Question

The nearest neighbor silver atoms in siver,crystals are $2.87\times {10}^{10}m$ apart what is the density id silver ? Silver ctrystallises in Fcc form. (Atomic mass of Ag$=108)$

Answer 1

For fcc $a=1.414d$ where d= distance between the nearest silver atoms.

$a=(1.414\times 287)pm$

$a=405.8pm=405.8\times {10}^{-10}cm$

Also for fcc $z=4$

$d=\frac{Z\times M}{a^3\times N_A}$

$d=\frac{4\times 107.87}{(405.8\times {10}^{-10}{)}^3\times 6.023\times {10}^{23}}$

$d=10.73gc{m}^{-3}$