Chapter 7: Solid State

Problem 1, Page no: 209

In [3]:
# Solution
print "Weiss indices  \t\t\t1/2, 2/3, infinity \t2/3, 2, 1/3"
print "Reciprocal of Weiss indices  2, 3/2, 1/infinity \t3/2, 1/2, 3"
print "Clear fractions\t\t\t4, 3, 0            \t3, 1, 6"
print "Miller indices \t\t\t  (430)            \t  (316)"

Weiss indices  			1/2, 2/3, infinity 	2/3, 2, 1/3
Reciprocal of Weiss indices  2, 3/2, 1/infinity 	3/2, 1/2, 3
Clear fractions			4, 3, 0            	3, 1, 6
Miller indices 			  (430)            	  (316)

Problem 2, Page no: 209

In [5]:
from math import sqrt


# Variable
a = 450                 # pm

# Solution
d220 = a / sqrt(2 ** 2 + 2 ** 2 + 0)
print "Interplanar spacing", int(d220)

Interplanar spacing 159

Problem 3, Page no: 209

In [7]:
# Solution
print "Intercept    \t(a, b ,c)\t\t(a, 2b, c) \t\t(a, 2b, 2c) \t\t(infi, b, -c)"
print "Weiss indices\t 1, 1, 1 \t\t 1, 2, 1   \t\t 1, 2, 2    \t\t infi, 1, -1"
print "Reciprocals  \t 1, 1, 1 \t\t 1, 1/2, 1 \t\t 1, 1/2, 1/2\t\t 0, 1, -1"
print "Miller indicec\t (111)  \t\t (212)    \t\t (211)        \t\t   (011)"

Intercept    	(a, b ,c)		(a, 2b, c) 		(a, 2b, 2c) 		(infi, b, -c)
Weiss indices	 1, 1, 1 		 1, 2, 1   		 1, 2, 2    		 infi, 1, -1
Reciprocals  	 1, 1, 1 		 1, 1/2, 1 		 1, 1/2, 1/2		 0, 1, -1
Miller indicec	 (111)  		 (212)    		 (211)        		   (011)

Problem 4, Page no: 209

In [9]:
# Variables
rNa = 0.98 * 10 ** - 10             # m
rCl = 1.81 * 10 ** - 10             # m

# Solution
rr = rNa / rCl
print "When the radius ration is", "{:.2f}".format(rr),
print ", the coordination number is 6."

When the radius ration is 0.54 , the coordination number is 6.

Problem 5, Page no: 210

In [2]:
# Variables
rLi = 68                 # pm
rF = 136.                 # pm

# Solution
rr = rLi / rF
print "Radius ratio =", rr
print "The structure of LiF is scc and C.N of Li+ = 6"

Radius ratio = 0.5
The structure of LiF is scc and C.N of Li+ = 6

Problem 6, Page no: 210

In [3]:
from math import sin


# Variables
lamda = 2 * 10 ** - 10          # m
theta = 30                      # degrees

# Solution
print "For first-order reflection"
d = lamda / (2 * sin(theta))
dist = 2 * d
print "Hence, distance between planes is", "{:.0e}".format(abs(dist)), "m"

For first-order reflection
Hence, distance between planes is 2e-10 m

Problem 7, Page no: 210

In [17]:
from math import sqrt


# Variables
r = 174.6                   # pm

# Solution
a = r * sqrt(8)
print "For 200 plane: h = 2, k = 0, l = 0"
d200 = a / sqrt(2 ** 2)
print "d200 =", "{:.1f}".format(d200), "pm"
print "For 200 plane: h = 2, k = 2, l = 0"
d220 = a / sqrt(2 ** 2 + 2 ** 2)
print "d220 =", d220, "pm"

For 200 plane: h = 2, k = 0, l = 0
d200 = 246.9 pm
For 200 plane: h = 2, k = 2, l = 0
d220 = 174.6 pm

Problem 8, Page no: 210

In [4]:
# Constant
N = 6.023 * 10 ** 23

# Variables
wt = 55.6
p = 0.29                # nm
n = 2

# Solution
print "For BCC pattern,"
print "number of atoms per unit cell = 2"
d = n * (wt * 10 ** -3) / (N * (p * 10 ** - 9) ** 3)
print "Density of the metal is", "{:.2e}".format(d), "kg / m^3"
print "Number of nearest enighbour for BCC = 8"

For BCC pattern,
number of atoms per unit cell = 2
Density of the metal is 7.57e+03 kg / m^3
Number of nearest enighbour for BCC = 8

Problem 9, Page no: 210

In [19]:
# Solution
print "Intercept    \t2a/2, 2b, c/3"
print "Weiss indices\t1, 2, 1/3"
print "Reciprocals  \t1, 1/2, 3"
print "Miller indices\t (216)"

Intercept    	2a/2, 2b, c/3
Weiss indices	1, 2, 1/3
Reciprocals  	1, 1/2, 3
Miller indices	 (216)

Problem 10, Page no: 211

In [23]:
# Constant
N = 6.023 * 10 ** 23

# Variables
D = 0.53                # g / cm ^ 3
MM = 6.94               # g / mol
n = 2

# Solution
print "For BCC pattern,"
print "number of atoms per unit cell = 2"
V = D * N / (n * MM)
V = 1 / V
print "Volume of a unit cell of lithium metal is", "{:.2e}".format(V), "cc"

For BCC pattern,
number of atoms per unit cell = 2
Volume of a unit cell of lithium metal is 4.35e-23 cc

Problem 11, Page no: 211

In [25]:
from math import sqrt


print "AB remain in BCC structure if the edge length is a then body diagonal",
print "is root(3)a"
print "root(3)a = 2(r+  +  r-)"
A = (sqrt(3) * 0.4123 - 2 * 0.81) / 2
print "A+ =", "{:.2f}".format(A), "nm"

AB remain in BCC structure if the edge length is a then body diagonal is root(3)a
root(3)a = 2(r+  +  r-)
A+ = -0.45 nm