z ) / L³p * Na ----- ( 2 )
where ; Z ( atomic mass of Pb ) = 207.19 g/mol, Lp ( lattice number ) = 0.4949 *10⁻⁷ cm, Na ( avogadro's constant ) = 6.022* 10²³ atoms/mol
Insert values into equation ( 2 )
p(density of lead atoms ) = 11.331 g/cm³
B) The number of vacancies per gram of Pb
Nv = ( number of vacancies per unit ) / L³p * p ----- ( 3 )
where ; number of vacancies per unit = 4/500 , Lp = 0.4949 *10⁻⁷ cm
p = 11.331 g/cm³
Insert values into equation ( 3 )
Nv = 5.8246 * 10¹⁸ vacancies
Hence we can conclude that The density of lead is : 11.33 g/cm³ and The number of vacancies per gram of Pb : 5.8246 * 10¹⁸ vacancies.
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