Chapter 1: Structure and Bonding

Problem: 1, Page no: 35

In [6]:
# Constant
c = 3 * 10 ** 10                # Velocity of light, cm/sec

# Variable
wavelength = 3500 * 10 ** -8    # Wavelength of radiation, cm

# Solution
print "v = c / wavelength"
print "v: Velocity, c: Speed of light"

v = c / wavelength

print "The frequency of radiation is", '{:.2e}'.format(v), "Heartz."

v = c / wavelength
v: Velocity, c: Speed of light
The frequency of radiation is 8.57e+14 Heartz.

Problem: 2, Page no: 36

In [7]:
# Constant
c = 3 * 10 ** 8                 # speed of light, m/sec

# Variable
f = 5 * 10 ** 16                # frequency, cycles/sec

# Solution
v_bar = f / c
print "The wave number is", '{:.2e}'.format(v_bar), "cycles/m."

The wave number is 1.67e+08 cycles/m.

Problem: 3, Page no: 36

In [8]:
# Constant
c = 3 * 10 ** 8                 # Speed of light, m/sec

# Variable
T = 2.4 * 10 ** -10             # Time period, sec

# Solution
f = 1 / T                       # Frequency, /sec
lamda = c / f                   # wavelength, m
v_bar = 1 / lamda               # wavenumber, /meter

print "Frequency:", '{:.2e}'.format(f), "/sec"
print "Wavelength:", '{:.2e}'.format(lamda), "m"
print "Wave number:", '{:.2e}'.format(v_bar), "/m"

Frequency: 4.17e+09 /sec
Wavelength: 7.20e-02 m
Wave number: 1.39e+01 /m

Problem: 4, Page no: 36

In [9]:
import math

# Constants
c = 3 * 10 ** 8                 # Speed of light, m/sec
m = 9.1 * 10 ** -31             # Mass of electron, kg
h = 6.626 * 10 ** -34           # Plank's constant, J.sec

# Variable
ke = 4.55 * 10 ** -25           # Kinetic Energy, J

# Solution
v = math.sqrt(ke * 2 / m)

lamda = h / (m * v)

print "The de Broglie wavelength is", '{:.2e}'.format(lamda), "m"

The de Broglie wavelength is 7.28e-07 m

Problem: 5, Page no: 36

In [10]:
# Constant
h = 6.626 * 10 ** -34           # Plank's constant, J.sec

# Variables
m = 10 * 10 ** -3               # Mass of the ball, kg
v = 10 ** 5                     # Velocity of ball, cm / sec

# Solution
lamda = (h * 10 ** 7) / (m * v)
print "The Wavelength of iron ball is", "{:.2}".format(lamda), "cm"

The Wavelength of iron ball is 6.6e-30 cm

Problem: 6, Page no: 37

In [12]:
# Constant
h = 6.626 * 10 ** -34           # Plank's constant, J.sec

# Variable
lamda = 2 * 10 ** -10           # wavelength, m

# Solution
p = h / lamda

print "The momentum of the particle is", "{:.2}".format(p), "kg.m/s"

The momentum of the particle is 3.3e-24 kg.m/s

Problem: 7, Page no: 37

In [20]:
# Constants
m = 9.1 * 10 ** -31             # Mass of electron, kg
h = 6.626 * 10 ** -34           # Plank's constant, J.sec
pi = 3.141                      # Pi

# Variable
delta_x = 1 * 10 ** -10         # uncertainty in velocity, m

# Solution
delta_v = h / (4 * pi * m * delta_x)

print "Uncertainty in position of electron >=",
print "{:.2}".format(delta_v), "m/s"

Uncertainty in position of electron >= 5.8e+05 m/s

Problem: 8, Page no: 37

In [19]:
# Constants
h = 6.626 * 10 ** -34           # Plank's constant, J.sec
pi = 3.141                      # Pi

# Variables
m = 10 ** -11                   # Mass of particle, g
v = 10 ** -4                    # Velocity of particle, cm/sec
delta_v = 0.1 / 100             # Uncertainty in velocity

# Solution
delta_v = v / 1000
delta_x = (h * 10 ** 7) / (4 * pi * delta_v * m)

print "Uncertainty in position >=",
print "{:.3e}".format(delta_x), "cm"

Uncertainty in position 5.27e-10 cm

Problem: 9, Page no: 37

In [22]:
# Constants
c = 3 * 10 ** 8                 # Speed of light, m/sec
h = 6.626 * 10 ** -34           # Plank's constant, J.sec

# Variable
lamda = 650 * 10 ** -12         # Wavelength of radiation, m

# Solution
E = h * c / lamda

print "Energy per photon", "{:.3e}".format(E), "J"

Energy per photon 3.058e-16 J

Problem: 10, Page no: 37

In [24]:
# Constant
h = 6.625 * 10 ** -34           # Plank's constant, J.sec

# Variables
v = 6.5 * 10 ** 7               # Velocity of particle, m/s
lamda = 5 * 10 ** -11           # Wavelength, m

# Solution
P = h / lamda

print "The momentum of the particle", "{:.2e}".format(P), "kg.m/s"

The momentum of the particle 1.33e-23 kg.m/s

Problem: 11, Page no: 38

In [26]:
import math

# Constants
c = 3 * 10 ** 8                 # Speed of light, m/sec
m = 9.1 * 10 ** -31             # Mass of electron, kg
h = 6.626 * 10 ** -34           # Plank's constant, J.sec

# Variables
lamda = 200 * 10 ** -7          # Wavelength, cm
wf = 6.95 * 10 ** -12           # Work function, erg

# Solution
E = (h * c) * 10 ** 9 / lamda

print "Energy of photon", "{:.3e}".format(E), "erg"

ke = E - wf

v = math.sqrt((2 * ke) / (m * 10 ** 3)) * 10 ** -2

print "The maximum velocity of electron", "{:.3e}".format(v), "m/sec"

Energy of photon 9.939e-12 erg
The maximum velocity of electron 8.105e+05 m/sec

Problem: 12, Page no: 38

In [34]:
# Constant
h = 6.626 * 10 ** -34           # Plank's constant, J.sec

# Variables
m = 150                         # Weight of ball, gm
v = 50                          # Velocity, m/sec

lamda = h / (m * v * 10 ** -8)
print "Wavelength of ball", "{:.3e}".format(lamda), "m"
print "Its wavelength is so short that it does not fall",
print "in visible range, so we cannot observe it."

Wavelength of ball 8.835e-30 m
Its wavelength is so short that it does not fall in visible range, so we cannot observe it.

Problem: 13, Page no: 39

In [35]:
# Constant
h = 6.626 * 10 ** -34           # Plank's constant, J.sec
pi = 3.141                      # Pi

# Variables
m = 0.1                         # Mass of base ball, kg
delta_x = 10 ** -10             # Uncertainty in position, m

# Solution
delta_v = h / (4 * pi * m * delta_x)

print "Uncertainty in velocity >=", "{:.2e}".format(delta_v), "m/s"

Uncertainty in velocity >= 5.27e-24 m/s

Problem: 14, Page no: 39

In [38]:
# Constant
t_v = 1.3 * 10 ** 15            # Threashold freq. Pt, /sec
h = 6.626 * 10 ** -34           # Plank's constant, J.sec


# Solution
print "The threshold frequency is the lowest frequency",
print "that photons may possess to produce the photoelectric",
print "effect."
E = h * t_v
print "The energy corresponding to this frequency is the minimum",
print "energy =", "{:.2e}".format(E), "erg"

The threshold frequency is the lowest frequency that photons may possess to produce the photoelectric effect.
The energy corresponding to this frequency is the minimum energy = 8.61e-19 erg

Problem: 15, Page no: 39

In [39]:
# Constants
m = 9.1 * 10 ** -31             # Mass of electron, kg
h = 6.626 * 10 ** -34           # Plank's constant, J.sec
e = 1.602 * 10 ** -19           # Charge of electron, C

# Variable
v = 1.87 * 10 ** 9              # Velocity of electron, m/sec

# Solution
V = m * v ** 2 / (2 * e)
lamda = h / (m * v)

print "The voltage is", "{:.2e}".format(V), "volt"
print "The de Broglie wavelength is", "{:.2e}".format(lamda), "m"

The voltage is 9.93e+06 volt
The de Broglie wavelength is 3.89e-13 m

Problem 16, Page no: 39

In [40]:
# Constants
m = 9.1 * 10 ** -31             # Mass of electron, kg
h = 6.626 * 10 ** -34           # Plank's constant, J.sec

# Variable
lamda = 4.8 * 10 ** -9          # Wavelength of electron, m

# Solution
ke = ((h / lamda) ** 2) / (2 * m)

print "The Kinetic Energy of moving electron is", "{:.2e}".format(ke),
print "J"

The Kinetic Energy of moving electron is 1.05e-20 J

Problem: 17, Page no: 39

In [44]:
# Constants
m = 9.1 * 10 ** -31             # Mass of electron, kg
h = 6.626 * 10 ** -34           # Plank's constant, J.sec
c = 3 * 10 ** 8                 # Speed of light, m/sec

# Variables
v = 6.46 * 10 ** 5              # Velocity of electron, m/sec
lamda = 200 * 10 ** -9          # Wavelength of light, m

# Solution
E = (h * c) / lamda
ke = m * v ** 2
w = E - ke

print "The workfunction of the metal surface is", "{:.3e}".format(w),
print "J"

The workfunction of the metal surface is 6.141e-19 J

Problem: 18, Page no: 40

In [45]:
import math

# Constants
e = 1.602 * 10 ** -19           # Charge of proton, C
m_p = 1.66 * 10 ** -27          # Mass of proton, kg
m_e = 9.1 * 10 ** -31           # Mass of electron, kg
h = 6.626 * 10 ** -34           # Plank's constant, J.sec

# Variable
V = 35                          # Acceleration potential, volt

# Solution
lamda_p = h / math.sqrt(2 * e * V * m_p)
lamda_e = h / math.sqrt(2 * e * V * m_e)

print "The wavelength of electron when accelerated with same",
print "potential is", "{:.3e}".format(lamda_e), "m"

The wavelength of electron when accelerated with same potential is 2.074e-10 m

Problem: 19, Page no: 41

In [51]:
B_O1 = (10 - 6) / 2              # Bond Order for O2
B_O2 = (10 - 7) / 2              # Bond Order for O2-

print "Bond length of O2- > O2 as Bond order of O2",
print "> Bond order of O2- :", B_O1 > B_O2
print "Both are paramagnetic, because they contain unpaired electrons."

Bond length of O2- > O2 as Bond order of O2 > Bond order of O2- : True
Both are paramagnetic, because they contain unpaired electrons.

Problem: 20, Page no: 41

In [8]:
B_O = (9 - 4) / 2.0         # Bond order of N2+

print "MO configuration of N2+ is"
print "σ(1s2)σ*(1s2)σ(2s2)σ*(2s2) [π(2px2) = π(2py2)] σ(2pz1)\n"
print "The Bond order of N2+, 1/2[Nb - Na] =", B_O

MO configuration of N2+ is
σ(1s2)σ*(1s2)σ(2s2)σ*(2s2) [π(2px2) = π(2py2)] σ(2pz1)

The Bond order of N2+, 1/2[Nb - Na] = 2.5

Problem: 21, Page no: 41

In [55]:
# Solution
v_n = 2 * 5             # number of valence e- in nitrogen
v_co = 4 + 6            # number of valence e- in CO

print "The number of valence electrons in N2", v_n
print "The number of valence electrons in CO", v_co

The number of valence electrons in N2 10
The number of valence electrons in CO 10

Problem: 22, Page no: 41

In [2]:
print "O2(+2) [B.O. = 1.0] < O2(-1) [B.O. = 1.5] <",
print "O2 [B.O. = 2.0] < O2(+1) [B.O. =2.5]"

O2(+2) [B.O. = 1.0] < O2(-1) [B.O. = 1.5] < O2 [B.O. = 2.0] < O2(+1) [B.O. =2.5]

Problem: 23, Page no: 41

In [6]:
print "The number of electrons in N2-", (7 + 8)
print "The molecular configuration of N2- is"
print "σ(1s2)σ*(1s2)σ(2s2)σ*(2s2) [π(2px2) = π(2py2)]",
print "σ(2pz2) [π*(2px1) = π*(2py0)] OR [π*(2px0) = π*(2py1)]"

The number of electrons in N2- 15
The molecular configuration of N2- is
σ(1s2)σ*(1s2)σ(2s2)σ*(2s2) [π(2px2) = π(2py2)] σ(2pz2) [π*(2px1) = π*(2py0)] OR [π*(2px0) = π*(2py1)]