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{
"metadata": {
"name": "",
"signature": "sha256:0de7fdc3a9080098412999030f35bc2fb5913b94ca2786fe525f30a16f00b7a8"
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 10: Molecules, Lasers and Solids"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 10.1, Page 342"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"\n",
"#Variable declaration\n",
"m = 2.33e-026; # Mass of a nitrogen atom, kg\n",
"R = 1.1e-010; # Interatomic separation between two nitrogen atoms, m\n",
"h = 6.626e-034; # Planck's constant, Js\n",
"e = 1.6e-019; # Energy equivalent of 1 eV, J\n",
"\n",
"#Calculations\n",
"h_bar = h/(2*math.pi); # Reduced Planck's constant, Js\n",
"I = m*R**2/2; # Momemt of rotational inertia of nitrogen gas molecule, kg-Sq.m\n",
"l = 1; # Rotational angular momentum quantum number\n",
"E_rot = h_bar**2*l*(l+1)/(2*I); # The energy for lowest rotational state of the nitrogen molecule, J\n",
"\n",
"#Result\n",
"print \"The energy for lowest rotational state of the nitrogen molecule = %4.2e eV\"%(E_rot/e)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The energy for lowest rotational state of the nitrogen molecule = 4.93e-04 eV\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 10.2, Page 343"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"\n",
"#Variable declaration\n",
"h = 6.626e-034; # Planck's constant, Js\n",
"e = 1.6e-019; # Energy equivalent of 1 eV, J\n",
"h_bar = h/(2*math.pi); # Reduced Planck's constant, Js\n",
"k = 1.38e-023; # Boltzmann constant, J/K\n",
"u = 1.67e-027; # Mass equivalent of 1 amu, kg\n",
"\n",
"#Calculations\n",
"m1 = 34.97*u; # Atomic mass of chlorine atom, kg\n",
"m2 = 1.008*u; # Atomic mass of hydrogen atom, kg\n",
"mu = m1*m2/(m1 + m2); # Reduced mass of the HCl system, kg\n",
"delta_E = 0.36; # Spacing between vibrational energy levels of the HCl molecule, eV\n",
"omega = delta_E*e/h_bar; # Angular frequency of vibration, rad/s\n",
"kapa = mu*omega**2; # Effective force constant for HCl molecule, N/m\n",
"T = delta_E*e/k; # Classical temperature associated with the rotational energy spacing, K\n",
"\n",
"#Results\n",
"print \"The effective force constant for HCl molecule = %3d N/m\"%(math.ceil(kapa))\n",
"print \"The classical temperature associated with the rotational energy spacing = %4d K\"%(math.ceil(T))\n",
"#answers differ due to rounding errors"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The effective force constant for HCl molecule = 489 N/m\n",
"The classical temperature associated with the rotational energy spacing = 4174 K\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 10.4, Page 358"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Variable declaration\n",
"e = 1.602e-019; # Charge on an electron, C\n",
"N_A = 6.023e+023; # Avogadro's number\n",
"alpha = 1.7476; # Madelung constant\n",
"E = -764.4e+003; # Dissociation energy of NaCl molecule, J/mol\n",
"V = E/N_A; # Repulsive potential energy, J\n",
"k = 8.988e+009; # Coulomb's constant, N-Sq.m/C^2\n",
"r0 = 0.282e-009; # Equilibrium separation for nearest neighbour in NaCl, m\n",
"\n",
"#Calculations\n",
"rho = r0*(1+r0*V/(k*alpha*e**2)); # Range parameter for NaCl, nm\n",
"\n",
"#Result\n",
"print \"The range parameter for NaCl = %6.4f nm\"%(rho/1e-009)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The range parameter for NaCl = 0.0316 nm\n"
]
}
],
"prompt_number": 3
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 10.5, Page 365"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Variable declaration\n",
"e = 1.602e-019; # Charge on an electron, C\n",
"r = 5.29e-011; # Orbital radius of electron equal to Bohr radius, m\n",
"B = 2.0; # Applied magnetic field, T\n",
"m = 9.11e-031; # Mass of an electron, kg\n",
"\n",
"#Calculations\n",
"delta_mu = e**2*r**2*B/(4*m); # Induced diamagnetic moment in the atom, J/T\n",
"\n",
"#Result\n",
"print \"The induced diamagnetic moment in the atom = %3.1e J/T\"%delta_mu"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The induced diamagnetic moment in the atom = 3.9e-29 J/T\n"
]
}
],
"prompt_number": 4
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 10.6, Page 366"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"\n",
"#Variable declaration\n",
"mu_B = 9.27e-024; # Bohr's magneton, J/T\n",
"B = 0.50; # Applied magnetic field, T\n",
"k = 1.38e-023; # Boltzmann constant, J/K\n",
"\n",
"#Calculations&Results\n",
"T = 10*mu_B*B/k; # Temperature at which mu*B = 0.1k*T, K\n",
"b_muB = mu_B*B/(k*T);\n",
"ratio = b_muB/math.tanh(b_muB); # Ratio of b_muB and tanh(b_muB)\n",
"if (ratio - 1) < 0.01:\n",
" print \"The value of T = %4.2f K is suitable as a classical temperature.\"%T\n",
"else:\n",
" print \"The value of T = %4.2f K is not suitable as a classical temperature.\"%T\n",
"\n",
"# For higher temperature\n",
"T = 100; # Given temperature\n",
"b_muB = mu_B*B/(k*T);\n",
"ratio = b_muB/math.tanh(b_muB); # Ratio of b_muB and tanh(b_muB)\n",
"if (ratio - 1) < 0.001:\n",
" print \"At the value of T = %4.2f K, the approximation is an excellent one.\"%T\n",
"else:\n",
" print \"At the value of T = %4.2f K, the approximation is not an excellent.\"%T"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The value of T = 3.36 K is suitable as a classical temperature.\n",
"At the value of T = 100.00 K, the approximation is an excellent one.\n"
]
}
],
"prompt_number": 5
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 10.7, Page 374"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Variable declaration\n",
"k = 1.38e-023; # Boltzmann constant, J/K\n",
"e = 1.602e-019; # Energy equivalent of 1 eV, J\n",
"h = 6.62e-034; # Planck's constant, Js\n",
"c = 3.00e+008; # Speed of light, m/s\n",
"T_c = 9.25; # Critical temperature for niobium, K\n",
"\n",
"#Calculations\n",
"E_g = 3.54*k*T_c; # Energy gap for niobium from BCS theory, J\n",
"lamda = h*c/E_g; # Minimum photon wavelength needed to break the Cooper pair, m\n",
"\n",
"#Results\n",
"print \"The energy gap for niobium = %4.2f meV\"%(E_g/(e*1e-003))\n",
"print \"The minimum photon wavelength needed to break the Cooper pair = %4.2e m\"%lamda"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The energy gap for niobium = 2.82 meV\n",
"The minimum photon wavelength needed to break the Cooper pair = 4.39e-04 m\n"
]
}
],
"prompt_number": 6
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 10.8, Page 382"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"\n",
"#Variable declaration\n",
"r = 1e-002; # Radius of the loop, m\n",
"phi0 = 2.068e-015; # Magnetic flux penetrating to the loop, T-Sq.m\n",
"\n",
"#Calculations\n",
"A = math.pi*r**2; # Area of the loop, Sq.m\n",
"B = phi0/A; # Magnetic field perpendicular to the loop, T\n",
"\n",
"#Result\n",
"print \"The magnetic field perpendicular to the loop = %4.2e T\"%B"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The magnetic field perpendicular to the loop = 6.58e-12 T\n"
]
}
],
"prompt_number": 7
}
],
"metadata": {}
}
]
}
|