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How do you calculate the packing fraction of hcp structure?
0.74
Volume of a HCP unit cell?
The volume of HCP is 8*pi*r^3 or 25.13*r^3
How do you calculate the volume of a HCP crystal?
Are you takling Material Science class? Volume of HCP crystal = (a^2) (c) cos30 Im taking Material Science and Engineering
Show that the atomic packing factor for HCP is 0.74?
To answer this question we must look at the HCP itself. We know it is a Hexagonal close packet (HCP) With a CN=12 and common elements are Ti, Mg, Cd etc.Now we have the backup theory, let's answer the question.APF (Atomic packing factor) will equal Vs/Vcwhere Vs is volume of the atoms contained within the unit cell. Vcbeing the volume of total unit cell. Now we look at the structure HCP:Imagine a hexagonal with 6 equilateral triangles contained.Firstly what is Vs? Easy: (volume or sphere?) =4/3(pie)R^3 x 6 (x6 because of 6 spheres per unit cell, remember? HCP has 6 atoms contained)Dissect one equilateral triangle off to work out it's area. The area of this equilateral triangle? Now the side lengths will be a (any value) with 60 degree angles at each edge (total of 180 degrees) therefore the are will be A=ax a x 1/2 x sin(60) =(Sqrt(3) x (a^2)) / 4Now multiply the area of the single triangle by 6 (remember, 6 triangles).Total Area of base of hexagonal =(Sqrt(3) x (3) x a^2) / 2Now remember the formula relating a with R =>a =2RSub R into our base area formula, therefore =6 x Sqrt(3) x R^2 (excluding working out)Now recall the c/a ratio of HCP? c/a=1.63Hence c =1.63 x a > Now sub R (a =2R...)c =3.26 x RNow recall the question? APF? Therefore look back at Vc/Vs, what are we missing? Vc,now Vc=c x base...Vc=3.26 x R x 10.392 x R^2 =33.878 x R^3thereforeAPF =Vs (recall from top working out) / Vc=(8 x (pie) x R^3) / 33.878 x R^3Vc=0.74 ( cancel the R^3)Easy eh? haha
How do you calculate density of compounds?
p = n x Mr / Vc x NAwhere n is the atoms/unit cell e.g. fcc packing n = 4 and for bcc packing n = 2Mr is the Atomic Mass in g/molVc is the volume/unit cell cm3 = a3 where a can be found by the radius of the atom and the packing used. e.g in bcc packing it is "a = 4r/1.732" . In Fcc packing it is "a= sin (4r)" or "a = cos (4r)"NA is avorgados constant, = 6.023 x1023
