# Chapter 2: Spectroscopy and Photochemistry¶

## Problem: 1, Page no: 65¶

In [2]:
# Constants
m_br79 = 78.9183            # Mass of 79Br, amu
m_br81 = 80.9163            # Mass of 91Br, amu
Na = 6.022 * 10 ** 23       # Mole constant, /mol
pi = 3.141                  # Pi
c = 3 * 10 ** 10            # Speed of light, cm /s

# Variable
wave_no = 323.2             # Wave no. of fund. vibration of 79Br - 81Br, /cm

# Solution
mu = (m_br79 * m_br81) / ((m_br79 + m_br81) * Na)

k = 4 * (pi * c * wave_no) ** 2 * mu * 10 ** -3

print "The force constant of the bond is", "{:.3e}".format(k), "N/m"

The force constant of the bond is 2.461e+02 N/m


## Problem: 2, Page no: 65¶

In [3]:
# Constants
Na = 6.022 * 10 ** 23           # Mole constant, /mol
pi = 3.141                      # Pi
c = 3 * 10 ** 10                # Speed of light, cm /s
h = 6.626 * 10 ** -34           # Plank's constant, J.sec

# Variables
b_l = 112.81 * 10 ** -12        # Equillibrium bond length, m
m1 = 12                         # Mass of Carbon, g /mol
m2 = 16                         # Mass of Oxygen, g /mol

# Solution
mu = m1 * m2 / ((m1 + m2) * Na)  # g
mu *= 10 ** -3                   # kg

B = h / (8 * pi ** 2 * mu * b_l ** 2 * c)
v2_3 = B * 6

print "The reduced mass of CO is", "{:.3e}".format(mu), "kg"
print "The frequency of 3->2 transition is", "{:.2f}".format(v2_3), "/cm"

The reduced mass of CO is 1.139e-26 kg
The frequency of 3->2 transition is 11.59 /cm


## Problem: 3, Page no: 66¶

In [4]:
# Constants
Na = 6.022 * 10 ** 23       # Mole constant, /mol

# Variables
d_NaCl = 2.36 * 10 ** -10           # Intermolecular dist. NaCl, m
m_Cl = 35 * 10 ** -3                # Atomic mass, kg /mol
m_Na = 23 * 10 ** -3                # Atomic mass, kg /mol

# Solution
mu = m_Na * m_Cl / ((m_Na + m_Cl) * 10 ** -3 * Na) * 10 ** -3

I = mu * d_NaCl ** 2

print "The reduced mass of NaCl is", "{:.3e}".format(mu), "kg"
print "The moment of inertia of NaCl is", "{:.3e}".format(I), "kg.m^2"

The reduced mass of NaCl is 2.305e-26 kg
The moment of inertia of NaCl is 1.284e-45 kg.m^2


## Problem: 4, Page no: 66¶

In [5]:
import math

# Constant
e = 4000                    # Extinction coeff., dm^3/mol/cm

# Variable
x = 3                       # Solution thickness, cm

# Solution
A = math.log10(1 / 0.3)     # Absorbance
C = A / (e * x)

print "The concentration of the solution is", "{:.2e}".format(C), "mol/dm^3"

The concentration of the solution is 4.36e-05 mol/dm^3


## Problem: 5, Page no: 67¶

In [6]:
# Constants
pi = 3.141                  # Pi
c = 3 * 10 ** 10            # Speed of light, cm /s

# Variables
v_bar = 2140                # Fundamental vibrating freq, /cm
m_C = 19.9 * 10 ** -27      # Atomic mass of C, kg
m_O = 26.6 * 10 ** -27      # Atomic mass of O, kg

# Solution
mu = m_O * m_C / (m_C + m_O)
k = 4 * (pi * c * v_bar) ** 2 * mu

print "The force constant of the molecule is", "{:.3e}".format(k), "N/m"

The force constant of the molecule is 1.852e+03 N/m


## Problem: 6, Page no: 67¶

In [8]:
print "a) Microwave < IR < UV-Visible < X-Ray."
print "b) HCl and NO because they possess permanent dipole moments, so they are rotationally active."

a) Microwave < IR < UV-Visible < X-Ray.
b) HCl and NO because they possess permanent dipole moments, so they are rotationally active.


## Problem: 7, Page no: 67¶

In [9]:
import math

# Constants
pi = 3.141                  # pi
c = 3 * 10 ** 10            # speed of light, cm /s
h = 6.626 * 10 ** -34       # Plank's constant, J.sec
Na = 6.022 * 10 ** 23       # Mole constant, /mol

# Variables
d = 20.7                    # Interspacing, /cm
m1 = 1                      # Mass of H, g / mol
m2 = 35.5                   # Masso f Cl, g / mol

# Solution
B = 0.1035 * 10 ** 2        # /m
I = h / (8 * pi ** 2 * B * c)
mu = m1 * m2 / ((m1 + m2) * Na)
mu *= 10 ** -3
r = math.sqrt(I / mu)

print "The intermolecular distance of HCl is", "{:.3e}".format(r), "m"
# Discrepency in value is due to error in calculation in the textbook

The intermolecular distance of HCl is 1.294e-10 m


## Problem: 8, Page no: 68¶

In [10]:
import math

# Constant
e = 8000                # Molar absorbtion coeff, dm^3 / mol / cm

# Variable
l = 2.5                 # Thickness of solution, cm

# Solution
C = math.log10(1 / 0.3) / (e * l)

print "The concentration of Solution from Lambert-Beer's Law is",
print "{:.2e}".format(C), "mol/dm^3"

The concentration of Solution from Lambert-Beer's Law is 2.61e-05 mol/dm^3


## Problem: 9, Page no: 68¶

In [13]:
print "a) In the visible-UV spectra, CH2 = CHOCHOCH3 exhibits"
print "a higher value of lambda(max) because it has two conjugated"
print "chromophores, that is, one double bond (C=C) and a carbonyl"
print "group."

print
print "b) Because of the symmetrical vibrations of C=C double bond and"
print "triple bond, ethylene and acetylene do not absorb IR energy."

a) In the visible-UV spectra, CH2 = CHOCHOCH3 exhibits
a higher value of lambda(max) because it has two conjugated
chromophores, that is, one double bond (C=C) and a carbonyl
group.

b) Because of the symmetrical vibrations of C=C double bond and
triple bond, ethylene and acetylene do not absorb IR energy.


## Problem: 10, Page no: 68¶

In [14]:
print "Because CO2 is a linear molecule."
v_deg = 3 * 3 - 5
print "The vibrational degree of freedom is", v_deg

Because CO2 is a linear molecule.
The vibrational degree of freedom is 4