Question

Tin atoms are introduced into an FCC copper ,producing an alloy with a lattice parameter of 4.7589×10-8cm and a density of 8.772g/cm3 .Cal the atomic percentage of tin present in the alloy

Answer 1

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:

$\mathbf{density (\rho) = \dfrac{(no \ of \ atoms/cell)(atomic \ mass )}{(lattice \ parameter )^3(6.022*10^{23} atoms/ mol)} }$

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

$\mathbf{8.772 g/cm^3 = \dfrac{(x)(118.69 \ g/mol) +(4-x)(63.54 \ g/mol)}{(4.7589*10^{-8} cm )^3(6.022*10^{23} atoms/ mol)} }$

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 % = $\frac{no \ of \ atoms \ per \ cell \ in \ tin }{no \ of \ atoms \ per \ cell \ in \ the \ metal}*100$

atomic % = $\frac{5.72 \ atoms / cell}{4 \ atoms/ cell} *100$

atomic % = 143 %