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* 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 = (m* Z)/(a^(3)* N_(A))` (1)

where,

`N_(A)` = Avagadro's number =
`6.23* 10_(23)`

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

`8.772 = (118.69* x* (4 - n)* 63.54)/((3.7589* 10^(-8))^(3)* 6.023* 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 =
`(no. of atoms in Tin/cell)/(Total no. of atoms in FCC lattice)`

atomic % Tin present in alloy =
`(0.479)/(4)` =
`0.1198* 100`

atomic % Tin present in alloy = 11.98%