{
"metadata": {
"name": ""
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 2: Spectroscopy and Photochemistry"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Problem: 1, Page no: 65"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Constants\n",
"m_br79 = 78.9183 # Mass of 79Br, amu\n",
"m_br81 = 80.9163 # Mass of 91Br, amu\n",
"Na = 6.022 * 10 ** 23 # Mole constant, /mol\n",
"pi = 3.141 # Pi\n",
"c = 3 * 10 ** 10 # Speed of light, cm /s\n",
"\n",
"# Variable\n",
"wave_no = 323.2 # Wave no. of fund. vibration of 79Br - 81Br, /cm\n",
"\n",
"# Solution\n",
"mu = (m_br79 * m_br81) / ((m_br79 + m_br81) * Na)\n",
"\n",
"k = 4 * (pi * c * wave_no) ** 2 * mu * 10 ** -3\n",
"\n",
"print \"The force constant of the bond is\", \"{:.3e}\".format(k), \"N/m\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The force constant of the bond is 2.461e+02 N/m\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Problem: 2, Page no: 65"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Constants\n",
"Na = 6.022 * 10 ** 23 # Mole constant, /mol\n",
"pi = 3.141 # Pi\n",
"c = 3 * 10 ** 10 # Speed of light, cm /s\n",
"h = 6.626 * 10 ** -34 # Plank's constant, J.sec\n",
"\n",
"# Variables\n",
"b_l = 112.81 * 10 ** -12 # Equillibrium bond length, m\n",
"m1 = 12 # Mass of Carbon, g /mol\n",
"m2 = 16 # Mass of Oxygen, g /mol\n",
"\n",
"# Solution\n",
"mu = m1 * m2 / ((m1 + m2) * Na) # g\n",
"mu *= 10 ** -3 # kg\n",
"\n",
"B = h / (8 * pi ** 2 * mu * b_l ** 2 * c)\n",
"v2_3 = B * 6\n",
"\n",
"print \"The reduced mass of CO is\", \"{:.3e}\".format(mu), \"kg\"\n",
"print \"The frequency of 3->2 transition is\", \"{:.2f}\".format(v2_3), \"/cm\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The reduced mass of CO is 1.139e-26 kg\n",
"The frequency of 3->2 transition is 11.59 /cm\n"
]
}
],
"prompt_number": 3
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Problem: 3, Page no: 66"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Constants\n",
"Na = 6.022 * 10 ** 23 # Mole constant, /mol\n",
"\n",
"# Variables\n",
"d_NaCl = 2.36 * 10 ** -10 # Intermolecular dist. NaCl, m\n",
"m_Cl = 35 * 10 ** -3 # Atomic mass, kg /mol\n",
"m_Na = 23 * 10 ** -3 # Atomic mass, kg /mol\n",
"\n",
"# Solution\n",
"mu = m_Na * m_Cl / ((m_Na + m_Cl) * 10 ** -3 * Na) * 10 ** -3\n",
"\n",
"I = mu * d_NaCl ** 2\n",
"\n",
"print \"The reduced mass of NaCl is\", \"{:.3e}\".format(mu), \"kg\"\n",
"print \"The moment of inertia of NaCl is\", \"{:.3e}\".format(I), \"kg.m^2\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The reduced mass of NaCl is 2.305e-26 kg\n",
"The moment of inertia of NaCl is 1.284e-45 kg.m^2\n"
]
}
],
"prompt_number": 4
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Problem: 4, Page no: 66"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"\n",
"# Constant\n",
"e = 4000 # Extinction coeff., dm^3/mol/cm\n",
"\n",
"# Variable\n",
"x = 3 # Solution thickness, cm\n",
"\n",
"# Solution\n",
"A = math.log10(1 / 0.3) # Absorbance\n",
"C = A / (e * x)\n",
"\n",
"print \"The concentration of the solution is\", \"{:.2e}\".format(C), \"mol/dm^3\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The concentration of the solution is 4.36e-05 mol/dm^3\n"
]
}
],
"prompt_number": 5
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Problem: 5, Page no: 67"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Constants\n",
"pi = 3.141 # Pi\n",
"c = 3 * 10 ** 10 # Speed of light, cm /s\n",
"\n",
"# Variables\n",
"v_bar = 2140 # Fundamental vibrating freq, /cm\n",
"m_C = 19.9 * 10 ** -27 # Atomic mass of C, kg\n",
"m_O = 26.6 * 10 ** -27 # Atomic mass of O, kg\n",
"\n",
"# Solution\n",
"mu = m_O * m_C / (m_C + m_O)\n",
"k = 4 * (pi * c * v_bar) ** 2 * mu\n",
"\n",
"print \"The force constant of the molecule is\", \"{:.3e}\".format(k), \"N/m\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The force constant of the molecule is 1.852e+03 N/m\n"
]
}
],
"prompt_number": 6
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Problem: 6, Page no: 67"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"print \"a) Microwave < IR < UV-Visible < X-Ray.\"\n",
"print \"b) HCl and NO because they possess permanent dipole moments, so they are rotationally active.\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"a) Microwave < IR < UV-Visible < X-Ray.\n",
"b) HCl and NO because they possess permanent dipole moments, so they are rotationally active.\n"
]
}
],
"prompt_number": 8
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Problem: 7, Page no: 67"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"\n",
"# Constants\n",
"pi = 3.141 # pi\n",
"c = 3 * 10 ** 10 # speed of light, cm /s\n",
"h = 6.626 * 10 ** -34 # Plank's constant, J.sec\n",
"Na = 6.022 * 10 ** 23 # Mole constant, /mol\n",
"\n",
"# Variables\n",
"d = 20.7 # Interspacing, /cm\n",
"m1 = 1 # Mass of H, g / mol\n",
"m2 = 35.5 # Masso f Cl, g / mol\n",
"\n",
"# Solution\n",
"B = 0.1035 * 10 ** 2 # /m\n",
"I = h / (8 * pi ** 2 * B * c)\n",
"mu = m1 * m2 / ((m1 + m2) * Na)\n",
"mu *= 10 ** -3\n",
"r = math.sqrt(I / mu)\n",
"\n",
"print \"The intermolecular distance of HCl is\", \"{:.3e}\".format(r), \"m\"\n",
"# Discrepency in value is due to error in calculation in the textbook\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The intermolecular distance of HCl is 1.294e-10 m\n"
]
}
],
"prompt_number": 9
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Problem: 8, Page no: 68"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"\n",
"# Constant\n",
"e = 8000 # Molar absorbtion coeff, dm^3 / mol / cm\n",
"\n",
"# Variable\n",
"l = 2.5 # Thickness of solution, cm\n",
"\n",
"# Solution\n",
"C = math.log10(1 / 0.3) / (e * l)\n",
"\n",
"print \"The concentration of Solution from Lambert-Beer's Law is\",\n",
"print \"{:.2e}\".format(C), \"mol/dm^3\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The concentration of Solution from Lambert-Beer's Law is 2.61e-05 mol/dm^3\n"
]
}
],
"prompt_number": 10
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Problem: 9, Page no: 68"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"print \"a) In the visible-UV spectra, CH2 = CHOCHOCH3 exhibits\"\n",
"print \"a higher value of lambda(max) because it has two conjugated\"\n",
"print \"chromophores, that is, one double bond (C=C) and a carbonyl\"\n",
"print \"group.\"\n",
"\n",
"print\n",
"print \"b) Because of the symmetrical vibrations of C=C double bond and\"\n",
"print \"triple bond, ethylene and acetylene do not absorb IR energy.\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"a) In the visible-UV spectra, CH2 = CHOCHOCH3 exhibits\n",
"a higher value of lambda(max) because it has two conjugated\n",
"chromophores, that is, one double bond (C=C) and a carbonyl\n",
"group.\n",
"\n",
"b) Because of the symmetrical vibrations of C=C double bond and\n",
"triple bond, ethylene and acetylene do not absorb IR energy.\n"
]
}
],
"prompt_number": 13
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Problem: 10, Page no: 68"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"print \"Because CO2 is a linear molecule.\"\n",
"v_deg = 3 * 3 - 5\n",
"print \"The vibrational degree of freedom is\", v_deg"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Because CO2 is a linear molecule.\n",
"The vibrational degree of freedom is 4\n"
]
}
],
"prompt_number": 14
}
],
"metadata": {}
}
]
}