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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.

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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]

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