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+//Example 3.15 (a)//

+rW=0.137;//nm //atomic radius of tungsten (From appendix 2)

+a=(4*rW)/(sqrt(3))//Body centered cubic

+mprintf("a = %f nm",a)

+l=sqrt(2)*a; // face diagonal length

+mprintf("\n l = %f nm",l)

+

+//The area of the  (111) plane within yhe unit cell

+c=sqrt(3);//given

+d=2;//given

+h=(c/d)*l

+//mprintf("h = %f ",h)

+A=(1/2)*l*h

+mprintf("\nA = %f nm^2",A)

+c1=3;//atoms

+d1=1/6;//atoms

+ad=(c1*d1)/A

+mprintf("\nad = %f atoms/nm^2",ad)

+

+//(b)

+// Following the calculations of sample problem 3.14b we find that the length of the body diagonal is

+b=0.143;// atomic radius of Aluminium

+a1=(4*b)/(sqrt(2))  //Face centered cubic

+//mprintf("\n a1 = %f nm",a1)

+l1=sqrt(2)*a1;

+mprintf("\nl1 = %f nm",l1)

+//the area of the (111) plane within the unit cell is

+A1=(1/2)*l1*(c/d)*l1

+mprintf("\nA1 = %f nm^2",A1)

+e1=(1/2);//atoms

+ad2=((c1*d1)+(c1*e1))/A1

+mprintf("\nad2 %f atoms/nm^2",ad2)