{
"cells": [
{
"cell_type": "markdown",
"metadata": {},
"source": [
"# chapter1: Quantum Mechanics"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## example 1.1;page:1.5"
]
},
{
"cell_type": "code",
"execution_count": 1,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Ex-1.1 : pg no:1.5\n",
"The energy of the particle from de Broglie wavelength in eV:\n"
]
},
{
"data": {
"text/plain": [
"(None, 0.019)"
]
},
"execution_count": 1,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"#cal of de broglie wavelenth\n",
"#intiation of all variables \n",
"#given that\n",
"lamda = 2.1e-10 \n",
"m = 1.67e-027 \n",
"h = 6.626e-034 \n",
"e = 1.6e-019 \n",
"# From de Broglie relation, lambda = h/math.sqrt(2*m*E)\n",
"E = h**2/(2*m*lamda**2*e)\n",
"print(\"Ex-1.1 : pg no:1.5\")\n",
"print(\"The energy of the particle from de Broglie wavelength in eV:\"),round(E,3)\n",
"#The energy of the particle from de Broglie wavelength = 1.863e-002 e"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## example 1.2;page:1.5"
]
},
{
"cell_type": "code",
"execution_count": 2,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Ex1.2: Page-1.5\n",
"The de Broglie wavelength of the particle in m:\n"
]
},
{
"data": {
"text/plain": [
"(None, 9.063964727801313e-17)"
]
},
"execution_count": 2,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"#cal of de brogle wavelength\n",
"#intiation of all variables \n",
"#given that\n",
"import math\n",
"m = 1.67e-027 # Mass of the particle, kg\n",
"h = 6.626e-034 # Planck's constant, Js\n",
"e = 1.6e-019 # Energy equivalent of 1 eV, J/eV\n",
"E = 1e+011*e # Energy of the particle, J\n",
"lamda=h/math.sqrt(2*m*E) # de Broglie wavelength of the particle\n",
"print(\"Ex1.2: Page-1.5\")\n",
"print(\"The de Broglie wavelength of the particle in m:\"),lamda\n",
"# Result \n",
"# The de Broglie wavelength of the particle = 9.06e-017 m "
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## example:1.3;page:1.5"
]
},
{
"cell_type": "code",
"execution_count": 3,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Ex1.3: Page-1.5\n",
"The de Broglie wavelength of the electron in angstrom\n"
]
},
{
"data": {
"text/plain": [
"(None, 0.09)"
]
},
"execution_count": 3,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"#cal of de brogle wavelength of the electron\n",
"#intiation of all variables \n",
"#Given that\n",
"import math\n",
"V = 20e+03; # Accelerating voltage of electron, V\n",
"lamda = 12.25/math.sqrt(V); # de Broglie wavelength of the accelerated electron, m\n",
"print(\"Ex1.3: Page-1.5\")\n",
"print(\"The de Broglie wavelength of the electron in angstrom\"),round (lamda,2)\n",
"# Result \n",
"# The de Broglie wavelength of the electron = 0.0866 angstrom "
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## example:1.4;page:1.6"
]
},
{
"cell_type": "code",
"execution_count": 4,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Ex1.4: Page-1.6\n",
"The energy of the electron from de Broglie wavelength in eV:\n"
]
},
{
"data": {
"text/plain": [
"(None, 5.5757694136484825e-14)"
]
},
"execution_count": 4,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"#cal of de brogle wavelength\n",
"#intiation of all variables \n",
"#Given that\n",
"lamda = 5.2e-03; # de Broglie wavelength of the electron, m\n",
"m = 9.1e-031; # Mass of the electron, kg\n",
"h = 6.626e-034; # Planck's constant, Js\n",
"e = 1.6e-019; # Energy equivalent of 1 eV, J/eV\n",
"# From de Broglie relation, lambda = h/sqrt(2*m*E), solving for E\n",
"E = h**2/(2*m*lamda**2*e); # Energy of the electron, eV\n",
"print(\"Ex1.4: Page-1.6\")\n",
"print(\"The energy of the electron from de Broglie wavelength in eV:\"), E\n",
"# Result \n",
"# The energy of the electron from de Broglie wavelength = 5.576e-014 eV "
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## example:1.5;page:1.6"
]
},
{
"cell_type": "code",
"execution_count": 5,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Ex1.5: Page-1.6\n",
"The velocity of the neutron in m/s:\n",
"The de Broglie wavelength of the neutron in m:\n"
]
},
{
"data": {
"text/plain": [
"(None, 2.8662773171280257e-13)"
]
},
"execution_count": 5,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"#cal of velocity and de brogle wavelength of the neutron\n",
"#intiation of all variables \n",
"#Given that\n",
"import math\t\n",
"m = 1.67e-027; # Mass of the neutron, kg\n",
"h = 6.626e-034; # Planck's constant, Js\n",
"e = 1.6e-019; # Energy equivalent of 1 eV, J/eV\n",
"E = 1e+04*e; # Energy of the neutron, J\n",
"# As E = 1/2*m*v**2, solving for v\n",
"v = math.sqrt(2*E/m); # Velocity of the neutron, m/s\n",
"lamda = h/(m*v); # de Broglie wavelength of the neutron, m\n",
"print(\"Ex1.5: Page-1.6\")\n",
"print(\"The velocity of the neutron in m/s:\"),round (v,2)\n",
"print(\"The de Broglie wavelength of the neutron in m:\"), lamda\n",
"# Result \n",
"# The velocity of the neutron = 1.38e+006 m/s\n",
"# The de Broglie wavelength of the neutron = 2.87e-013 m "
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## example:1.6;page:1.6"
]
},
{
"cell_type": "code",
"execution_count": 6,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Ex1.6: Page-1.6\n",
"The de Broglie wavelength of the thermal neutrons in angstrom:\n"
]
},
{
"data": {
"text/plain": [
"(None, 1.45)"
]
},
"execution_count": 6,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"#cal of de brogle wavelength of the thermal neutrons\n",
"#intiation of all variables \n",
"#Given that\n",
"import math\t\t\n",
"m = 1.67e-027; # Mass of the neutron, kg\n",
"k = 1.38e-023; # Boltzmann constant, J/mol/K\n",
"T = 27+273; # Room temperature, K\n",
"h = 6.626e-034; # Planck's constant, Js\n",
"e = 1.6e-019; # Energy equivalent of 1 eV, J/eV\n",
"v = math.sqrt(3*k*T/m); # Velocity of the neutron, m/s\n",
"lamda = h/(m*v); # de Broglie wavelength of the neutron, m\n",
"print(\"Ex1.6: Page-1.6\")\n",
"print(\"The de Broglie wavelength of the thermal neutrons in angstrom:\"),round (lamda/1e-010,2)\n",
"# Result \n",
"# The de Broglie wavelength of the thermal neutrons = 1.45 angstrom "
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## example:1.7;page:1.6"
]
},
{
"cell_type": "code",
"execution_count": 7,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Ex1.7: Page-1.6\n",
"The angle of deviation for first order diffraction maxim in degrees:\n"
]
},
{
"data": {
"text/plain": [
"(None, 0.044)"
]
},
"execution_count": 7,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"#cal of Angle of deviation\n",
"#intiation of all variables \n",
"#given that\n",
"import math\n",
"m = 9.1e-031; # Mass of the electron, kg\n",
"e = 1.6e-019; # Energy equivalent of 1 eV, J/eV\n",
"h = 6.626e-034; # Planck's constant, Js\n",
"E = 20e+03*e; # Energy of the electron, J\n",
"# As 1/2*m*v**2 = E, solving for v\n",
"v = math.sqrt(2*E/m); # Velocity of the electron, m/s\n",
"lamda = h/(m*v); # de Broglie wavelength of the electron, m\n",
"n = 1; # First order diffraction\n",
"d = 9.8e-011; # Atomic spacing for thin gold foil, m\n",
"# Using Bragg's equation, 2*d*sin(theta) = n*lamda and solving for theta\n",
"theta = math.sin(n*lamda/(2*d)); # Angle of deviation for first order diffraction maxima, degree\n",
"print(\"Ex1.7: Page-1.6\")\n",
"print(\"The angle of deviation for first order diffraction maxim in degrees:\"),round (theta,3)\n",
"# Result \n",
"# The angle of deviation for first order diffraction maxima = 2.54 degrees "
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## example:1.8;page:1.7"
]
},
{
"cell_type": "code",
"execution_count": 8,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Ex1.8: Page-1.7\n",
"The de Broglie wavelength of the electron in angstrom:\n"
]
},
{
"data": {
"text/plain": [
"(None, 5.49)"
]
},
"execution_count": 8,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"#cal of de brogle wavelength of the electron\n",
"#intiation of all variables \n",
"#given that\n",
"import math\t\t\t\t\t\n",
"m = 9.1e-031; # Mass of the electron, kg\n",
"e = 1.6e-019; # Energy equivalent of 1 eV, J/eV\n",
"h = 6.626e-034; # Planck's constant, Js\n",
"E = 5*e; # Energy of the electron, J\n",
"# As 1/2*m*v**2 = E, solving for v\n",
"v = math.sqrt(2*E/m); # Velocity of the electron, m/s\n",
"lamda = h/(m*v); # de Broglie wavelength of the electron, m\n",
"print(\"Ex1.8: Page-1.7\")\n",
"print(\"The de Broglie wavelength of the electron in angstrom:\"),round (lamda/1e-010,2)\n",
"# Result \n",
"# The de Broglie wavelength of the electron = 5.5 angstrom"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## example:1.9;page:1.7"
]
},
{
"cell_type": "code",
"execution_count": 9,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Ex1.9: Page-1.7\n",
"The de Broglie wavelength of the neutron in angstrom\n"
]
},
{
"data": {
"text/plain": [
"(None, 0.29)"
]
},
"execution_count": 9,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"#cal of de brogle wavelength of the neutron\n",
"#intiation of all variables \n",
"#given that\n",
"import math\n",
"m = 1.67e-027; # Mass of the neutron, kg\n",
"e = 1.6e-019; # Energy equivalent of 1 eV, J/eV\n",
"h = 6.626e-034; # Planck's constant, Js\n",
"E = 1*e; # Energy of the electron, J\n",
"lamda = h/math.sqrt(2*m*E); # de Broglie wavelength of the neutron, m\n",
"print(\"Ex1.9: Page-1.7\")\n",
"print(\"The de Broglie wavelength of the neutron in angstrom\"),round (lamda/1e-010,2)\n",
"# Result \n",
"# The de Broglie wavelength of the neutron = 0.29 angstrom "
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## example:1.10;page:1.8"
]
},
{
"cell_type": "code",
"execution_count": 10,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Ex1.10 Page-1.8\n",
"The de Broglie wavelength associated with moving proton in m:\n"
]
},
{
"data": {
"text/plain": [
"(None, 5.95)"
]
},
"execution_count": 10,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"#cal of de brogle wavelength\n",
"#intiation of all variables \n",
"#given that\n",
"m = 1.67e-027; # Mass of the proton, kg\n",
"c = 3e+08; # Speed of light, m/s\n",
"v = 1./20.*c; # Velocity of the proton, m/s\n",
"h = 6.626e-034; # Planck's constant, Js\n",
"lamda = h/m*v; # de Broglie wavelength of the neutron,\n",
"print(\"Ex1.10 Page-1.8\")\n",
"print(\"The de Broglie wavelength associated with moving proton in m:\"),round (lamda,2)\n",
"# Result \n",
"# The de Broglie wavelength associated with moving proton = 2.645e-14 m"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## example:1.11;page:1.8"
]
},
{
"cell_type": "code",
"execution_count": 11,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Ex1.11: Page-1.8\n",
"The wavelength of matter wave associated with moving proton in m:\n"
]
},
{
"data": {
"text/plain": [
"(None, 1.9838323353293413e-15)"
]
},
"execution_count": 11,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"#cal of wavelength of matter wave\n",
"#intiation of all variables \n",
"#given that\n",
"m = 1.67e-027; # Mass of the proton, kg\n",
"v = 2e+08; # Velocity of the proton, m/s\n",
"h = 6.626e-034; # Planck's constant, Js\n",
"lamda = h/(m*v); # de Broglie wavelength of the neutron, m\n",
"print(\"Ex1.11: Page-1.8\")\n",
"print(\"The wavelength of matter wave associated with moving proton in m:\"),lamda\n",
"# Result \n",
"# The wavelength of matter wave associated with moving proton = 1.984e-15 m "
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## example:1.12;page:1.17"
]
},
{
"cell_type": "code",
"execution_count": 12,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Ex1.12: Page-1.17\n",
"The de Broglie wavelength of the electron accelearted through a given potential in m:\n"
]
},
{
"data": {
"text/plain": [
"(None, 1.736483832029858e-10)"
]
},
"execution_count": 12,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"#cal of de brogle wavelength of the electron\n",
"#intiation of all variables \n",
"#given that\n",
"import math\n",
"m = 9.1e-031; # Mass of the electron, kg\n",
"q = 1.6e-019; # Charge on an electron, C\n",
"V = 50; # Accelearting potential, V\n",
"E = q*V; # Energy gained by the electron, J\n",
"h = 6.626e-034; # Planck's constant, Js\n",
"lamda = h/math.sqrt(2*m*E); # de Broglie wavelength of the electron, m\n",
"print(\"Ex1.12: Page-1.17\")\n",
"print(\"The de Broglie wavelength of the electron accelearted through a given potential in m:\"), lamda\n",
"# Result \n",
"# The de Broglie wavelength of the electron accelearted through a given potential = 1.736e-10 m "
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## example:1.13;page:1.17"
]
},
{
"cell_type": "code",
"execution_count": 13,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Ex1.13: Page-1.17\n",
"The interplanar spacing of the crystal in angstrom:\n"
]
},
{
"data": {
"text/plain": [
"(None, 0.855)"
]
},
"execution_count": 13,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"#cal of interplanar spacing of the crystal\n",
"#intiation of all variables \n",
"#given that\n",
"import math\n",
"theta = 45; # Diffraction angle, degrees\n",
"h = 6.626e-034; # Planck's constant\n",
"m = 1.67e-027; # Mass of a neutron, kg\n",
"n = 1; # Order of diffraction\n",
"k = 1.38e-023; # Boltzmann constant, J/mol/K\n",
"T = 27+273; # Absolute room temperature, K\n",
"E = 3/2*k*T; # Energy of the neutron, J\n",
"lamda = h/math.sqrt(2*m*E); # de-Broglie wavelength of neutrons, m\n",
"# From Bragg's law, 2*d*sin(theta) = n*lamda, solving for d\n",
"d = n*lamda/(2*math.sin(theta));\n",
"print(\"Ex1.13: Page-1.17\")\n",
"print(\"The interplanar spacing of the crystal in angstrom:\"), round(d/1e-010,3)\n",
"# Result \n",
"# The interplanar spacing of the crystal = 1.03 angstrom "
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## example:1.14;page:1.18"
]
},
{
"cell_type": "code",
"execution_count": 14,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Ex1.14: Page-1.18\n",
"The interplanar spacing of the crystal in m:\n"
]
},
{
"data": {
"text/plain": [
"(None, 2.508685000852822e-11)"
]
},
"execution_count": 14,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"#cal of interplanar spacing of the crystal\n",
"#intiation of all variables \n",
"#given that\n",
"import math\n",
"theta = 70; # Glancing angle at which reflection occurs, degrees\n",
"h = 6.626e-034; # Planck's constant\n",
"m = 9.1e-031; # Mass of a electron, kg\n",
"e = 1.6e-019; # Electronic charge, C\n",
"V = 1000; # Accelerating potential, V\n",
"n = 1; # Order of diffraction\n",
"E = e*V; # Energy of the electron, J\n",
"lamda = h/math.sqrt(2*m*E); # de-Broglie wavelength of electron, m\n",
"# From Bragg's law, 2*d*sin(theta) = n*lamda, solving for d\n",
"d = n*lamda/(2*math.sin(theta)); # Interplanar spacing, m\n",
"print(\"Ex1.14: Page-1.18\")\n",
"print(\"The interplanar spacing of the crystal in m:\"), d\n",
"# Result \n",
"# The interplanar spacing of the crystal = 2.0660e-11 m "
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## example:1.15;page:1.18"
]
},
{
"cell_type": "code",
"execution_count": 15,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Ex1.15: Page-1.18\n",
"The de-Broglie wavelength of electron accelerated at V volts in angstrom:\n"
]
},
{
"data": {
"text/plain": [
"(None, 12.23)"
]
},
"execution_count": 15,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"#cal of de-Broglie wavelength of electron\n",
"#intiation of all variables \n",
"#given that\n",
"import math\n",
"h = 6.6e-034; # Planck's constant\n",
"m = 9.1e-031; # Mass of a electron, kg\n",
"e = 1.6e-019; # Electronic charge, C\n",
"V = 1; # For simplicity the accelerating potential is assumed to be unity, V\n",
"E = e*V; # Energy of the electron, J\n",
"lamda = h/math.sqrt(2*m*E); # de-Broglie wavelength of electron, m\n",
"print(\"Ex1.15: Page-1.18\")\n",
"print(\"The de-Broglie wavelength of electron accelerated at V volts in angstrom:\"),round (lamda/1e-010,2)\n",
"# Result \n",
"# de-Broglie wavelength of electron accelerated at V volts = 12.23/sqrt(V) angstrom "
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## example:1.16;page:1.18"
]
},
{
"cell_type": "code",
"execution_count": 16,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Ex1.16: Page-1.18\n",
"de-Broglie wavelength of electron accelerated at volts in m:\n"
]
},
{
"data": {
"text/plain": [
"(None, 1.223061372490817e-10)"
]
},
"execution_count": 16,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"#cal of de-Broglie wavelength of electron\n",
"#intiation of all variables \n",
"#given that\n",
"import math\n",
"h = 6.6e-034; # Planck's constant\n",
"m = 9.1e-031; # Mass of a electron, kg\n",
"e = 1.6e-019; # Electronic charge, C\n",
"V = 100; # Accelerating potential for electron, V\n",
"E = e*V; # Energy of the electron, J\n",
"lamda = h/math.sqrt(2*m*E); # de-Broglie wavelength of electron, m\n",
"print(\"Ex1.16: Page-1.18\")\n",
"print(\"de-Broglie wavelength of electron accelerated at volts in m:\") ,lamda\n",
"# Result \n",
"# de-Broglie wavelength of electron accelerated at 100 volts = 1.2231e-10 m "
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## example:1.17;page:1.19"
]
},
{
"cell_type": "code",
"execution_count": 17,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Ex1.17: Page-1.19\n",
"The wavelength associated with mass moving with velocity m/s in m:\n"
]
},
{
"data": {
"text/plain": [
"(None, 5.999999999999999e-34)"
]
},
"execution_count": 17,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"#cal of wavelength\n",
"#intiation of all variables \n",
"#given that\n",
"m = 10e-03; # Mass of the body, kg\n",
"v = 110; # Velocity of the mass, m/s\n",
"h = 6.6e-034; # Planck's constant\n",
"lamda = h/(m*v); # de-Broglie wavelength of electron, m\n",
"print(\"Ex1.17: Page-1.19\")\n",
"print(\"The wavelength associated with mass moving with velocity m/s in m:\"),lamda\n",
"# Result \n",
"# The wavelength associated with mass moving with velocity 110 m/s = 6e-34 m "
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## example:1.18;page:1.19"
]
},
{
"cell_type": "code",
"execution_count": 18,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Ex1.18: Page-1.19\n",
"The wavelength associated with moving electron in angstrom:\n"
]
},
{
"data": {
"text/plain": [
"(None, 1.37)"
]
},
"execution_count": 18,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"#cal of the wavelength\n",
"#intiation of all variables \n",
"#given that\n",
"import math\n",
"m = 9.1e-031; # Mass of the electron, kg\n",
"Ek = 1.27e-017; # Kinetic energy of electron, J\n",
"h = 6.6e-034; # Planck's constant\n",
"lamda = h/math.sqrt(2*m*Ek); # de-Broglie wavelength of electron, m\n",
"print(\"Ex1.18: Page-1.19\")\n",
"print(\"The wavelength associated with moving electron in angstrom:\"),round (lamda/1e-010,2)\n",
"# Result \n",
"# The wavelength associated with moving electron = 1.37 angstrom "
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## example:1.19;page:1.19"
]
},
{
"cell_type": "code",
"execution_count": 19,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Ex1.19: Page-1.19\n",
"The kinetic energy of electron having wavelength in eV:\n"
]
},
{
"data": {
"text/plain": [
"(None, 9.1e-12, 18063.99131601402)"
]
},
"execution_count": 19,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"#cal of kinetic energy of electron\n",
"#intiation of all variables \n",
"#given that\n",
"m = 9.1e-031; # Mass of the electron, kg\n",
"h = 6.6e-034; # Planck's constant\n",
"e = 1.6e-019; # Energy equivalent of 1 eV, J/eV\n",
"lamda = 9.1e-012; # de-Broglie wavelength of electron, m\n",
"# We have lamda = h/(m*v), solving for v\n",
"v = h/(m*lamda); # Velocity of the electron, m/s\n",
"K = 1/2*m*v**2; # Kinetic energy of electron, J\n",
"print(\"Ex1.19: Page-1.19\")\n",
"print(\"The kinetic energy of electron having wavelength in eV:\"), lamda, K/e\n",
"# Result \n",
"# The kinetic energy of electron having wavelength 9.1e-12 m = 1.81e+04 eV "
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## example:1.20;page:1.19"
]
},
{
"cell_type": "code",
"execution_count": 20,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Ex1.20 : Page-1.19\n",
"The speed of proton for an equivalent wavelength of that of electron:\n",
"Ratio of kinetic energies of electron and proton, therefore Ke > Kp:\n"
]
},
{
"data": {
"text/plain": [
"(None, 1835.16)"
]
},
"execution_count": 20,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"#cal of interplanar spacing of the crystal\n",
"#intiation of all variables \n",
"#given that\n",
"m_e = 9.1e-031; # Mass of the electron, kg\n",
"m_p = 1.67e-027; # Mass of the proton, kg\n",
"v_e = 1; # For simplicity assume velocity of electron to be unity, m/s\n",
"# From de-Broglie relation, \n",
"# lamda_p = lamda_e = h(m*v_p), solving for v_p\n",
"v_p = m_e*v_e/m_p; # Velocity of the proton, m/s\n",
"# As lamda_e = h/sqrt(2*m_e*K_e) and lamda_p = h/sqrt(2*m_p*K_p), solving for K_e/K_p\n",
"K_ratio = m_p/m_e; # Ratio of kinetic energies of electron and proton\n",
"print(\"Ex1.20 : Page-1.19\")\n",
"print(\"The speed of proton for an equivalent wavelength of that of electron:\"),round(v_p,4)\n",
"print(\"Ratio of kinetic energies of electron and proton, therefore Ke > Kp:\"), round(K_ratio,2)\n",
"# Result \n",
"# The speed of proton for an equivalent wavelength of that of electron = 5.4e-04 ve\n",
"# Ratio of kinetic energies of electron and proton = 1.8e+03, therefore Ke > Kp "
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## example:1.21;page:1.20"
]
},
{
"cell_type": "code",
"execution_count": 21,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Ex1.21: Page-1.20\n",
"de-Broglie wavelength of the electron in angstrom:\n"
]
},
{
"data": {
"text/plain": [
"(None, 1.73)"
]
},
"execution_count": 21,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"#cal of de-Broglie wavelength of the electron\n",
"#intiation of all variables \n",
"#given that\n",
"import math\n",
"V = 50; # Potential difference, V\n",
"m = 9.1e-031; # Mass of the electron, kg\n",
"e = 1.6e-019; # Electronic charge, C\n",
"h = 6.6e-034; # Planck's constant, Js\n",
"lamda = h/math.sqrt(2*m*e*V); # From de-Broglie relation,\n",
"print(\"Ex1.21: Page-1.20\")\n",
"print(\"de-Broglie wavelength of the electron in angstrom:\"),round (lamda/1e-010,2)\n",
"# Result \n",
"# de-Broglie wavelength of the electron = 1.73 angstrom"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## example:1.23;page:1.31"
]
},
{
"cell_type": "code",
"execution_count": 22,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Ex1.23:: Page-1.31\n",
"The minimum accuracy to locate the position of an electron in m:\n"
]
},
{
"data": {
"text/plain": [
"(None, 0.0002)"
]
},
"execution_count": 22,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"#cal of locate the position of an electron\n",
"#intiation of all variables \n",
"#given that\n",
"v = 740; # Speed of the electron, m/s\n",
"m = 9.1e-031; # Mass of the electron, kg\n",
"h = 6.6e-034; # Planck's constant, Js\n",
"p = m*v; # Momentum of the electron, kg-m/s\n",
"frac_v = 0.05/100; # Correctness in the speed \n",
"delta_p = p*frac_v; # Uncertainty in momentum, kg-m/s\n",
"delta_x = h/(4*3.14)*1/delta_p; # Uncertainty in position, m\n",
"print(\"Ex1.23:: Page-1.31\")\n",
"print(\"The minimum accuracy to locate the position of an electron in m:\"),round (delta_x,4)\n",
"# Result \n",
"# The minimum accuracy to locate the position of an electron = 1.56e-04 m "
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## example:1.24;page:1.31"
]
},
{
"cell_type": "code",
"execution_count": 23,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Ex1.24 : Page-1.31\n",
"The uncertainty in energy of an emitted photon in eV:\n"
]
},
{
"data": {
"text/plain": [
"(None, 3e-06)"
]
},
"execution_count": 23,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"#cal of the uncertainty in energy of an emitted photon\n",
"#intiation of all variables \n",
"#given that\n",
"h = 6.6e-034; # Planck's constant, Js\n",
"h_cross = h/(2*3.14); # Reduced Planck's constant, Js\n",
"delta_t = 1e-010; # Uncertainty in time, s\n",
"# From Energy-time uncertainty, \n",
"# delta_E*delta_t = h_cross/2, solving for delta_E\n",
"delta_E = h_cross/(2*delta_t); # Uncertainty in energy of an emitted photon, J \n",
"print(\"Ex1.24 : Page-1.31\")\n",
"print(\"The uncertainty in energy of an emitted photon in eV:\"),round (delta_E/1.6e-019,6)\n",
"# Result \n",
"# The uncertainty in energy of an emitted photon = 3.283e-06 eV"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## example:1.25;page:1.31"
]
},
{
"cell_type": "code",
"execution_count": 24,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Ex1.25: : Page-1.31\n",
"The minimum uncertainty in velocity of electron in m/s:\n"
]
},
{
"data": {
"text/plain": [
"(None, 7252.75)"
]
},
"execution_count": 24,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"#cal of minimum uncertainty\n",
"#intiation of all variables \n",
"#given that\n",
"h = 6.6e-034; # Planck's constant, Js\n",
"delta_x_max = 1e-007; # Uncertainty in length, m\n",
"m = 9.1e-031; # Mass of an electron, kg\n",
"# From Position-momentum uncertainty, \n",
"# delta_p_min = m*delta_v_min = h/delta_x_max, solving for delta_v_min\n",
"delta_v_min = h/(delta_x_max*m); # Minimum uncertainty in velocity of electron, m/s\n",
"print(\"Ex1.25: : Page-1.31\")\n",
"print(\"The minimum uncertainty in velocity of electron in m/s:\"),round (delta_v_min,2)\n",
"# Result \n",
"# The minimum uncertainty in velocity of electron = 7.25e+03 m/s "
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## example:1.26;page:1.32"
]
},
{
"cell_type": "code",
"execution_count": 25,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Ex1.26: Page-1.32\n",
"The minimum uncertainty in momemtum of proton in kg-m/s:\n",
"The kinetic energy of proton in eV:\n"
]
},
{
"data": {
"text/plain": [
"(None, 112819.34)"
]
},
"execution_count": 25,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"#cal of minimum uncertainty and kinetic energy of proton\n",
"#intiation of all variables \n",
"#given that\n",
"h = 6.6e-034; # Planck's constant, Js\n",
"delta_x_max = 8.5e-014; # Uncertainty in length, m\n",
"m = 1.67e-027; # Mass of proton, kg\n",
"# From Position-momentum uncertainty, \n",
"# delta_p_min*delta_x_max = h, solving for delta_p_min\n",
"delta_p_min = h/delta_x_max; # Minimum uncertainty in momentum of electron, kg-m/s\n",
"p_min = delta_p_min; # Minimum momentum of the proton, kg.m/s\n",
"delta_E = p_min**2/(2*m); \n",
"print(\"Ex1.26: Page-1.32\")\n",
"print(\"The minimum uncertainty in momemtum of proton in kg-m/s:\"), p_min\n",
"print(\"The kinetic energy of proton in eV:\"),round (delta_E/1.6e-019,2)\n",
"# Result \n",
"# The minimum uncertainty in momemtum of proton = 7.76e-21 kg-m/s\n",
"# The kinetic energy of proton = 1.128e+05 eV "
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## example:1.27;page:1.32"
]
},
{
"cell_type": "code",
"execution_count": 26,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Ex1.27:: Page-1.32\n",
"The percentage uncertainty in momentum of electron = 2d percent:\n"
]
},
{
"data": {
"text/plain": [
"(None, 15.90167880419624)"
]
},
"execution_count": 26,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"#cal of percentage uncertainty\n",
"#intiation of all variables \n",
"#given that\n",
"h = 6.6e-034; # Planck's constant, Js\n",
"e = 1.6e-019; # Energy equivalent of 1 eV, J/eV\n",
"E = 0.15*1e+03*e; # Energy of the electron, J\n",
"m = 9.1e-031; # Mass of electron, kg\n",
"delta_x = 0.5e-010; # Position uncertainty of electron, m\n",
"p = (2*m*E)**(1/2); # Momentum of the electron, kg-m/s\n",
"# delta_x*delta_p = h/(4*%pi), solving for delta_p\n",
"delta_p = h/(4*3.14*delta_x); # Uncertainty in momentum of electron, kg-m/s\n",
"frac_p = delta_p/p*100; # Percentage uncertainty in momentum of electron, kg-m/s\n",
"print(\"Ex1.27:: Page-1.32\")\n",
"print(\"The percentage uncertainty in momentum of electron = 2d percent:\"), frac_p\n",
"# Result \n",
"# The percentage uncertainty in momentum of electron = 15 percent "
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## example:1.28;page:1.33"
]
},
{
"cell_type": "code",
"execution_count": 27,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Ex1.28:: Page-1.33\n",
"The position uncertainty of particle in m:\n"
]
},
{
"data": {
"text/plain": [
"(None, 2.787680143269864e-14)"
]
},
"execution_count": 27,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"#cal of position uncertainty of particle\n",
"#intiation of all variables \n",
"#given that\n",
"h = 6.6e-034; # Planck's constant, Js\n",
"e = 1.6e-019; # Energy equivalent of 1 eV, J/eV\n",
"delta_v = 7.54e-015; # Uncertainty in velocity of the particle, m/s\n",
"m = 0.25e-06; # Mass of particle, kg\n",
"# delta_x*delta_p = h/(4*%pi), solving for delta_x\n",
"delta_x = h/(4*3.14*m*delta_v); # Position uncertainty of particle, m\n",
"print(\"Ex1.28:: Page-1.33\")\n",
"print(\"The position uncertainty of particle in m:\"), delta_x\n",
"# Result \n",
"# The position uncertainty of particle = 2.79e-14 m "
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## example:1.29;page:1.33"
]
},
{
"cell_type": "code",
"execution_count": 28,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Ex1.29 : Page-1.33\n",
"The position uncertainty of moving electron in m:\n"
]
},
{
"data": {
"text/plain": [
"(None, 0.0003)"
]
},
"execution_count": 28,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"#cal of position uncertainty of moving electron\n",
"#intiation of all variables \n",
"#given that\n",
"h = 6.6e-034; # Planck's constant, Js\n",
"e = 1.6e-019; # Energy equivalent of 1 eV, J/eV\n",
"v = 450; # Velocity of the electron, m/s\n",
"delta_v = v*0.05/100; # Uncertainty in velocity of the particle, m/s\n",
"m = 9.1e-031; # Mass of electron, kg\n",
"# delta_x*delta_p = h/(4*%pi), solving for delta_x\n",
"delta_x = h/(4*3.14*m*delta_v); # Position uncertainty of particle, m\n",
"print(\"Ex1.29 : Page-1.33\")\n",
"print(\"The position uncertainty of moving electron in m:\"),round (delta_x,4)\n",
"# Result \n",
"# The position uncertainty of moving electron = 2.57e-04 m "
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## example:1.30;page:1.33"
]
},
{
"cell_type": "code",
"execution_count": 29,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Ex1.30:: Page-1.33\n",
"The smallest possible uncertainty in position of the electron in angstrom:\n"
]
},
{
"data": {
"text/plain": [
"(None, 0.02)"
]
},
"execution_count": 29,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"#cal of smallest possible uncertainty\n",
"#intiation of all variables \n",
"#given that\n",
"import math\n",
"h = 6.6e-034; # Planck's constant, Js\n",
"e = 1.6e-019; # Energy equivalent of 1 eV, J/eV\n",
"c = 3e+08; # Speed of light, m/s\n",
"v = 3e+07; # Velocity of the electron, m/s\n",
"m0 = 9.1e-031; # Rest mass of electron, kg\n",
"m = m0/math.sqrt(1-v**2/c**2); # Mass of moving electron, kg\n",
"delta_p_max = m*v; # Maximum uncertainty in momentum of the particle, m/s\n",
"# delta_x_min*delta_p_max = h/(4*%pi), solving for delta_x_min\n",
"delta_x_min = h/(4*3.14*delta_p_max); # Minimum position uncertainty of particle, m\n",
"print(\"Ex1.30:: Page-1.33\")\n",
"print(\"The smallest possible uncertainty in position of the electron in angstrom:\"),round (delta_x_min/1e-010,2)\n",
"# Result \n",
"# The smallest possible uncertainty in position of the electron = 0.019 angstrom"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## example:1.31;page:1.44"
]
},
{
"cell_type": "code",
"execution_count": 30,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Ex1.31: : Page-1.44\n",
"The energy difference between the neighbouring levels of Na at the highest state in:\n"
]
},
{
"data": {
"text/plain": [
"(None, 1.0592022109560034e-07)"
]
},
"execution_count": 30,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"#cal of energy difference\n",
"#intiation of all variables \n",
"#given that\n",
"import math\n",
"h = 6.6e-034; # Planck's constant, Js\n",
"m = 9.1e-031; # Electronic mass, kg\n",
"e = 1.6e-019; # Energy equivalent of 1 eV, J/eV\n",
"l = 2e-002; # Length of the side of the cube, m\n",
"E_F = 9*e; # Fermi energy, J\n",
"# As E_F = h**2/(8*m*l**2)*(nx**2 + ny**2 + nz**2) and nx = ny = nz for a cube, solving for nx\n",
"nx = math.sqrt(E_F*(8*m*l**2)/(3*h**2)); # Value of integer for a cube\n",
"E = h**2/(8*m*l**2)*3*nx**2; # Fermi energy, J\n",
"E1 = h**2/(8*m*l**2)*((nx-1)**2 + nx**2 + nx**2); # Energy of the level just below the fermi level, J\n",
"delta_E = E - E1; # Difference in the energy between the neighbouring levels of Na at the highest state, J\n",
"print(\"Ex1.31: : Page-1.44\")\n",
"print(\"The energy difference between the neighbouring levels of Na at the highest state in:\"), delta_E/e\n",
"# Result \n",
"# The energy difference between the neighbouring levels of Na at the highest state = 1.06e-07 eV "
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## example:1.32;page:1.45"
]
},
{
"cell_type": "code",
"execution_count": 31,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Ex1.32:: Page-1.45\n",
"The energy of the neutron confined in a nucleus in eV:\n"
]
},
{
"data": {
"text/plain": [
"(None, 6113398.2)"
]
},
"execution_count": 31,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"#cal of energy of the neutron\n",
"#intiation of all variables \n",
"#given that\n",
"h = 6.6e-034; # Planck's constant, Js\n",
"m = 1.67e-027; # Electronic mass, kg\n",
"e = 1.6e-019; # Energy equivalent of 1 eV, J/eV\n",
"nx = 1; # Principle quantum numbers in 3D corresponding to the longest energy state\n",
"ny = 1;\n",
"nz = 1;\n",
"lx = 1e-014; # Dimensions of the box to which the neutron is confined, m\n",
"ly = 1e-014;\n",
"lz = 1e-014;\n",
"E = h**2/(8*m)*(nx**2/lx**2+ny**2/ly**2+nz**2/lz**2); # Energy of the neutron confined in the nucleus, J\n",
"print(\"Ex1.32:: Page-1.45\")\n",
"print(\"The energy of the neutron confined in a nucleus in eV:\"),round (E/e,2)\n",
"# Result \n",
"# The energy of the neutron confined in a nucleus = 6.11e+06 eV "
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## example:1.33;page:1.46"
]
},
{
"cell_type": "code",
"execution_count": 32,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Ex1.33:: Page-1.46\n",
"The energy of the electron moving in one dimensional infinitely high potential box in eV:\n"
]
},
{
"data": {
"text/plain": [
"(None, 8.48)"
]
},
"execution_count": 32,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"#cal of energy of the electron\n",
"#intiation of all variables \n",
"#given that\n",
"h = 6.6e-034; # Planck's constant, Js\n",
"m = 9.1e-031; # Electronic mass, kg\n",
"e = 1.6e-019; # Energy equivalent of 1 eV, J/eV\n",
"n = 1; # For simplicity assume principle quantum number to be unity\n",
"l = 2.1e-010; # Length of one dimensional potential box, m\n",
"E = h**2*n**2/(8*m*l**2); # Energy of the electron, J\n",
"print(\"Ex1.33:: Page-1.46\")\n",
"print(\"The energy of the electron moving in one dimensional infinitely high potential box in eV:\"),round (E/e,2)\n",
"# Result \n",
"# The energy of the electron moving in one dimensional infinitely high potential box = 8.48 n**2 eV "
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## example:1.34;page:1.46"
]
},
{
"cell_type": "code",
"execution_count": 33,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Ex1.34:: Page-1.46\n",
"The lowest energy of the electron in a one dimensional force free region in eV:\n"
]
},
{
"data": {
"text/plain": [
"(None, 3.05)"
]
},
"execution_count": 33,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"#cal of lowest energy of the electron\n",
"#intiation of all variables \n",
"#given that\n",
"h = 6.6e-034; # Planck's constant, Js\n",
"m = 9.1e-031; # Electronic mass, kg\n",
"e = 1.6e-019; # Energy equivalent of 1 eV, J/eV\n",
"n = 1; # The lowest energy state of electron\n",
"l = 3.5e-010; # Length of one dimensional potential box, m\n",
"E = h**2*n**2/(8*m*l**2); # Energy of the electron in the lowest state, J\n",
"print(\"Ex1.34:: Page-1.46\")\n",
"print(\"The lowest energy of the electron in a one dimensional force free region in eV:\"),round (E/e,2)\n",
"# Result \n",
"# The lowest energy of an electron in a one dimensional force free region = 3 eV"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## example:1.35;page:1.46"
]
},
{
"cell_type": "code",
"execution_count": 34,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Ex1.35:: Page-1.46\n",
"The energy of the electron in first state in J:\n",
"The energy of the electron in second state in J:\n",
"The energy of the electron in third state in J:\n"
]
},
{
"data": {
"text/plain": [
"(None, 5.966941645612005e-19)"
]
},
"execution_count": 34,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"#cal of energy of the electron\n",
"#intiation of all variables \n",
"#given that\n",
"h = 6.6e-034; # Planck's constant, Js\n",
"m = 9.1e-031; # Electronic mass, kg\n",
"e = 1.6e-019; # Energy equivalent of 1 eV, J/eV\n",
"l = 9.5e-010; # Length of one dimensional potential box, m\n",
"# First energy level\n",
"n = 1; # The first energy state of electron\n",
"E1 = h**2*n**2/(8*m*l**2); # Energy of the electron in first state, J\n",
"# Second energy level\n",
"n = 2; # The second energy state of electron\n",
"E2 = h**2*n**2/(8*m*l**2); # Energy of the electron in second state, J\n",
"# Third energy level\n",
"n = 3; # The third energy state of electron\n",
"E3 = h**2*n**2/(8*m*l**2); # Energy of the electron in third state, J\n",
"print(\"Ex1.35:: Page-1.46\")\n",
"print(\"The energy of the electron in first state in J:\"), E1\n",
"print(\"The energy of the electron in second state in J:\"), E2\n",
"print(\"The energy of the electron in third state in J:\"), E3\n",
"# Result \n",
"# The energy of the electron in first state = 6.6e-20 J\n",
"# The energy of the electron in second state = 2.7e-19 J\n",
"# The energy of the electron in third state = 6.0e-19 J "
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## example:1.36;page:1.47"
]
},
{
"cell_type": "code",
"execution_count": 35,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Ex1.36:: Page-1.47\n",
"The energy of the electron in lowest state in eV:\n",
"The energy of the electron in second state in eV:\n"
]
},
{
"data": {
"text/plain": [
"(None, 23.93)"
]
},
"execution_count": 35,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"#cal of energy of the electron\n",
"#intiation of all variables \n",
"#given that\n",
"h = 6.6e-034; # Planck's constant, Js\n",
"m = 9.1e-031; # Electronic mass, kg\n",
"e = 1.6e-019; # Energy equivalent of 1 eV, J/eV\n",
"l = 2.5e-010; # Length of one dimensional potential box, m\n",
"# First energy level\n",
"n = 1; # The lowest energy state of electron\n",
"E1 = h**2*n**2/(8*m*l**2); # Energy of the electron in first state, J\n",
"# Second energy level\n",
"n = 2; # The second energy state of electron\n",
"E2 = h**2*n**2/(8*m*l**2); # Energy of the electron in second state, J\n",
"print(\"Ex1.36:: Page-1.47\")\n",
"print(\"The energy of the electron in lowest state in eV:\"),round (E1/e,2)\n",
"print(\"The energy of the electron in second state in eV:\"),round (E2/e,2)\n",
"# Result \n",
"# The energy of the electron in lowest state = 5.98 eV\n",
"# The energy of the electron in second state = 23.93 eV "
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## example:1.37;page:1.47"
]
},
{
"cell_type": "code",
"execution_count": 36,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Ex1.37:: Page-1.47\n",
"The lowest energy of the neutron confined to the nucleus in MeV:\n"
]
},
{
"data": {
"text/plain": [
"(None, 0.05)"
]
},
"execution_count": 36,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"#cal of lowest energy of the neutron\n",
"#intiation of all variables \n",
"#given that\n",
"h = 6.6e-034; # Planck's constant, Js\n",
"m = 1.67e-027; # Electronic mass, kg\n",
"e = 1.6e-019; # Energy equivalent of 1 eV, J/eV\n",
"l = 2.5e-010; # Length of one dimensional potential box, m\n",
"delta_x = 1e-014; # Uncertainty in position of neutron, m\n",
"# From uncertainty principle, \n",
"# delta_x*delta_p = h/(4*%pi), solving for delta_p\n",
"delta_p = h/(4*3.14*delta_x); # Uncertainty in momentum of neutron, kg-m/s\n",
"p = delta_p; # Momemtum of neutron in the box, kg-m/s\n",
"KE = p**2/(2*m); # Kinetic energy of neutron in the box, J\n",
"print(\"Ex1.37:: Page-1.47\")\n",
"print(\"The lowest energy of the neutron confined to the nucleus in MeV:\"),round (KE/(e*1e+06),2)\n",
"# Result \n",
"# The lowest energy of the neutron confined to the nucleus = 0.05 MeV "
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## example:1.38;page:1.56"
]
},
{
"cell_type": "code",
"execution_count": 37,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Ex1.38: : Page-1.56\n",
"The energy of scattered X-ray:\n",
"The energy of recoil electron:\n"
]
},
{
"data": {
"text/plain": [
"(None, 4.88)"
]
},
"execution_count": 37,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"#cal of energy of scattered X-ray and recoil electron\n",
"#intiation of all variables \n",
"#given that\n",
"import math\n",
"h = 6.6e-034; # Planck's constant, Js\n",
"m0 = 9.1e-031; # Electronic mass, kg\n",
"c = 3e+08; # Speed of light, m/s\n",
"e = 1.6e-019; # Energy equivalent of 1 eV, J/eV\n",
"phi = 45; # Scattering angle of X-rays, degrees\n",
"E = 75; # Incident energy of X-rays, keV\n",
"# As from Compton shift formula\n",
"# 1/E_prime - 1/E = 1/(m0*c**2)*(1-cosd(phi))\n",
"# Solving for E_prime\n",
"E_prime = 1/((1/(m0*c**2/(e*1e+03)))*(1-math.cos(phi))+1/E); # Energy of scattered photon, keV\n",
"E_recoil = E - E_prime; # Energy of recoil electron, keV\n",
"print(\"Ex1.38: : Page-1.56\")\n",
"print(\"The energy of scattered X-ray:\"),round (E_prime,2)\n",
"print(\"The energy of recoil electron:\"),round (E_recoil,2)\n",
"# Result \n",
"# The energy of scattered X-ray = 71.9 keV\n",
"# The energy of recoil electron = 3.1 keV "
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## example:1.39;page:1.57"
]
},
{
"cell_type": "code",
"execution_count": 38,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Ex1.39: : Page-1.57\n",
"The wavelength of scattered X-ray in angstrom:\n"
]
},
{
"data": {
"text/plain": [
"(None, 0.25)"
]
},
"execution_count": 38,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"#cal of wavelength of scattered X-ray\n",
"#intiation of all variables \n",
"#given that\n",
"import math\n",
"h = 6.6e-034; # Planck's constant, Js\n",
"m0 = 9.1e-031; # Electronic mass, kg\n",
"c = 3e+08; # Speed of light, m/s\n",
"e = 1.6e-019; # Energy equivalent of 1 eV, J/eV\n",
"phi = 60; # Scattering angle of X-rays, degrees\n",
"E = 75; # Incident energy of X-rays, keV\n",
"# As from Compton shift formula\n",
"delta_L = h/(m0*c)*(1-math.cos(3.14)); # Change in photon wavelength, m\n",
"lamda = 0.198e-010; # Wavelength of incident photon, m\n",
"lamda_prime = (lamda+delta_L)/1e-010; # Wavelength of scattered X-ray, angstrom \n",
"print(\"Ex1.39: : Page-1.57\")\n",
"print(\"The wavelength of scattered X-ray in angstrom:\"),round (lamda_prime,2)\n",
"# Result \n",
"# The wavelength of scattered X-ray = 0.2101 angstrom"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## example:1.40;page:1.57"
]
},
{
"cell_type": "code",
"execution_count": 39,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Ex1.40:: Page-1.57\n",
"The wavelength of scattered X-ray at degrees view in angstrom:\n",
"The recoil energy of electron scattered through degrees:\n"
]
},
{
"data": {
"text/plain": [
"(None, 60, 84.82)"
]
},
"execution_count": 39,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"#cal of wavelength of scattered X-ray and recoil energy of electron\n",
"#intiation of all variables \n",
"#given that\n",
"import math\n",
"h = 6.6e-034; # Planck's constant, Js\n",
"m0 = 9.1e-031; # Electronic mass, kg\n",
"c = 3e+08; # Speed of light, m/s\n",
"e = 1.6e-019; # Energy equivalent of 1 eV, J/eV\n",
"pi = 180; # Scattering angle of X-rays, degrees\n",
"lamda = 1.78; # Wavelength of incident photon, m\n",
"lamda_prime = 1.798; # Wavelength of scattered X-ray, angstrom \n",
"# As from Compton shift formula\n",
"# lamda_prime - lamda = h/(m0*c)*(1-cosd(pi)), Change in photon wavelength, m\n",
"# Or we may write, lamda_prime - lamda = k*(1-cosd(pi))\n",
"# solving for k\n",
"k = (lamda_prime - lamda)/(1-math.cos(pi)); # k = h/(m0*c) value, angstrom\n",
"# For pi = 60\n",
"pi = 60; # New angle of scattering, degrees\n",
"lamda_prime = lamda + k*(1-math.cos(pi)); # Wavelength of scattered radiation at 60 degree angle, angstrom\n",
"print(\"Ex1.40:: Page-1.57\")\n",
"print(\"The wavelength of scattered X-ray at degrees view in angstrom:\"), pi,round (lamda_prime,2)\n",
"# Recoil energy of electron\n",
"E = h*c*(1/lamda - 1/lamda_prime)*1e+010; # Recoil energy of electron, joule\n",
"print(\"The recoil energy of electron scattered through degrees:\"), pi,round (E/e,2) \n",
"# Result \n",
"# The wavelength of scattered X-ray at 60 degrees view = 1.7845 angstrom\n",
"# The recoil energy of electron scattered through 60 degrees = 17.5 eV "
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## example:1.41;page:1.58"
]
},
{
"cell_type": "code",
"execution_count": 40,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Ex1.41:: Page-1.58\n",
"The wavelength of scattered X-ray as viewed atdegrees in m\n",
"The recoil energy of electron scattered through degrees in eV:\n",
"The direction of emission of recoil electron in degrees:\n"
]
},
{
"data": {
"text/plain": [
"(None, 0.32)"
]
},
"execution_count": 40,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"#cal of wavelength of scattered X-ray and recoil energy of electron\n",
"#intiation of all variables \n",
"#given that\n",
"import math\n",
"h = 6.6e-034; # Planck's constant, Js\n",
"m0 = 9.1e-031; # Electronic mass, kg\n",
"c = 3e+08; # Speed of light, m/s\n",
"e = 1.6e-019; # Energy equivalent of 1 eV, J/eV\n",
"phi = 90; # Scattering angle of X-rays, degrees\n",
"E = 510*1e+03*e; # Energy of incident photon, J\n",
"# As E = h*c/lamda, solving for lamda\n",
"lamda = h*c/E; # Wavelength of incident photon, m\n",
"# As from Compton shift formula\n",
"# lamda_prime - lamda = h/(m0*c)*(1-math.cosd(phi)), solving for lamda_prime\n",
"lamda_prime = lamda + h/(m0*c)*(1-math.cos(phi)); # Wavelength of scattered X-ray, m \n",
"print(\"Ex1.41:: Page-1.58\")\n",
"print(\"The wavelength of scattered X-ray as viewed atdegrees in m\"), phi,lamda_prime\n",
"# Recoil energy of electron\n",
"E = h*c*(1/lamda - 1/lamda_prime); # Recoil energy of electron, joule\n",
"print(\"The recoil energy of electron scattered through degrees in eV:\"),round (E/e,2)\n",
"# Direction of recoil electron\n",
"theta = math.tan(lamda*math.sin(phi)/(lamda_prime-lamda*math.cos(phi))); # Direction of recoil electron, degrees\n",
"print(\"The direction of emission of recoil electron in degrees:\"),round (theta,2)\n",
"# Result \n",
"# The wavelength of scattered X-ray as viewed at 90 degrees = 4.84e-12 m\n",
"# The recoil energy of electron scattered through 90 degrees = 2.55e+05 eV\n",
"# The direction of emission of recoil electron = 26.61 degrees \n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## example:1.42;page:1.59"
]
},
{
"cell_type": "code",
"execution_count": 41,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Ex1.42: : Page-1.59\n",
"The number of electrons striking the target per sec in electrons/sec:\n",
"The maximum speed of the electrons when they strike in m/s:\n"
]
},
{
"data": {
"text/plain": [
"(None, 66033624.47)"
]
},
"execution_count": 41,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"#cal of number of electrons and maximum speed of the electrons\n",
"#intiation of all variables \n",
"#given that\n",
"import math\n",
"m = 9.1e-031; # Electronic mass, kg\n",
"c = 3e+08; # Speed of light, m/s\n",
"e = 1.6e-019; # Charge on the electron, C\n",
"V = 12.4e+03; # Potential diffeence applied across the X-ray tube, V\n",
"i = 2e-03; # Current through the X-ray tube, A\n",
"t = 1; # Time for which the electrons strike the target material, s\n",
"N = i*t/e; # Number of electrons striking the target per sec, per sec\n",
"v_max = math.sqrt(2*e*V/m); # Maximum speed of the electrons, m/s\n",
"print(\"Ex1.42: : Page-1.59\")\n",
"print(\"The number of electrons striking the target per sec in electrons/sec:\"), N\n",
"print(\"The maximum speed of the electrons when they strike in m/s:\"),round (v_max,2)\n",
"# Result \n",
"# The number of electrons striking the target per sec = 1.25e+16 electrons/sec\n",
"# The maximum speed of the electrons when they strike = 6.6e+07 m/s"
]
}
],
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"anaconda-cloud": {},
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"display_name": "Python [Root]",
"language": "python",
"name": "Python [Root]"
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"name": "ipython",
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