Question

Tin atoms are introduced into a FCC copper crystal, producing an alloy with a lattice pa- rameter of 3.7589 x 10-8 cm and a density of 8.772 g/cm3. Calculate the atomic percentage of tin present in the alloy.

Answer 1

Given:

Lattice parameter, a = $3.7589\times 10^{-8}cm$

$density, \rho = 8.772g/cm^{3}$

Solution:

We know that  for FCC, total no. of atoms in a crystal lattice = 4

let the number of atoms in Tin for alloying be 'n'

⇒ Total no. of Copper atoms in the alloy,  = 4 - n

Also mass of Copper, m = 63.54 g/mol

atomic mass of Tin = 118.69 g/mol

We Know density of the crystal lattice is given by the formula:

$\rho = \frac{m\times Z}{a^{3}\times N_{A}}$                 (1)

where,

$N_{A}$ = Avagadro's number =  $6.23\times 10_{23}$

Putting all the values in eqn (1), we get

$8.772 = \frac{118.69\times x\times (4 - n)\times 63.54}{(3.7589\times 10^{-8})^{3}\times 6.023\times 10^{23}}$

280.6 = 55.15n +254.16

n = 0.479 atoms/cell

Now to calculate the atomic percentage of Tin present in the alloy:

atomic percentage = $\frac{no. of atoms in Tin/cell}{Total no. of atoms in FCC lattice}$

atomic % Tin present in alloy = $\frac{0.479}{4}$ = $0.1198\times 100$

atomic % Tin present in alloy = 11.98%