Answer:
[tex]APF(atomic packing factor)=0.31736[/tex]
Explanation:
From the question we are told that
Density of metal [tex]\rho=11.6 g/cm3[/tex]
Atomic weight of [tex]W=157.6 g/mol[/tex]
Atomic radius of [tex]r= 0.127 nm[/tex]
Lattice parameters=>[tex]x=0.523nm[/tex] and [tex]y=0.330 nm[/tex]
Generally the equation for atomic packing factor is mathematically given as
[tex]APF(atomic packing factor)=\frac{Spere's\ volume}{unit\ cell\ volume}[/tex]
[tex]APF(atomic packing factor)=\frac{N*VN}{VC}[/tex]
Generally the equation for number of atoms N is mathematically given as
[tex]N=\frac{\rho }{atomic raduis*Avacados constant}[/tex]
[tex]N=\frac{11.6 }{0.127*10^{-9}*6.02214086*10^{23} mol-1}[/tex]
[tex]N=4[/tex]
Therefore APF(atomic packing factor)
[tex]APF(atomic packing factor)=\frac{N*VN}{VC}[/tex]
[tex]APF(atomic packing factor)=\frac{4*\frac{4}{3} \pi (0.127)^3}{(0.523)^2 *0.330}[/tex]
[tex]APF(atomic packing factor)=0.31736[/tex]
