Question

Consider a hypothetical metal that has a density of 11.6 g/cm3, an atomic weight of 157.6 g/mol, and an atomic radius of 0.127 nm. Compute the atomic packing factor if the unit cell has tetragonal symmetry; values for the and lattice parameters are 0.523 nm and 0.330 nm, respectively.

Answer 1

Answer:

$APF(atomic packing factor)=0.31736$

Explanation:

From the question we are told that

Density of metal $\rho=11.6 g/cm3$

Atomic weight of $W=157.6 g/mol$

Atomic radius of $r= 0.127 nm$

Lattice parameters=>$x=0.523nm$  and $y=0.330 nm$

Generally the equation for atomic packing factor is mathematically given as

$APF(atomic packing factor)=\frac{Spere's\ volume}{unit\ cell\ volume}$

$APF(atomic packing factor)=\frac{N*VN}{VC}$

Generally the equation for number of atoms N is mathematically given as

$N=\frac{\rho }{atomic raduis*Avacados constant}$

$N=\frac{11.6 }{0.127*10^{-9}*6.02214086*10^{23} mol-1}$

$N=4$

Therefore APF(atomic packing factor)

$APF(atomic packing factor)=\frac{N*VN}{VC}$

$APF(atomic packing factor)=\frac{4*\frac{4}{3} \pi (0.127)^3}{(0.523)^2 *0.330}$

$APF(atomic packing factor)=0.31736$