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authorprashantsinalkar2017-10-10 12:27:19 +0530
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+//Example 3.5//
+a=0.741;//nm //unit cell dimensions
+b=0.494;//nm //unit cell dimensions
+c=0.255;//nm //unit cell dimensions
+v=a*b*c
+mprintf("v = %f nm^3",v)
+a1=12.01;//gram //atomic mass of carbon
+b1=1.008;//gram // atomic mass of Hydogen
+c1=0.6023*10^24;//atoms // Avogardo's number
+d1=2;//Number of electrons
+e1=4;//Number of electrons
+m=((d1*a1)+(e1*b1))/c1
+mprintf("\nm = (%e n)g",m)
+//Therefore, the unit cell density is,
+d=10^7;//nm/cm
+p=(m/v)*d^3
+mprintf("\n p = %f g/cm^3 (As answer in the textbook is calculated wrong)",p)
+//solving for n gives
+n=2
+//Aa a result, there are
+mprintf("\n4(=2n)C atoms + 8(=4n)H atoms per unit cell.")