In [2]:

```
from math import sqrt
#Variable declaration
#Assuming r=1 for simpliciy in calculations
r = 1
#Calculations
a = (4*r)/sqrt(3)
#Let R be the radius of interstitial sphere that can fit into the void,therefore,
R = (a-2*r)/2
#Result
print "The maximum radius of interstitial sphere is",round(R,3),"r"
```

In [3]:

```
from math import sqrt
#Variable declaration
r1 = 1.258*10**-10 #atomic radius(m)
r2 = 1.292*10**-10 #atomic radius(m)
#Calculations
#In BCC
a_bcc = (4*r1)/sqrt(3)
v_bcc = a_bcc**3 #volume of unit cell(m^3)
n1 = ((1./8.)*8.)+1
V1 = v_bcc/n1 #volume occupied by 1 atom(m^3)
#In FCC
a_fcc = 2*sqrt(2)*r2
v_fcc = a_fcc**3 #volume of unit cell(m^3)
n2 = ((1./2.)*6.)+((1./8.)*8.)
V2 = v_fcc/n2 #volume occupied by 1 atom(m^3)
del_v = ((V1-V2)/V1)*100 #change in volume
#Result
print "During the conversion of iron from BCC to FCC, the decrease in volume is",round(del_v,1),"%"
```

In [4]:

```
from math import sqrt
#Variable declaration
a = 0.27*10**-9 #nearest neighbour distance(m)
c = 0.494*10**-9 #height of unit cell(m)
M = 65.37 #atomic weight of zinc
N = 6.023*10**26 #Avogadro's number(k/mol)
#Calculations
V = (3*sqrt(3)*a**2*c)/2 #volume of unit cell
rho = (6*M)/(N*V) #density of crystal
#Results
print "Volume of unit cell =",round((V/1E-29),2),"*10^29 m^3"
print "Density of zinc =",round(rho),"kg/m^3"
print "\nThe solution differs because of rounding-off of the digits in the textbook"
```

In [5]:

```
from math import sqrt
#Varaible declaration
#Let r be the radius of atom and R be the radius of sphere
#For simplicity in calculations, let us assume r =1
r = 1
#Calculations
#For FCC structure
a = (4*r)/sqrt(2)
R = (a/2)-r
print "Maximum radius of sphere =",round(R,3),"r"
```

In [58]:

```
#Variable declaration
a = 0.356*10**-9 #cube edge(m)
M = 12.01 #atomic weight of carbon
N = 6.023*10**26 #Avogadro's number(k/mol)
na = 1.77*10**29 #no. of atoms per meter cube
#Calculations
#Diamond has 2 interpenetrating FCC lattices. Since each FCC unit cell has 4 atoms, the total no. of atoms per unit cell is 8
n = 8/a**3
m = M/N
rho = m*na
#Result
print "Number of atoms =",round((n/1E+29),3),"*10^29"
print "The density of diamond is",round(rho,1),"kg/m^3(Calculation mistake in the textbook)"
```

In [4]:

```
#Variable declaration
rho = 2.18 #density of NaCl(gm/cm^3)
N = 6.023*10**23 #Avogadro's number(/mol)
#Caculations
w = 23+35.5 #molecular weight of NaCl
m = w/N #mass of NaCl molecule
nm = rho/m #no. of molecules per unit volume(molecule/cm^3)
#Since NaCl is diatomic
n = 2*nm
#Let a be the distance between adjacent atoms in NaCl and
# n be the no. of atoms along the edge of the cube
#length of an edge = na
#volume of unit cube = n^3*a^3
a = (1/n)**(1./3.)
#Result
print "The distance between two adjacent atoms is",round((a/1E-8),2),"A"
```

In [3]:

```
from math import sqrt
#Variable declaration
w = 63.5 #atomic weight of copper
r = 1.278*10**-8 #atomic rdius(m)
N = 6.023*10**23 #Avogadro's number(/mol)
#Calculations
m = w/N #mass of each copper atom(gm)
#Since copper has FCC structure lattice constant
a = (4*r)/sqrt(2)
n = 4 #no. of atoms in unit cell of FCC structure
M = n*m #mass of unit cell
rho = M/a**3 #density
#Result
print "Density of copper =",round(rho,2),"gm/cm^3"
```