Mumbai university > FE > SEM 1 > Applied Physics 1

Marks: 7M

Year: May 2015

Given:

Cu has FCC structure.

$d = 2.08 A \ \ or \ \ 2.08 × 10^{-10}$m

$M=63.54$

To find: density (ρ ) & diameter (d) of Cu=?

Solution: plane is (1 1 1)

$ d= \frac{a}{\sqrt{1^2 + 1^2 + 1^2}}$

$ a = 2.08 × 10 ^{-10} ×\sqrt{3}$

$a =3.603 ×10^{-10} m$

For FCC structure-

$r= \frac{a\sqrt{2}}{4}= \frac{3.603 × 10^{-10} × \sqrt{2}}{4} =1.274 × 10^{-10}m = 1.274A$

$ d = 2r = 2 × 1.274 × 10^{-10} = 2.548 × 10^{-10}m \ \ or \ \ 2.548A$

density= Mass/volume

$density( ρ)= \frac{M}{a^3} \ \ \ where \ \ a^3 - volume$

$= \frac{63.54}{(3.603 × 10^{-10})^3}$

$density( ρ) = 1.358 × 10^{30} kg/m^3$