Answer:
atomic percentage = 143 %
Explanation:
Let x be the number of tin atoms and there are 4 atoms / cell in the FCC structure , then 4 -x be the number of copper atoms . Therefore, the value of x can be determined by using the density equation as shown below:
[tex]\mathbf{density (\rho) = \dfrac{(no \ of \ atoms/cell)(atomic \ mass )}{(lattice \ parameter )^3(6.022*10^{23} atoms/ mol)} }[/tex]
where;
the lattice parameter is given as : 4.7589 × 10⁻⁸ cm
The atomic mass of tin is 118.69 g/mol
The atomic mass of copper is 63.54 g/mol
The density is 8.772 g/cm³
[tex]\mathbf{8.772 g/cm^3 = \dfrac{(x)(118.69 \ g/mol) +(4-x)(63.54 \ g/mol)}{(4.7589*10^{-8} cm )^3(6.022*10^{23} atoms/ mol)} }[/tex]
569.32 = 118.69x + 254.16-63.54x
569.32 - 254.16 = 118.69x - 63.54 x
315.16 = 55.15x
x = 315.16/55.15
x = 5.72 atoms/cell
As there are four atoms per cell in FCC structure for the metal, thus, the atomic percentage of the tin is calculated as follows :
atomic % = [tex]\frac{no \ of \ atoms \ per \ cell \ in \ tin }{no \ of \ atoms \ per \ cell \ in \ the \ metal}*100[/tex]
atomic % = [tex]\frac{5.72 \ atoms / cell}{4 \ atoms/ cell} *100[/tex]
atomic % = 143 %
