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| author | Jovina Dsouza | 2014-07-07 16:34:28 +0530 |
|---|---|---|
| committer | Jovina Dsouza | 2014-07-07 16:34:28 +0530 |
| commit | fffcc90da91b66ee607066d410b57f34024bd1de (patch) | |
| tree | 7b8011d61013305e0bf7794a275706abd1fdb0d3 /Concepts_Of_Modern_Physics | |
| parent | 299711403e92ffa94a643fbd960c6f879639302c (diff) | |
| download | Python-Textbook-Companions-fffcc90da91b66ee607066d410b57f34024bd1de.tar.gz Python-Textbook-Companions-fffcc90da91b66ee607066d410b57f34024bd1de.tar.bz2 Python-Textbook-Companions-fffcc90da91b66ee607066d410b57f34024bd1de.zip | |
adding book
Diffstat (limited to 'Concepts_Of_Modern_Physics')
| -rwxr-xr-x | Concepts_Of_Modern_Physics/Chapter_1.ipynb | 356 | ||||
| -rwxr-xr-x | Concepts_Of_Modern_Physics/Chapter_10.ipynb | 152 | ||||
| -rwxr-xr-x | Concepts_Of_Modern_Physics/Chapter_11.ipynb | 314 | ||||
| -rwxr-xr-x | Concepts_Of_Modern_Physics/Chapter_12.ipynb | 440 | ||||
| -rwxr-xr-x | Concepts_Of_Modern_Physics/Chapter_2.ipynb | 334 | ||||
| -rwxr-xr-x | Concepts_Of_Modern_Physics/Chapter_3.ipynb | 470 | ||||
| -rwxr-xr-x | Concepts_Of_Modern_Physics/Chapter_4.ipynb | 336 | ||||
| -rwxr-xr-x | Concepts_Of_Modern_Physics/Chapter_5.ipynb | 142 | ||||
| -rwxr-xr-x | Concepts_Of_Modern_Physics/Chapter_6.ipynb | 62 | ||||
| -rwxr-xr-x | Concepts_Of_Modern_Physics/Chapter_7.ipynb | 184 | ||||
| -rwxr-xr-x | Concepts_Of_Modern_Physics/Chapter_8.ipynb | 163 | ||||
| -rwxr-xr-x | Concepts_Of_Modern_Physics/Chapter_9.ipynb | 271 | ||||
| -rwxr-xr-x | Concepts_Of_Modern_Physics/README.txt | 10 | ||||
| -rwxr-xr-x | Concepts_Of_Modern_Physics/screenshots/3_4.png | bin | 0 -> 23632 bytes | |||
| -rwxr-xr-x | Concepts_Of_Modern_Physics/screenshots/9_4.png | bin | 0 -> 19042 bytes | |||
| -rwxr-xr-x | Concepts_Of_Modern_Physics/screenshots/9_7.png | bin | 0 -> 22906 bytes |
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diff --git a/Concepts_Of_Modern_Physics/Chapter_1.ipynb b/Concepts_Of_Modern_Physics/Chapter_1.ipynb new file mode 100755 index 00000000..f4a61f1b --- /dev/null +++ b/Concepts_Of_Modern_Physics/Chapter_1.ipynb @@ -0,0 +1,356 @@ +{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 1:Relativity"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.1,Page no:9"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math \n",
+ "\n",
+ "#Varible declaration\n",
+ "t0= 3600 # time interval on Earth, seconds\n",
+ "t= 3601.0 #time interval for spacecraft as measured from Earth, seconds\n",
+ "\n",
+ "#Calculation\n",
+ "c= 2.998 *(10**8) #speed of light, m/s\n",
+ "v=c*math.sqrt((1-((t0/t)**2))) #relative velocity of spacecraft, m/s\n",
+ "\n",
+ "#Result\n",
+ "print\"The speed of the Spacecraft relative to Earth is:%.2g \"%v,\"m/s\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The speed of the Spacecraft relative to Earth is:7.1e+06 m/s\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.2,Page no:13"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Varible declaration\t\n",
+ "fg= 5.6*(10**14) #frequency of green color, Hz\n",
+ "fr= 4.8*(10**14) #frequency of red color, Hz\n",
+ "c= 3.0*(10**8) #velocity of light, m/s\n",
+ "\n",
+ "#Calculation\n",
+ "v= c*((fg**2 - fr**2)/(fg**2 + fr**2)) #longitudinal speed of observer, m/s\n",
+ "v= v*3.6 #convert to km/h\n",
+ "R= 1.0 #rate at which fine is to be imposed per km/h, $\n",
+ "l= 80.0 #speed limit upto which no fine is to be imposed, km/h\n",
+ "fine= v-l # fine to be imposed, $\n",
+ "\n",
+ "#Result\n",
+ "print\"The fine imposed is:\",fine,\"$\\n\"\n",
+ "\n",
+ "print\"NOTE:Approx value of v is taken in book as 1.65*10^8,which is very less precise.\\nTherefore,chnge in final answer\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The fine imposed is: 165176390.588 $\n",
+ "\n",
+ "NOTE:Approx value of v is taken in book as 1.65*10^8,which is very less precise.\n",
+ "Therefore,chnge in final answer\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.3,Page no:14"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Varible declaration\n",
+ "v= 6.12*(10**7) #receding velocity with respect to Earth, m/s\n",
+ "c= 3.0*(10**8) #velocity of light, m/s\n",
+ "L0= 500.0 #initial wavelength of spectral line, nm\n",
+ "\n",
+ "#Calculation\n",
+ "L= L0*math.sqrt(((1+(v/c))/(1-(v/c)))) #final wavelength of spectral light, nm\n",
+ "Ls= L-L0 #shift in wavelength, nm\n",
+ "\n",
+ "#Varible declaration\n",
+ "print\"Shift in Green spectral line is: \",round(Ls),\"nm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Shift in Green spectral line is: 115.0 nm\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.4,Page no:19"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Varible declaration\n",
+ "StartingAge= 20 #starting age for both Dick and Jane\n",
+ "c= 3*(10**8) #velocity of light, m/s\n",
+ "v= 0.8*c #rate of separation of Dick and Jane, m/s\n",
+ "t0= 1 #interval for emission of signals, yr\n",
+ "\n",
+ "#Calculation\n",
+ "t1= t0*((1+v/c)/(1-v/c)) #interval for reception of signals on outward journey, yr\n",
+ "t1= t0*(math.sqrt((1+v/c)/(1-v/c))) #interval for reception of signals on outward journey, yr\n",
+ "t2= t0*(math.sqrt((1-v/c)/(1+v/c))) #interval for reception of signals on return trip, yr\n",
+ "#Dick's frame of reference\n",
+ "Tout1= 15 #duration of outward trip, yr\n",
+ "Tin1= 15 #duration of return trip, yr\n",
+ "JaneAge= StartingAge+(Tout1/t1)+(Tin1/t2) #Jane's age according to Dick\n",
+ "#Jane's frame of reference\n",
+ "Tout2= 25 #duration of outward trip, yr\n",
+ "d= 20 #delay in transmission of signal to Jane, caused by distance of the star, yr\n",
+ "Tin2= 5 #duration of return trip\n",
+ "DickAge= StartingAge+((Tout2+d)/t1)+(Tin2/t2) #Dick's age according to JAne\n",
+ "\n",
+ "#Result\n",
+ "print\"According to Dick, age of Jane is:\",JaneAge,\"years\"\n",
+ "print\"According to Jane, age of Dick is:\",DickAge,\"years\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "According to Dick, age of Jane is: 70.0 years\n",
+ "According to Jane, age of Dick is: 50.0 years\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.6,Page no:27"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math \n",
+ "\n",
+ "#Varible declaration\n",
+ "mf= 1 #mass of each entity, kg\n",
+ "c= 3*(10**8) #velocity of light, m/s\n",
+ "v= 0.6*c #velocity of fragments relative to original body, m/s\n",
+ "\n",
+ "#Calculation\n",
+ "E0= 2*((mf*(c**2))/math.sqrt(1-((v/c)**2))) #Total energy of fragments\n",
+ "m= E0/(c**2) #mass of original body, kg\n",
+ "\n",
+ "#Result\n",
+ "print\"The total mass of the stationary body is: \",m,\"kg\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The total mass of the stationary body is: 2.5 kg\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.7,Page no:28"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Varible declaration\n",
+ "r=1.4 # Rate of arrival of Solar Energy at erath, kW per square meter\n",
+ "R=1.5*(10**11) #Radius of Earth, m\n",
+ "\n",
+ "#Calculation\n",
+ "P=r*(4*math.pi*(R**2)) #Total power recieved by Earth, kW\n",
+ "P= P*(10**3) #W\n",
+ "C= 3*(10**8) #Velocity of light, m/s\n",
+ "E=P #Energy lost by Sun, J\n",
+ "m= E/(C**2) #Mass of Sun lost per second as energy, kg\n",
+ "\n",
+ "#Result\n",
+ "print\"Mass lost by sun per second, is:%.2g\"%m,\"kg\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Mass lost by sun per second, is:4.4e+09 kg\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.8,Page no:32"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math \n",
+ "\n",
+ "#Varible declaration\n",
+ "c= 3*(10**8) #Velocity of light, m/s\n",
+ "me= 0.511/(c**2) #mass of electron, MeV\n",
+ "mp=0 #mass of proton, MeV\n",
+ "p= 2.000/c #momenta for both particles, MeV\n",
+ " \n",
+ "#Calculation \n",
+ "##Using Eq. 1.24 and 1.25, Page 31 \n",
+ "Ee=math.sqrt(((me**2)*(c**4))+((p**2)*(c**2))) #Total energy of electron, MeV\n",
+ "Ep= p*c #Total energy of proton, MeV\n",
+ "\n",
+ "#Result\n",
+ "print\"Total energy of Electron is: \",round(Ee,3),\"MeV\"\n",
+ "print\"Total energy of Photon is: \",Ep,\"MeV\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Total energy of Electron is: 2.064 MeV\n",
+ "Total energy of Photon is: 2.0 MeV\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.11,Page no:44"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Varible declaration \n",
+ "c=3*(10**8) #velocity of light, m/s\n",
+ "VaE= 0.90*c #velocity of spacecraft alpha w.r.t Earth, m/s\n",
+ "VbA= 0.50*c #velocity of spacecraft beta w.r.t. Alpha, m/s\n",
+ "\n",
+ "#Calculation\n",
+ "VbE= (VaE+VbA)/(1+((VaE*VbA)/(c**2))) #velocity of beta w.r.t Earth, m/s\n",
+ "VbE=VbE/c #Converting to percent of c\n",
+ "\n",
+ "#Result\n",
+ "print\"The required velocity of spacecraft Beta w.r.t. Earth is:\",round(VbE,2),\"c\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The required velocity of spacecraft Beta w.r.t. Earth is: 0.97 c\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Concepts_Of_Modern_Physics/Chapter_10.ipynb b/Concepts_Of_Modern_Physics/Chapter_10.ipynb new file mode 100755 index 00000000..a9c68145 --- /dev/null +++ b/Concepts_Of_Modern_Physics/Chapter_10.ipynb @@ -0,0 +1,152 @@ +{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 10:The Solid state"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 10.1,Page no:342"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration \n",
+ "import math\n",
+ "Ro= 0.281 #equilibrium distance between ions, nm\n",
+ "alpha= 1.748 #Madelung constant\n",
+ "n= 9 #exponent, from observed compressibilities of NaCl\n",
+ "e= 1.6*(10**(-19)) #charge of an electron, C\n",
+ "Po= 8.85*(10**(-12)) #Permittivity of free space, F/m\n",
+ "\n",
+ "#Calculation\n",
+ "K=1.0/(4*(math.pi)*Po) #constant, N.m**2/C**2\n",
+ "Uo= -(K*alpha*(e**2)*(1.0-(1.0/n)))/(Ro*(10**(-9))) #Potential energy per ion pair, J\n",
+ "Uo= Uo/e #converting to eV\n",
+ "E1= 5.14 #Ionisation energy for Na, eV\n",
+ "E2= -3.61 #electron affinity of Cl, eV\n",
+ "E= E1+E2 #Electron transfer energy, eV\n",
+ "Ecohesive = (Uo +E) #per electron pair, eV\n",
+ "Ecohesive= Ecohesive/2.0 #for each ion, eV\n",
+ "\n",
+ "#Result\n",
+ "print\"The cohesive energy in NaCl is: \",round(Ecohesive,2),\"eV\"\n",
+ "print\"\\nWhich is not far from experimental value of -3.28 eV\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The cohesive energy in NaCl is: -3.21 eV\n",
+ "\n",
+ "Which is not far from experimental value of -3.28 eV\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 10.2,Page no:350"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "\n",
+ "#Variable declaration \n",
+ "A= 1.0 #cross-sectional area of wire, mm**2\n",
+ "I= 1.0 #current in wire, A\n",
+ "n= 8.5*(10**28) # electrons/m**3\n",
+ "e= 1.6*(10**(-19)) #charge of an electron, C\n",
+ "\n",
+ "#Calculation\n",
+ "Vdrift= I/(n*(A*(10**(-6)))*e) #m/s\n",
+ "\n",
+ "#Result\n",
+ "print\"The drift velocity of electrons in the copper wire is:%.2g\"%Vdrift,\"m/s\"\n",
+ "print\"\\nNOTE:Calculation mistake in book.Wrongly written as 7.4*10^-4\"\n",
+ " \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The drift velocity of electrons in the copper wire is:7.4e-05 m/s\n",
+ "\n",
+ "NOTE:Calculation mistake in book.Wrongly written as 7.4*10^-4\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 10.3,Page no:353"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration \n",
+ "n= 8.48*(10**28) #free electron density, m**(-3)\n",
+ "Vfermi= 1.57*(10**6) #Fermi Velocity, m/s\n",
+ "rho= 1.72*(10**(-8)) #resistivity, ohm\n",
+ "e= 1.6*(10**(-19)) #charge of an electron, C\n",
+ "Me= 9.1*(10**(-31)) #mass of electron, kg\n",
+ "\n",
+ "#Calculation\n",
+ "lamda= Me*Vfermi/(n*(e**2)*rho) #m\n",
+ "lamda= lamda*(10**9) #converting to nm\n",
+ "\n",
+ "#Result\n",
+ "print\"The mean free path is:\",round(lamda,1),\"nm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The mean free path is: 38.3 nm\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Concepts_Of_Modern_Physics/Chapter_11.ipynb b/Concepts_Of_Modern_Physics/Chapter_11.ipynb new file mode 100755 index 00000000..c03fdd42 --- /dev/null +++ b/Concepts_Of_Modern_Physics/Chapter_11.ipynb @@ -0,0 +1,314 @@ +{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 11:Nuclear Structure"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:11.1,Page no:393"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration \n",
+ "u= 1.66*(10**(-27)) #atomic mass unit, kg\n",
+ "Mc= 12*u # atomic mass of Carbon-12, kg\n",
+ "R= 2.7 #radius of nucleus, fm\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "R=R*(10**(-15)) #converting to m\n",
+ "density= Mc/((4.0/3.0)*(math.pi)*(R**3)) # kg/m**3\n",
+ "\n",
+ "#Calculation\n",
+ "print\"Density of Carbon 12 nucleus is:%.2g\"%density,\"kg/m**3\"\n",
+ " \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Density of Carbon 12 nucleus is:2.4e+17 kg/m**3\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:11.2,Page no:393"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "r= 2.4 #distance between centre of the protons, fm\n",
+ "r= r*(10**(-15)) #converting to m\n",
+ "e= 1.6*(10**(-19)) #charge of an electron, C\n",
+ "Po= 8.85*(10**(-12)) #Permittivity of free space, F/m\n",
+ "\n",
+ "#Calculation\n",
+ "K=1/(4*(math.pi)*Po) #constant, N.m**2/C**2\n",
+ "F= K*(e**2)/(r**2) #N\n",
+ "\n",
+ "#Result\n",
+ "print\"The repulsive force is: \",round(F),\"N\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The repulsive force is: 40.0 N\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:11.3,Page no:395"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "B= 1.0 #strength of magnetic field, T\n",
+ "Mneutron= 3.152*(10**(-8)) #Magnetic moment for neutron, eV/T\n",
+ "\n",
+ "#Calculation\n",
+ "#Part (a)\n",
+ "Mproton= 2.793*Mneutron #Magnetic moment for proton, eV/T\n",
+ "dE= 2*Mproton*B #eV\n",
+ "#Part (b)\n",
+ "h= 4.13*(10**(-15)) #Planck's constant, eV.s\n",
+ "Flarmor= dE/h #Hz\n",
+ "Flarmor= Flarmor/(10**6) #converting to MHz\n",
+ "\n",
+ "#Result\n",
+ "print\"(a).The energy difference is:%.4g\"%dE,\"eV\"\n",
+ "print\"(b).The Larmor frequency for a proton in the field is:\",round(Flarmor,1),\"MHz(APPROX)\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a).The energy difference is:1.761e-07 eV\n",
+ "(b).The Larmor frequency for a proton in the field is: 42.6 MHz(APPROX)\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:11.4,Page no:401"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration \n",
+ "Ebinding= 160.647 #binding nergy, MeV\n",
+ "Mh= 1.007825 #Mass of H1 atom, u\n",
+ "Mn= 1.008665 #Mass of neutron, u\n",
+ "Z=10 #number of protons\n",
+ "N=10 #number of neutrons\n",
+ "\n",
+ "#Calculation\n",
+ "Mneon= ((Z*Mh)+(N*Mn))-(Ebinding/931.49) #using Eqn 11.7\n",
+ "\n",
+ "#Result\n",
+ "print\"The atomic mass of Neon 10 isotope is: \",round(Mneon,3),\"u\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The atomic mass of Neon 10 isotope is: 19.992 u\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:11.5,Page no:402"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "\n",
+ "#Part(a)\n",
+ "Ca_42=41.958622 #[u] Mass of 42Ca20 \n",
+ "\n",
+ "#from table\n",
+ "M_neutron=1.008665 #[u] Mass of a free neutron\n",
+ "Ca_41=40.962278 #[u] Mass of 41Ca20 after removal of 1 neutron\n",
+ "#Part(b):\n",
+ "#from table\n",
+ "M_proton=1.007276466812 #[u] Mass of a free proton\n",
+ "K_19=40.96237 #[u] Mass of Potasium Isotope 41K19\n",
+ "\n",
+ "\n",
+ "#Calculation\n",
+ "#Part (a):\n",
+ "n_total=Ca_41+M_neutron\n",
+ "p_total=K_19+M_proton\n",
+ "neutron_BE=931.49*(n_total-Ca_42)#Binding energy in MeV\n",
+ "proton_BE=931.49*(p_total-Ca_42)\n",
+ "print \"(a).Binding energy of missing neutron is: \",round(neutron_BE,2),\"MeV\"\n",
+ "print\"(b).Binding energy for missing proton=\",round(proton_BE,2),\"MeV\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a).Binding energy of missing neutron is: 11.48 MeV\n",
+ "(b).Binding energy for missing proton= 10.27 MeV\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:11.6,Page no:407"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration \n",
+ "Z= 30.0 #proton number\n",
+ "N=34.0 #Neutron number\n",
+ "\n",
+ "#Calculation\n",
+ "#Using Eqn 11.7\n",
+ "Mh= 1.007825 #Mass of H1 atom, u\n",
+ "Mn= 1.008665 #Mass of neutron, u\n",
+ "Mzinc= 63.929 #atomic mass of zinc, u\n",
+ "Ebinding= ((Z*Mh)+(N*Mn)-Mzinc)*931.49 #MeV\n",
+ "#Using semiempirical formula, Eqn 11.18, Page 407\n",
+ "a1= 14.1 #Mev\n",
+ "a2= 13.0 #MeV\n",
+ "a3= 0.595 #Mev\n",
+ "a4= 19.0 #MeV\n",
+ "a5= 33.5 #MeV\n",
+ "A= Z+N \n",
+ "E2= ((a1*A)-(a2*(A**(2.0/3.0)))-(a3*Z*(Z-1)/(A**(1.0/3.0)))-(a4*((A-2*Z)**2)/A)+(a5/(A**(3.0/4.0)))) #MeV\n",
+ "\n",
+ "#Result\n",
+ "print\"Binding energy of Zinc 64 isotope is: \",round(Ebinding,1),\"MeV\"\n",
+ "print\"The binding energy using semi-empirical formula, in MeV, is: \",round(E2,1),\"MeV\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Binding energy of Zinc 64 isotope is: 559.2 MeV\n",
+ "The binding energy using semi-empirical formula, in MeV, is: 561.7 MeV\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:11.7,Page no:408"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "A=25\n",
+ "\n",
+ "#Calculation\n",
+ "#Derivation part\n",
+ "#dEb/dZ=-a3*(2*z-1)/A^(1/3)+4*a4*(A-2*Z)/A\n",
+ "#Z=a3*A**-1/3+4*a4/(2*a3*A^-1/3+8*a4*A**-1)\n",
+ "Z=(0.595*A**(-1.0/3.0)+76)/(1.19*A**(-1.0/3.0)+(152*A**-1))\n",
+ "print \"For A=25,Z=\",round(Z,1),\"\\nfor which we conclude that Z=\",round(Z),\"should be the atomic no. of most stable isobar of A=25\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "For A=25,Z= 11.7 \n",
+ "for which we conclude that Z= 12.0 should be the atomic no. of most stable isobar of A=25\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Concepts_Of_Modern_Physics/Chapter_12.ipynb b/Concepts_Of_Modern_Physics/Chapter_12.ipynb new file mode 100755 index 00000000..45956077 --- /dev/null +++ b/Concepts_Of_Modern_Physics/Chapter_12.ipynb @@ -0,0 +1,440 @@ +{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 12:Nuclear Transformations"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:12.2,Page no:425"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "Thalf= 3.82 #half-life in days, d\n",
+ "Lambda= 0.693/Thalf #decay constant\n",
+ "p= 0.6 # 60.0 percent of sample\n",
+ "\n",
+ "#Calculation\n",
+ "import numpy\n",
+ "import math\n",
+ "No= numpy.poly([1]) #Number of undecayed nuclei, at time t=0\n",
+ "N= (1.0-p)*No #Number of undecayed nuclei, at time t\n",
+ "t= (1.0/Lambda)*(np.log((N/No))) #decay time in days, d\n",
+ "t= t*(-1.0) \n",
+ "\n",
+ "#Result\n",
+ "print\"The decay time for Radon is: \",round(t[0],2),\"d\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The decay time for Radon is: 5.05 d\n"
+ ]
+ }
+ ],
+ "prompt_number": 20
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:12.3,Page no:427"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration \n",
+ "Thalf= 3.82 #half-life in days, d\n",
+ "Lambda= 0.693/(Thalf*86400.0) #decay constant, s**(-1)\n",
+ "Wradon= 1.00 #weight of sample, mg\n",
+ "MRadon= 222.0 #atomic mass of sample, u\n",
+ "\n",
+ "#Calculation\n",
+ "N= Wradon*(10**(-6))/(MRadon*(1.66*(10**(-27)))) #number of atoms\n",
+ "R= Lambda*N #decays/sec\n",
+ "R_tbq=round(R/10**12,2) #in TBq\n",
+ "R_Ci=R_tbq*27.15 #Ci Calories\n",
+ "\n",
+ "\n",
+ "#Result\n",
+ "print\"The activity of the sample is: %.2g\"%R,\"decays/sec=\",R_tbq,\"TBq=\",round(R_Ci),\"Ci\"\n",
+ " \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The activity of the sample is: 5.7e+12 decays/sec= 5.7 TBq= 155.0 Ci\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:12.4,Page no:427"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "Ro= 155 #initial activity, Ci\n",
+ "Lambda= 2.11*(10**(-6)) #decay constant, s**(-1)\n",
+ "t= 7 #days\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "t= t*86400 #converting to s\n",
+ "R= Ro*((math.exp(-(Lambda*t)))) #final activity, Ci\n",
+ "\n",
+ "#Result\n",
+ "print\"The activity after one week is: \",round(R),\"Ci\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The activity after one week is: 43.0 Ci\n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:12.5,Page no:428"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration \n",
+ "R= 13.0 #present activity,\n",
+ "Ro= 16.0 #activity of live wood\n",
+ "Thalf= 5760.0 #half life of radiocarbon, y\n",
+ "\n",
+ "#Calculation\n",
+ "Lambda= 0.693/(Thalf) #decay constant, y**(-1)\n",
+ "t= (1.0/Lambda)*(math.log(Ro/R)) #age of sample, y\n",
+ "\n",
+ "#Result\n",
+ "print\"The age of the wooden sample is: %.2g\"%t,\"y\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The age of the wooden sample is: 1.7e+03 y\n"
+ ]
+ }
+ ],
+ "prompt_number": 12
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:12.6,Page no:432"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration \n",
+ "Thalf1= 2.5*(10**5) #half-life of U-234, y\n",
+ "AtomicRatio= 1.8*(10**4) #atomic ratio of u-238 and U-234 in the sample\n",
+ "\n",
+ "#Calculation\n",
+ "Thalf2= AtomicRatio*Thalf1 #using Eqn12.9\n",
+ "\n",
+ "#Result\n",
+ "print\"The half-life of Uranium-238 is: %.2g\"%Thalf2,\"years\" \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The half-life of Uranium-238 is: 4.5e+09 years\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:12.7,Page no:433"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration \n",
+ "Npolonium= 84 #atomic number of polonium\n",
+ "Nalpha= 2 #atomic number of alpha particle\n",
+ "Z= Npolonium-Nalpha #atomic number of daughter nuclide\n",
+ "Mpolonium= 209.9829 #mass number of polonium, u\n",
+ "Malpha= 4.0026 #mass number of alpha particle, u\n",
+ "\n",
+ "#Calculation\n",
+ "A= Mpolonium-Malpha #mass number of daughter nuclide\n",
+ "# The daughter nuclide has atomic number: \n",
+ "# 82. \n",
+ "\n",
+ "# and mass number: \n",
+ "# 205.9803. \n",
+ " \n",
+ "Ealpha= 5.3 #energy of alpha particle, MeV\n",
+ "Q= Mpolonium*Ealpha/A #disintegration energy, MeV\n",
+ "Mq= Q/931 #mass equivalent for Q, u\n",
+ "Mnuclide= Mpolonium-Malpha-Mq #u\n",
+ "\n",
+ "#Result\n",
+ "print\"(a)The daughter nuclide has atomic number: \",Z\n",
+ "print\"and mass number: \",round(A)\n",
+ "print\"(b).The atomic mass of the daughter nuclide is: \",round(Mnuclide,4)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a)The daughter nuclide has atomic number: 82\n",
+ "and mass number: 206.0\n",
+ "(b).The atomic mass of the daughter nuclide is: 205.9745\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:12.8,Page no:444"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration \n",
+ "CrossSection= 2*(10**4) # capture cross section of Cadmium-113\n",
+ "CrossSection= CrossSection*(10**(-28)) # converting to m**2\n",
+ "Mcadmium= 112.0 #mean atomic mass of cadmium, u\n",
+ "density= 8.64*(10**3) #density of cadmium sheet used, kg/m**3\n",
+ "\n",
+ "#Calculation\n",
+ "#Part (a)\n",
+ "import math\n",
+ "t= 0.1 #hickness of sheet used, mm\n",
+ "t= t*10**(-3) #converting to m\n",
+ "p= 12.0 #percent of Cadmium-113 in natural cadmium\n",
+ "u= 1.66*(10**(-27)) #atomic mass unit, kg\n",
+ "n= (p/100.0)*density/(Mcadmium*u) #number of atoms, atoms/m**3\n",
+ "Fabsorbed= 1.0- (math.exp((-n)*(CrossSection)*(t))) #absorbed fraction\n",
+ "#Part (b)\n",
+ "t2= (-(math.log(0.01)))/(n*CrossSection) #required thickness, m\n",
+ "t2= t2*10**(3.0) #converting to mm\n",
+ "\n",
+ "\n",
+ "#Result\n",
+ "print\"(a).The fraction of incident beam absorbed is: \",round(Fabsorbed,2)\n",
+ "print\"(b).The thickness required to absorb 99 percent of incident thermal neutrons is: \",round(t2,2),\"mm\"\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a).The fraction of incident beam absorbed is: 0.67\n",
+ "(b).The thickness required to absorb 99 percent of incident thermal neutrons is: 0.41 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:12.9,Page no:445"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "CrossSection= 2.0*(10**4) # capture cross section of Cadmium-113, b\n",
+ "CrossSection= CrossSection*(10**(-28)) # converting to m**2\n",
+ "\n",
+ "#Calculation\n",
+ "n= (12.0/100.0)*(8.64*10**3)/(112.0*(1.66*10**(-27))) #number of atoms, atoms/m**3\n",
+ "Lambda= 1.0/(n*CrossSection) #mean free path, m\n",
+ "Lambda= Lambda*10**3 #converting to, mm\n",
+ "\n",
+ "#Result\n",
+ "print\"The mean free path is: \",round(Lambda,4),\"mm\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The mean free path is: 0.0897 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:12.10,Page no:446"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration \n",
+ "Thalf= 2.69 #half life of gold,d\n",
+ "Lambda= 0.693/(Thalf*86400.0) #decay constant, s**(-1)\n",
+ "R= 200.0 #required activity, mCi\n",
+ "R= R*10**(-6) #converting to Ci\n",
+ "\n",
+ "#Calculation\n",
+ "dN= R/(Lambda/(3.70*10**(10))) #atoms\n",
+ "Wgold= 10.0 #mass of foil, mg\n",
+ "u= 1.66*(10**(-27)) #atomic mass unit, kg\n",
+ "Mgold= 197.0 # u\n",
+ "n2= Wgold*10.0**(-6)/(Mgold*u) #total no. of atoms\n",
+ "phi= 2.0*10**(16) #flux, neutrons/m**2\n",
+ "CrossSection= 99.0*10**(-28) #m**2\n",
+ "dT= dN/(phi*n2*CrossSection) #s\n",
+ "dT_m=divmod(dT,60)\n",
+ "#Result\n",
+ "print\"The irradiation period required is: \",round(dT,1),\"s=\",round(dT_m[0],1),\"min\",round(dT_m[1],1),\"seconds(Approx)\"\n",
+ " "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The irradiation period required is: 409.9 s= 6.0 min 49.9 seconds(Approx)\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:12.11,Page no:450"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration \n",
+ "m1= 14.00307 #u\n",
+ "m2= 4.00260 #u\n",
+ "m3= 1.00783 #u\n",
+ "m4= 16.99913 #u\n",
+ "\n",
+ "#Calculation\n",
+ "k= m1+m2-m3-m4 # difference in total mass of reactants and products, u\n",
+ "Q= k*931.5 #energy exchanged, MeV\n",
+ "KEcm= -Q #minimum KE needed in centre of mass system, MeV\n",
+ "KElab= KEcm*(m2+m1)/m1 #minimum KE in laboratory system\n",
+ "\n",
+ "#Result\n",
+ "print\"The minimum KE required by the alpha particle is: \",round(KElab,3),\"MeV(APPROX)\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The minimum KE required by the alpha particle is: 1.545 MeV(APPROX)\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Concepts_Of_Modern_Physics/Chapter_2.ipynb b/Concepts_Of_Modern_Physics/Chapter_2.ipynb new file mode 100755 index 00000000..dfc67c62 --- /dev/null +++ b/Concepts_Of_Modern_Physics/Chapter_2.ipynb @@ -0,0 +1,334 @@ +{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 2:Particle properties of waves"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.1,Page no:61"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration \n",
+ "Ft= 660 #frequency of tuning fork, Hz\n",
+ "Fo= 5.00*(10**14) #frquency of atomic oscillator, Hz\n",
+ "Ef= 0.04 #vibrational energy of tuning fork, J\n",
+ "h= 6.63*(10**(-34)) #Planck's constant, J.s\n",
+ "\n",
+ "#Calculation\n",
+ "E1= h*Ft #Total energy of tuning fork, J\n",
+ "E2= h*Fo #Total energy of atomic oscillator, J\n",
+ "E2= E2/(1.60*(10**(-19))) #converting to eV\n",
+ "\n",
+ "#Result\n",
+ "print\"(a).Energy of tuning fork is:%.3g\"%E1,\"J\"\n",
+ "print\"(b).Energy of atomic oscillator is:\",round(E2,2),\"eV(approx)\"\n",
+ " "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a).Energy of tuning fork is:4.38e-31 J\n",
+ "(b).Energy of atomic oscillator is: 2.07 eV(approx)\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2,Page no:66"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration \n",
+ "l= 350 #Wavelength of UV light, nm\n",
+ "i= 1.00 #intensity of UV light, W/m**2\n",
+ "\n",
+ "#Part (a)\n",
+ "l= l*10**(-9) #converting to m\n",
+ "Ep= (1.24*(10**(-6)))/l #energy of photon, using Eqn (2.11) on Page 66, e.V\n",
+ "t= 2.2 #work function of Potassium surface, eV\n",
+ " \n",
+ "#Calculation\n",
+ "KEmax= Ep-t #Max KE of the phototelectrons, eV\n",
+ "\n",
+ "#Part (b) \n",
+ "A= 1.00 #Surface area, cm**2\n",
+ "A= A* 10**(-4) #converting to m**2\n",
+ "E= 5.68*(10**(-19)) #Photon energy, J\n",
+ "Np= i*A/E #number of incident photon, per second\n",
+ "Ne= (0.0050)*Np #number of photoeectrons emitted, per second\n",
+ "\n",
+ "#Result\n",
+ "print\"(a).Maximum KE of photoelectrin is: \",round(KEmax,1),\"eV\"\n",
+ "print\"(b).Rate of emission of photoelectrons is:%.2g\"%Ne,\"photoelectrons/s\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a).Maximum KE of photoelectrin is: 1.3 eV\n",
+ "(b).Rate of emission of photoelectrons is:8.8e+11 photoelectrons/s\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.3,Page no:72"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration \n",
+ "AP= 50000 #Accelerating potential of the x-ray machine, V\n",
+ "l= (1.24*(10**(-6)))/AP*(10**(9)) #Minimum wavelength, nm\n",
+ "\n",
+ "#Calculation\n",
+ "Fmax= 3*(10**8)/(l*(10**(-9))) #Maximum frequency, Hz\n",
+ "\n",
+ "#Result\n",
+ "print\"Minimum wavelength possible is: \",l,\"nm\"\n",
+ "print\"Maximum frequency possible is: %.3g\"%Fmax,\"Hz\"\n",
+ " "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Minimum wavelength possible is: 0.0248 nm\n",
+ "Maximum frequency possible is: 1.21e+19 Hz\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.4,Page no:78"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration \n",
+ "#part (a)\n",
+ "l= 10.0 #wavelength of x-ray, pm\n",
+ "r= 45.0 #angle of scattered articles, degree\n",
+ "lc= 2.426*(10**(-12)) #Compton wavelength for electron, m\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "k= math.cos(math.radians(45)) \n",
+ "lc= lc* 10.0**12 # converting to pm\n",
+ "print \n",
+ "l2= l+ lc*(1.0-k) #using Eqn 2.23\n",
+ "#Part (b)\n",
+ "lmax= l+(lc*2) #for (1-k)=2\n",
+ "#Part (c)\n",
+ "h= 6.63*(10**(-34)) #Planck's constant, J.s\n",
+ "c= 3*10**8 #velocity of light, m/s\n",
+ "c=c*10**12 #converting to pm/s\n",
+ "KEmax= (h*c)*((1/l)-(1/lmax)) #J\n",
+ "\n",
+ "#Result\n",
+ "print\"(a):The wavelength of scattered x-ray is: \",round(l2,1),\"pm\"\n",
+ "print\"(b):Maximum wavelength is: \",round(lmax,1),\"pm\"\n",
+ "print\"(c):The maximum KE of recoil electrons is:%.3g\"%KEmax,\"J\"\n",
+ "print\"which is equal to \",round(KEmax/1.6021773e-16,1),\"keV(approx)\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "(a):The wavelength of scattered x-ray is: 10.7 pm\n",
+ "(b):Maximum wavelength is: 14.9 pm\n",
+ "(c):The maximum KE of recoil electrons is:6.5e-15 J\n",
+ "which is equal to 40.6 keV(approx)\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.6,Page no:82"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration \n",
+ "c=3.0*10**8 #velocity of light, m/s\n",
+ "v= 0.5*c #velocity of electron and positron, m/s\n",
+ "y= 1.0/math.sqrt(1.0-(v/c**2)) #gamma, for relativistic momentum\n",
+ "m=0.511/c**2 #MeV/c**2\n",
+ "\n",
+ "#Calculation\n",
+ "p1_p2=((2*m*c*c**2)*(v/c**2))/(math.sqrt(1-(v/c)**2))\n",
+ "p1_plus_p2=(2.0*m*(c**2))/math.sqrt(1.0-(v/c)**2.0)\n",
+ "p1=(p1_p2+p1_plus_p2)/2.0\n",
+ "p2=p1_plus_p2-p1\n",
+ "E1=p1 #MeV\n",
+ "E2=p2 #MeV\n",
+ "\n",
+ "#Result\n",
+ "print\"The Energy of first photon is,E1=\",round(E1,3),\"MeV\"\n",
+ "print\"The Energyof second photon is:,E2=\",round(E2,3),\"MeV\"\n",
+ "\n",
+ "print\"NOTE:There is a mistake in the formula given in the book for p1_p2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Energy of first photon is,E1= 0.885 MeV\n",
+ "The Energyof second photon is:,E2= 0.295 MeV\n",
+ "NOTE:There is a mistake in the formula given in the book for p1_p2\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.7,Page no:85"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration \n",
+ "M= 4.9 #Linear attenuation coefficient for gamma ray in water, m**(-1)\n",
+ "I= 2.0 #Original intensity of gamma ray, MeV\n",
+ "\n",
+ "#Calculation\n",
+ "#Part (a)\n",
+ "x= 10.0 #distance travelled under water, cm\n",
+ "x= x/100.0 #converting to m\n",
+ "Irel= math.exp(-(M*x)) #Relative intensity\n",
+ "#Part(b)\n",
+ "Ip= I/100 #Present intensity, 1 percent of Original, MeV\n",
+ "x2= math.log(I/Ip)/M #distance travelled, m\n",
+ "\n",
+ "#Result\n",
+ "print\"(a)Relative intensity of the beam is: \",round(Irel,2)\n",
+ "print\"(b)The distance travelled by the beam is:\",round(x2,2),\"m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a)Relative intensity of the beam is: 0.61\n",
+ "(b)The distance travelled by the beam is: 0.94 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.8,Page no:86"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration \n",
+ "H= 22.5 #Height of fall, m\n",
+ "F= 7.3*(10**14) #Original frequency, Hz\n",
+ "c= 3*(10**8) #velocity of light, m/s\n",
+ "g= 9.8 #Acceleration due to gravity, m/s**2\n",
+ "\n",
+ "#Calculation\n",
+ "Frel= g*H*F/(c**2) #Change in frequency, Hz\n",
+ "\n",
+ "#Result\n",
+ "print\"The change in frquency of a photon fallin through 22.5 m is: \",round(Frel,1), \"Hz\"\n",
+ " "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The change in frquency of a photon fallin through 22.5 m is: 1.8 Hz\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Concepts_Of_Modern_Physics/Chapter_3.ipynb b/Concepts_Of_Modern_Physics/Chapter_3.ipynb new file mode 100755 index 00000000..749a4c27 --- /dev/null +++ b/Concepts_Of_Modern_Physics/Chapter_3.ipynb @@ -0,0 +1,470 @@ +{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 3:Wave propoerties of particles"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:3.1,Page no:94"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "m= 46.0 #mass, gms\n",
+ "v=30.0 #velocity, m/s\n",
+ "h= 6.63*(10**(-34)) #Planck's constant, J.s\n",
+ "m=m/1000.0 #convert to kgs\n",
+ "\n",
+ "#Calculation\n",
+ "a=h/(m*v) #de Broglie wavelength, m\n",
+ "print\"The de Broglie wavelength of the golf ball (in m) is:%.2g\"%a,\"m\"\n",
+ "#Part(b)\n",
+ "m= 9.1*(10**(-31)) #mass, kg\n",
+ "v=10**7 #velocity, m/s\n",
+ "h= 6.63*(10**(-34)) #Planck's constant, J.s\n",
+ "a=h/(m*v) #de Broglie wavelength, mts\n",
+ "\n",
+ "#Result\n",
+ "print\"de Broglie wavelength for the electron (in m) is:%.2g\"%a,\"m\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The de Broglie wavelength of the golf ball (in m) is:4.8e-34 m\n",
+ "de Broglie wavelength for the electron (in m) is:7.3e-11 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:3.2,Page no:94"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "a=10**(-15) #de Broglie wavelength, mts\n",
+ "Eo= 0.938 #proton rest energy, GeV\n",
+ "h= 4.136*(10**(-15)) #Planck's constant, eV.s\n",
+ "c= 2.998*(10**8) #velocity of light, m/s\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "p= h/a # p is momentum, kg.m/s \n",
+ "pc= (h*c)/a #Photon's energy, ev\n",
+ "pc=pc*(10**(-9)) #convert to GeV\n",
+ "\t\t\t\t#pc>E0, relativistic calculation\n",
+ "E= math.sqrt((Eo**2) + (pc**2)) #total energy, GeV\n",
+ "KE = E-Eo #Kinetic energy, GeV\n",
+ "KE= KE*1000 # convert to MeV\n",
+ "\n",
+ "#Result\n",
+ "print\"Kinetic Energy of the proton (in MeV) is: \",round(KE),\"MeV\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Kinetic Energy of the proton (in MeV) is: 617.0 MeV\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:3.3,Page no:103"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "\n",
+ "#Variable declaration\n",
+ "a= 2*(10**(-12)) #de Broglie wavelength, mts\n",
+ "h= 4.136*(10**(-15)) #Planck's constant, eV.s\n",
+ "c= 3*(10**8) #velocity of light, m/s\n",
+ "\n",
+ "#Calculation\n",
+ "pc= (h*c)/a #p is momentum, pc is electron's energy, eV\n",
+ "pc= pc/1000 #convert to keV\n",
+ "Eo= 511 #rest energy, keV\n",
+ "E= math.sqrt((Eo**2)+(pc**2)) #Total Energy, keV\n",
+ "KE= E-Eo #Kinetic energy, keV\n",
+ "\n",
+ "#Result\n",
+ "print\"kinetic energy of the electron (in keV) is,KE: \",round(KE,1),\"keV\"\n",
+ "vg= c*(math.sqrt(1-(Eo**2/E**2))) #group velocity, m/s\n",
+ "vp= c**2/vg #phase velocity, m/s\n",
+ "print\"group velocity of the electron (in m/s) is,vg: \",round(vg/c,3),\"c\"\n",
+ "print\"phase velocity of the electron (in m/s) is,vp:\",round(vp/c,1),\"c\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "kinetic energy of the electron (in keV) is,KE: 292.8 keV\n",
+ "group velocity of the electron (in m/s) is,vg: 0.772 c\n",
+ "phase velocity of the electron (in m/s) is,vp: 1.3 c\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:3.4,Page no:107"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "%pylab inline\n",
+ "#Variable declaration\n",
+ "m= 9.1*(10**(-31)) #mass, kg\n",
+ "L= 0.10 #length of box, nm\n",
+ "L= L*(10**(-9)) #convert to m\n",
+ "h= 6.63*(10**(-34)) #Planck's constant, J.s\n",
+ "\n",
+ "#Calculation\n",
+ "En=range(4)\n",
+ "p=range(5)\n",
+ "import matplotlib.pyplot as plt\n",
+ "#from numpy import random\n",
+ "import numpy as np\n",
+ "for n in range(1,5):\n",
+ " En[n-1]=(n**2)*(h**2)/(8*m*(L**2)) #Permitted energies, J\n",
+ " print\"For level:\",n\n",
+ " print\"Permitted ernergis :%.2g\"%En[n-1],\"J\"\n",
+ " En[n-1]=38*(n**2) \n",
+ " plt.plot([En[n-1]])\n",
+ " print\"Permitted energies (in eV) :\",En[n-1], \"eV\"\n",
+ " p[n]=plt.hlines(En[n-1],0,5, colors=np.random.rand(3,1), linestyles='solid', label='d', hold=None,)\n",
+ " p[n].axes.get_xaxis().set_visible(False)\n",
+ "plt.legend([p[1],p[2],p[3],p[4]],[\"n=1\",\"n=2\",\"n=3\",\"n=4\"])\n",
+ "plt.ylabel(\"Energy,eV\")\n",
+ "plt.title(\"Energy levels of an e- confined to a box 1nm wide\")\n",
+ "\n",
+ "plt.show()\n",
+ "\n",
+ "#Result\n",
+ "print\"\\nANS::The minimum energy the electron can have is:\",min(En),\"n**2 eV\\n\\n\"\n",
+ "\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Populating the interactive namespace from numpy and matplotlib\n",
+ "For level: 1\n",
+ "Permitted ernergis :6e-18 J\n",
+ "Permitted energies (in eV) : 38 eV\n",
+ "For level:"
+ ]
+ },
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " 2\n",
+ "Permitted ernergis :2.4e-17 J\n",
+ "Permitted energies (in eV) : 152 eV\n",
+ "For level: 3\n",
+ "Permitted ernergis :5.4e-17 J\n",
+ "Permitted energies (in eV) : 342 eV\n",
+ "For level: 4\n",
+ "Permitted ernergis :9.7e-17 J\n",
+ "Permitted energies (in eV) : 608 eV\n"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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4OPz3f/83OnXqhLCwMEyfPr056quD5wiIiIz32OcI/Pz88NlnnyE2NhY3b97E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\nAAAAAElFTkSuQmCC\n",
+ "text": [
+ "<matplotlib.figure.Figure at 0x79807f0>"
+ ]
+ },
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "ANS::The minimum energy the electron can have is: 38 n**2 eV\n",
+ "\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:3.5,Page no:107"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "m= 10.0 #mass, gms\n",
+ "m= m/1000.0 #convert to kgs\n",
+ "L= 10.0 #Length of box, cms\n",
+ "L= L/100.0 #convert to mts\n",
+ "h= 6.63*(10**(-34)) #Planck's constant, J.s\n",
+ "\n",
+ "#Calculation\n",
+ "En=range(5)\n",
+ "for n in range(1,5):\n",
+ " En[n-1]=(n**2)*(h**2)/(8*m*(L**2)) #Permitted energies, J\n",
+ " print\"for level:\",n\n",
+ " print\"Permitted ernergis :%.2g\"%En[n-1],\"J\"\n",
+ " \n",
+ "#Result\n",
+ "print \"Minimum energy is :En=%.2g\"%min(En),\"n**2 J,corresponding to n=1\"\n",
+ "\n",
+ " \n",
+ "#corresponding kinetic energy is very low, hence Quantum effects are imperceptible, and Newtonian mechanics is dominant"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "for level: 1\n",
+ "Permitted ernergis :5.5e-64 J\n",
+ "for level: 2\n",
+ "Permitted ernergis :2.2e-63 J\n",
+ "for level: 3\n",
+ "Permitted ernergis :4.9e-63 J\n",
+ "for level: 4\n",
+ "Permitted ernergis :8.8e-63 J\n",
+ "Minimum energy is :En=5.5e-64 n**2 J,corresponding to n=1\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:3.6,Page no:112"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "Xo= 10**(-11) #uncertainty at time t=o, mts\n",
+ "hb= 1.054*(10**(-34)) #h-bar, reduced Planck's constant, J.s\n",
+ "t= 1 #time, s\n",
+ "m= 1.672*(10**(-27)) #mass, kg\n",
+ "\n",
+ "#Calculation\n",
+ "x1= hb*t/(2*m*Xo) #uncertainty at time t=1, mts\n",
+ "\n",
+ "#Result\n",
+ "print\"accuracy in position of proton after 1.00 seconds is ,delta_x>=:%.3g\"%x1,\"m\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "accuracy in position of proton after 1.00 seconds is ,delta_x>=:3.15e+03 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:3.7,Page no:114"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "r= 5*(10**(-15)) #radius of nucleus, mts\n",
+ "Xo= 5*(10**(-15)) #assumed initial uncertainty, mts\n",
+ "hb= 1.054*(10**(-34)) #reduced Planck's constant, J.s\n",
+ "\n",
+ "#Calculation\n",
+ "p= hb/(2*Xo) #uncertainty in momentum, kg.m/s\n",
+ "c= 3*(10**8) #velocity of light, m/s\n",
+ "KE= p*c #minimum kinetic energy required, J\n",
+ "\n",
+ "#Result\n",
+ "print\"Uncertainty in momentum of the electron is :%.2g\"%p,\"kg.m/s\"\n",
+ "print\"The minimum energy required is :%.2g\"%KE,\"J \\n or\"\n",
+ "KE= KE/(1.6*(10**(-19))) #convert to eV\n",
+ "KE= KE/(10**6) #convert to MeV\n",
+ "print\"The minimum energy required is : \",round(KE),\"MeV\"\n",
+ " \n",
+ " "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Uncertainty in momentum of the electron is :1.1e-20 kg.m/s\n",
+ "The minimum energy required is :3.2e-12 J \n",
+ " or\n",
+ "The minimum energy required is : 20.0 MeV\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:3.8,Page no:115"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "r= 5.3*(10**(-11)) #radius of atom, mts\n",
+ "Xo= 5.3*(10**(-11)) #uncertainty in position, mts\n",
+ "hb= 1.054*(10**(-34)) #Reduced planck Constant, J.s\n",
+ "\n",
+ "#Calculation\n",
+ "p= hb/(2*Xo) #uncertainty in momentum, kg.m/s\n",
+ "m= 9.1*(10**(-31)) #mass, kg\n",
+ "KE= p**2/(2*m) # minimum kinetic energy, J\n",
+ "\n",
+ "#Result\n",
+ "print\"The minimum possible kinetic energy for an electron in the atom is :%.2g\"%KE,\"J\"\n",
+ "KE= KE/(1.6*(10**(-19))) #convert to eV\n",
+ "print\"OR,in eV:KE=\",round(KE,1),\"eV\"\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The minimum possible kinetic energy for an electron in the atom is :5.4e-19 J\n",
+ "OR,in eV:KE= 3.4 eV\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:3.9,Page no:116"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "t= 10**(-8) #time period between excitation and radiation, s\n",
+ "hb= 1.054*(10**(-34)) #Reduced Planck's constant, J.s\n",
+ "\n",
+ "#Calculation\n",
+ "Eo= hb/(2*t) #uncertainty in photon energy, J\n",
+ "h=hb*(2*(math.pi)) #Planck's constant\n",
+ "Fo= Eo/h #uncertainty in frequency of light, Hz\n",
+ "\n",
+ "#Result\n",
+ "print\"Photon energy is uncertain by (in J) :%.2g\"%Eo,\"J\"\n",
+ "print\"Frquency of photon is uncertain by (in Hz) :%.2g\"%Fo,\"Hz\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Photon energy is uncertain by (in J) :5.3e-27 J\n",
+ "Frquency of photon is uncertain by (in Hz) :8e+06 Hz\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Concepts_Of_Modern_Physics/Chapter_4.ipynb b/Concepts_Of_Modern_Physics/Chapter_4.ipynb new file mode 100755 index 00000000..36d4670d --- /dev/null +++ b/Concepts_Of_Modern_Physics/Chapter_4.ipynb @@ -0,0 +1,336 @@ +{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 4:Atomic Structure"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:4.1,Page no:125"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration \n",
+ "E= -13.6; #Energy required to separate electron and proton, eV\n",
+ "e= 1.6*(10**(-19)); #charge of an electron, C\n",
+ "E= E*e; #converting to J\n",
+ "Po= 8.85*(10**(-12)); #Permittivity of free space, F/m\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "r= e**2/(8*(math.pi)*Po*E); #radius, m\n",
+ "r= -r;\n",
+ "m= 9.1*(10**(-31)); #mass of electron, kg\n",
+ "v=e/math.sqrt(4*(math.pi)*Po*m*r); #velocity, m/s\n",
+ "\n",
+ "#Result\n",
+ "print\"The orbital radius of the electron is:%.2g\"%r,\"m\"\n",
+ "print\"The velocity of electron is:%.2g\"%v,\"m/s\"\n",
+ " \n",
+ " "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The orbital radius of the electron is:5.3e-11 m\n",
+ "The velocity of electron is:2.2e+06 m/s\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:4.2,Page no:135"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration \n",
+ "n1=1.0; #initial state\n",
+ "n2=3.0; #final state\n",
+ "E= -13.6; #energy in ground state, eV\n",
+ "\n",
+ "#Calculation\n",
+ "dE= E*((1/n2**2)-(1/n1**2)); #Change in energy, eV\n",
+ "\n",
+ "#Result\n",
+ "print\"The energy change of Hydrogen atom is: \",round(dE,1),\"eV\"\n",
+ " "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The energy change of Hydrogen atom is: 12.1 eV\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:4.3,Page no:135"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "#Part(a)\n",
+ "Rn= 10.0**(-5); #radius of Rydberg atom, m\n",
+ "Ao= 5.29*(10**(-11)); #Bohr radius, m\n",
+ "\n",
+ "#Calculation\n",
+ "n= math.sqrt(Rn/Ao); #Quantum number\n",
+ "E1= -13.6; #Ground state energy level, eV\n",
+ "En= E1/n**2.0; #Nth state energy level, eV\n",
+ "\n",
+ "#Result\n",
+ "print\"(a).The quantum number of the Rydberg atom is: \",round(n)\n",
+ "print\"(b).The energy ofthe rydberg atom is:%.3g\"%En,\"eV\"\n",
+ " "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a).The quantum number of the Rydberg atom is: 435.0\n",
+ "(b).The energy ofthe rydberg atom is:-7.19e-05 eV\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:4.4,Page no:138"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "n1= 3.0; #initial state\n",
+ "n2= 2.0; #final state\n",
+ "R= 1.097*(10**7); #Rydberg's constant, m**(-1)\n",
+ "\n",
+ "#Calculation\n",
+ "k= (1/n2**2)-(1/n1**2);\n",
+ "l= 1/(k*R); #longest wavelength, m\n",
+ "l= l*(10**9); #converting to nm\n",
+ "\n",
+ "#Result\n",
+ "print\"The longest in Balmer series of Hydrogen, in nm, is: \",round(l),\"nm\"\n",
+ " "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The longest in Balmer series of Hydrogen, in nm, is: 656.0 nm\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:4.5,Page no:139"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "n1=1.0; #initial state\n",
+ "n2=2.0; #final state\n",
+ "E1= 2.18*(10**(-18)); #Rydberg's constant, J\n",
+ "h= 6.63*(10**(-34)); #Planck's constant, J.s\n",
+ "\n",
+ "#Calculation\n",
+ "f1= (E1/h)*(2.0/n1**3); #Frequency for first orbit, rev/s\n",
+ "f2= (E1/h)*(2.0/n2**3); #Frequency for second orbit, rev/s\n",
+ "print\"Ans (A):Frequency of revolution for orbit n=1 is,f1: %.3g\"%f1,\"rev/s\"\n",
+ "print\"Frequency of revolution for orbit n=2 is,f2:%.2e\"%f2,\"rev/s\"\n",
+ "print\"which is equivalent to 0.823*10**15 rev/s 'without' any scientific notation\\n\"\n",
+ "#Part (b)\n",
+ "n1=2.0; #initial orbit\n",
+ "n2=1.0; #final orbit\n",
+ "f= (E1/(h))*((1.0/(n2**2))-(1.0/n1**3)); #frequency, Hz\n",
+ "print\"Ans(B):Frequency of emitted photon is: %.3g\"%f,\"Hz\\n\"\n",
+ "#Part (c)\n",
+ "n= 2.0; #orbit\n",
+ "f= f2; #from part (a)\n",
+ "dt= 10.0**(-8); # time duration, s\n",
+ "N= f*dt; #Number of revolutions\n",
+ "#Result\n",
+ "print\"Ans(C):Number of revolutions the electron makes is:%.3g\"%N\n",
+ " "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Ans (A):Frequency of revolution for orbit n=1 is,f1: 6.58e+15 rev/s\n",
+ "Frequency of revolution for orbit n=2 is,f2:8.22e+14 rev/s\n",
+ "which is equivalent to 0.823*10**15 rev/s 'without' any scientific notation\n",
+ "\n",
+ "Ans(B):Frequency of emitted photon is: 2.88e+15 Hz\n",
+ "\n",
+ "Ans(C):Number of revolutions the electron makes is:8.22e+06\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:4.7,Page no:142"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "#Part (a) \n",
+ "Me= 9.1*(10**(-31)); #mass of electron, kg\n",
+ "m= 207*Me; #mass of muon, kg\n",
+ "\n",
+ "#Calculation\n",
+ "Mp= 1836*Me; #mass of proton, kg\n",
+ "Mreduced= (m*Mp)/(m+Mp); #reduced mass, kg\n",
+ "Ao= 5.29*(10**(-11)); #Bohr's orbit for n=1, m\n",
+ "r1= Ao; #expected orbit for atom, m\n",
+ "r2= (Me/Mreduced)*r1; #reduced radius of orbit, m\n",
+ "#Part (b)\n",
+ "E=-13.6; # energy for elctron in n=1, eV\n",
+ "Ereduced= (Mreduced/Me)*E; #energy for eectron in mounic atom, eV\n",
+ "Ereduced= Ereduced/(10**3);#converting to keV\n",
+ "\n",
+ "#Result\n",
+ "print\"(A)Radius of the mounic atom formed, in m, is:%.3g\"%r2,\"m\"\n",
+ "print\"(B)Ionisation energy for the muonic atom is: \",round(Ereduced,2),\"keV\"\n",
+ " "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(A)Radius of the mounic atom formed, in m, is:2.84e-13 m\n",
+ "(B)Ionisation energy for the muonic atom is: -2.53 keV\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:4.8,Page no:156"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration \n",
+ "I= 7.7; #Intensity of beam, MeV\n",
+ "Dgold= 1.93*(10**4); #density of gold foil used, kg/m**3\n",
+ "u= 1.66*(10**(-27)); #atomic mass unit, kg\n",
+ "Mgold= 197*u; #atomic mass of gold, per atom\n",
+ "\n",
+ "#Calculation\n",
+ "n= Dgold/Mgold; #number of atoms per unit volume, atoms/m**3\n",
+ "Zgold= 79; #atomic number of gold\n",
+ "e= 1.6*(10**(-19)); #electronis charge, C\n",
+ "KE= (I*e)/(10**(-6)); #converting to J\n",
+ "angle= 45; #degree\n",
+ "p=1/math.tan(math.radians(angle/2));\n",
+ "Po= 8.85*(10**(-12)); #Permittivity of free space, F/m\n",
+ "t= 3*(10**(-7)); #thickness of foil, m\n",
+ "f= (math.pi)*n*t*(((Zgold*(e**2))/(4*(math.pi)*Po*KE))**2)*(p**2) #using Rutherford scattering formula\n",
+ "\n",
+ "#Result\n",
+ "print\"f=%.g\"%f\n",
+ "print\"Fraction of the beam scattered through 45 degree or more is: \",round(f*100,3),\"%\"\n",
+ " "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "f=7e-05\n",
+ "Fraction of the beam scattered through 45 degree or more is: 0.007 %\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Concepts_Of_Modern_Physics/Chapter_5.ipynb b/Concepts_Of_Modern_Physics/Chapter_5.ipynb new file mode 100755 index 00000000..99006788 --- /dev/null +++ b/Concepts_Of_Modern_Physics/Chapter_5.ipynb @@ -0,0 +1,142 @@ +{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 5:Quantum Mechanics"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.4,Page no:180"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration \n",
+ "L= 1.0; #assuming Length L of box to be 1, this would not affect the probability\n",
+ "x1=0.45; #lower bound\n",
+ "x2=0.55; #upper bound\n",
+ "\n",
+ "from scipy.integrate import quad\n",
+ "import math\n",
+ "\n",
+ "#Calculation \n",
+ "def f(x):\n",
+ " y=(math.sin(n*(math.pi)*x))**2\n",
+ " return(y)\n",
+ "n=1.0;\n",
+ "I1=quad(f,x1,x2) #for ground state\n",
+ "P1=(2/L*I1[0])\n",
+ "n=2.0;\n",
+ "I2=quad(f,x1,x2) #for ground state\n",
+ "P2=(2/L*I2[0])\n",
+ "\n",
+ "\n",
+ "\n",
+ "#Result\n",
+ "print\"The probability n ground state is: \",round(P1,3),\"=\",round(P1,3)*100,\"percent\"\n",
+ "print\"The probability in first excited state is: \",round(P2,4),\"=\",round(P2,4)*100,\"percent\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The probability n ground state is: 0.198 = 19.8 percent\n",
+ "The probability in first excited state is: 0.0065 = 0.65 percent\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.6,Page no:186"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "#Part (a)\n",
+ "E1= 1.0 #energy of first electron, eV\n",
+ "E2= 2.0 #energy of second electron, eV\n",
+ "Eb= 10.0 #height of barrier, eV\n",
+ "Wb= 0.50 #width of barrier, nm\n",
+ "Wb= Wb* 10**(-9) #converting to m\n",
+ "hbar= 1.054*10**(-34) #reduced Planck's conctaant, J.s\n",
+ "Me= 9.1*10**(-31) #mass of electron, kg\n",
+ "e= 1.6*10**(-19) #charge of an electron, J/eV\n",
+ "import math\n",
+ "#Calculation\n",
+ "\n",
+ "k2= (math.sqrt(2*(Me)*(Eb-E1)*(e)))/(hbar) #for 1.0 eV e-, m**(-1)\n",
+ "k1= (math.sqrt(2*Me*(Eb-E2)*e))/hbar #for first electron, m**(-1)\n",
+ "T1= math.exp((-2)*k2*Wb) #transmission probability for first electron\n",
+ "T2= math.exp((-2)*k1*Wb) #for second electron\n",
+ "print\"(A.)\\nTransmission probability for electrons with energy 1.0 eV is:%.2g\"%T1\n",
+ "print\"Transmission probability for electrons with energy 2.0 eV is: %.2g\"%T2\n",
+ "\n",
+ "#Part (b)\n",
+ "Wb= Wb*2; #Barrier width doubled\n",
+ "T11= math.exp((-2)*k2*Wb) # changed transmission probability for first electron\n",
+ "T22= math.exp((-2)*k1*Wb) #for second electron\n",
+ "\n",
+ "#Result\n",
+ "print\"\\n\\n(B.):After the barrier width is doubled:\\n\"\n",
+ "print\"Transmission probability for electrons with energy 1.0 eV is:%.2g\"%T11\n",
+ "print\"Transmission probability for electrons with energy 2.0 eV is: %.2g\"%T22\n",
+ "print\"\\n\\nNOTE:Calculation mistake in book in the calculation of k2,\\nit is wrongly written as 1.6e+10,\\nTHAT's Why a change in final answer\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(A.)\n",
+ "Transmission probability for electrons with energy 1.0 eV is:2.1e-07\n",
+ "Transmission probability for electrons with energy 2.0 eV is: 5.1e-07\n",
+ "\n",
+ "\n",
+ "(B.):After the barrier width is doubled:\n",
+ "\n",
+ "Transmission probability for electrons with energy 1.0 eV is:4.6e-14\n",
+ "Transmission probability for electrons with energy 2.0 eV is: 2.6e-13\n",
+ "\n",
+ "\n",
+ "NOTE:Calculation mistake in book in the calculation of k2,\n",
+ "it is wrongly written as 1.6e+10,\n",
+ "THAT's Why a change in final answer\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Concepts_Of_Modern_Physics/Chapter_6.ipynb b/Concepts_Of_Modern_Physics/Chapter_6.ipynb new file mode 100755 index 00000000..3e924e61 --- /dev/null +++ b/Concepts_Of_Modern_Physics/Chapter_6.ipynb @@ -0,0 +1,62 @@ +{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chaptet 6:Quantum theory of the Hydrogen Atom"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.4,Page no:226"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "B= 0.300 #magnetic field, T\n",
+ "Lambda= 450 #wavelength, nm\n",
+ "Lambda= Lambda*(10**(-9)) #converting to m\n",
+ "e= 1.6*(10**(-19)) #charge of an electron, C\n",
+ "Me= 9.1*(10**(-31)) #mass of electron, kg\n",
+ "c= 3 *(10**8) #speed of light, m/s\n",
+ "import math\n",
+ "#Calculation\n",
+ "dLambda= e*B*(Lambda**2)/(4*(math.pi)*Me*c) #m\n",
+ "dLambda= dLambda*(10**9) #converting to nm\n",
+ "\n",
+ "#Result\n",
+ "print\"The separation between Zeeman components is: \",round(dLambda,5),\"nm\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The separation between Zeeman components is: 0.00283 nm\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Concepts_Of_Modern_Physics/Chapter_7.ipynb b/Concepts_Of_Modern_Physics/Chapter_7.ipynb new file mode 100755 index 00000000..54659b71 --- /dev/null +++ b/Concepts_Of_Modern_Physics/Chapter_7.ipynb @@ -0,0 +1,184 @@ +{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 7:Many electron Atoms"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.1,Page no:230"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration \n",
+ "r= 5*(10**(-17)) #radius of spherical electron, m\n",
+ "Me= 9.1*(10**(-31)) #mass of electron, kg\n",
+ "h= 6.63*(10**(-34)) #Planck's constant, J.s\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "hbar= h/(2*(math.pi)) #reduced Planck's constant, J.s\n",
+ "v= (5*math.sqrt(3)/4)*(hbar/(Me*r)) #using Eqn 7.1, Page 230\n",
+ "c= 3*(10**8) #velocity of light, m/s\n",
+ "v= v/c #converting in terms of c, m/s\n",
+ "\n",
+ "#Result\n",
+ "print\"The velocity of electron in times of c is:%.3g\"%v,\"c\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The velocity of electron in times of c is:1.67e+04 c\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.2,Page no:241"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration \n",
+ "n= 2 #outer (2s) orbit of lithium\n",
+ "E2= -5.39 #Ionisation energy of lithium, for n=2 eV\n",
+ "E1= -13.6 #for n=1, eV\n",
+ "\n",
+ "#Calculation\n",
+ "Z= n*(math.sqrt(E2/E1)) #modification factor for effective charge\n",
+ "e= 1.6*(10**(-19)) #charge of an electron, C\n",
+ "Ceffective = Z*e \n",
+ " \n",
+ "#Result \n",
+ "print\"The effective charge is: \",round(Ceffective/e,2),\"e or%.3g\"%Ceffective,\"C\"\n",
+ " \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The effective charge is: 1.26 e or2.01e-19 C\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.3,Page no:248"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration \n",
+ "n= 2 #for 2p state\n",
+ "Ao= 5.29*(10**(-11)) #Bohr's orbit for n=1, m\n",
+ "r= (n**2)*Ao #orbital radius, m\n",
+ "f= 8.4*(10**14) #frequency of revolution, Hz ,using Eqn 4.4\n",
+ "\n",
+ "#Calculation\n",
+ "Mo= 4*(math.pi)*(10**(-7)) #Magnetic constant, T.m/A\n",
+ "e= 1.6*(10**(-19)) #charge of an electron, C\n",
+ "B= (Mo*f*e)/(2*r) #Magnetic field, T\n",
+ "Mb= 9.27*(10**(-24)) #Bohr Magneton, J/T\n",
+ "Um= Mb*B #Magnetic energy, J\n",
+ "Um= Um/e #converting to eV\n",
+ "\n",
+ "#Result\n",
+ "print\"The magnetic energy for electron is:%.2g\"%Um,\"eV\"\n",
+ "print\"\\nThe energy difference is twice this,which is:%.2g\"%(2*Um),\"eV\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The magnetic energy for electron is:2.3e-05 eV\n",
+ "\n",
+ "The energy difference is twice this,which is:4.6e-05 eV\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.8,Page no:257"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration \n",
+ "l= 0.180 #wavelength, nm\n",
+ "l= l* 10**(-9) #converting to m\n",
+ "c= 3*(10**8) #velocity of light, m/s\n",
+ "\n",
+ "#Calculation\n",
+ "f= c/l #frequency, Hz\n",
+ "R= 1.097*(10**7) #Rydberg's constant, per m\n",
+ "Z= 1+(math.sqrt((4*f)/(3*c*R))) #using Eqn 7.21\n",
+ "\n",
+ "#Result\n",
+ "print\"The element has atomic number: \",round(Z)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The element has atomic number: 27.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Concepts_Of_Modern_Physics/Chapter_8.ipynb b/Concepts_Of_Modern_Physics/Chapter_8.ipynb new file mode 100755 index 00000000..55544617 --- /dev/null +++ b/Concepts_Of_Modern_Physics/Chapter_8.ipynb @@ -0,0 +1,163 @@ +{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 8:Molecules"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Exampnle no:8.1,Page no:283"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration \n",
+ "r= 0.113 #bond length, nm\n",
+ "Mc= 1.99*(10**(-26)) #mass of C12, kg\n",
+ "Mo= 2.66*(10**(-26)) #mass of O16, kg\n",
+ "\n",
+ "#Calculation\n",
+ "#Part (a)\n",
+ "import math\n",
+ "Mco= (Mc*Mo)/(Mc+Mo) #mass of CO, kg\n",
+ "I= Mco*((r*(10.0**(-9.0)))**2.0) #moment of inertia, kg.m**2\n",
+ "J=1.0 #lowest rotational state\n",
+ "h= 6.63*(10**(-34)) #Planck's constant, J.s\n",
+ "hbar= h/(2.0*(math.pi)) #reduced Planck's constant, J.s\n",
+ "E1= (J*(J+1)*(hbar**2))/(2.0*I) #energy corresponding to state J=1, J\n",
+ "e= 1.6*(10**(-19)) #charge of an electron, C\n",
+ "E1= E1/e #converting to eV\n",
+ "#Part(b)\n",
+ "w= math.sqrt((2.0*7.61*10**-23)/(1.46*10**-46)) #angular velocity, rad/s\n",
+ "\n",
+ "#Result\n",
+ "print\"(a).The energy of CO molecule is:%.2e\"%E1,\"eV(approx)\"\n",
+ "print\"(b).The angular velocity is:%.3g\"%w,\"rad/sec\\n\"\n",
+ "\n",
+ "print\"NOTE:(Calculation mistake):Incorrect answer in book for (b) part\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a).The energy of CO molecule is:4.79e-04 eV(approx)\n",
+ "(b).The angular velocity is:1.02e+12 rad/sec\n",
+ "\n",
+ "NOTE:(Calculation mistake):Incorrect answer in book for (b) part\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Exampnle no:8.2,Page no:285"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "Ji=0 #initial state\n",
+ "Jf=1 #final state\n",
+ "f= 1.15*(10**11) #frequency for the absorption, Hz\n",
+ "h= 6.63*(10**(-34)) #Planck's constant, J.s\n",
+ "\n",
+ "#Calculation\n",
+ "hbar= h/(2*(math.pi)) #reduced Planck's constant, J.s\n",
+ "Ico= hbar*Jf/(2*(math.pi)*f) #moment of inertia, kg.m**2\n",
+ "Mco= 1.14*(10**(-26)) #Mass of CO, refer Exa 8.1\n",
+ "r= math.sqrt(Ico/Mco) #bond length, m\n",
+ "r= r*(10**9) #converting to nm\n",
+ "\n",
+ "#Result\n",
+ "print\"The bond length of CO molecule is: \",round(r,3),\"nm\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The bond length of CO molecule is: 0.113 nm\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Exampnle no:8.3,Page no:288"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "f= 6.42*(10**13) #frequency of absorbed radiation, Hz\n",
+ "Mco= 1.14*(10**(-26)) #mass of CO, kg\n",
+ "h= 6.63*(10**(-34)) #Planck's constant, J.s\n",
+ "\n",
+ "#Calculation\n",
+ "#Part (a)\n",
+ "k= 4*((math.pi)**2)*(f**2)*Mco #using Eqn 8.15, Page 287\n",
+ "#Part (b)\n",
+ "dE= h*f #separation, J\n",
+ "etoJ=dE*6.24*10**18 \n",
+ "\n",
+ "#Result\n",
+ "print\"(a).The force constant for the bond in CO molecule is: %.3g\"%k,\"N/m\"\n",
+ "print\"\\nNOTE:Slight error in calculation of book,'k' is written as 1.86*10**3\\n\\n\"\n",
+ "print\"(b).The separation in its vibrational eergy levels is:%.3g\"%dE,\"J=\",round(etoJ,3),\"eV\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a).The force constant for the bond in CO molecule is: 1.85e+03 N/m\n",
+ "\n",
+ "NOTE:Slight error in calculation of book,'k' is written as 1.86*10**3\n",
+ "\n",
+ "\n",
+ "(b).The separation in its vibrational eergy levels is:4.26e-20 J= 0.266 eV\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Concepts_Of_Modern_Physics/Chapter_9.ipynb b/Concepts_Of_Modern_Physics/Chapter_9.ipynb new file mode 100755 index 00000000..bfd0fdbb --- /dev/null +++ b/Concepts_Of_Modern_Physics/Chapter_9.ipynb @@ -0,0 +1,271 @@ +{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 9:Statistical Mechanics"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:9.1,Page no:299"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Variable declaration\n",
+ "k= 8.617*10**(-5) #Boltzmann constant, eV/K\n",
+ "To=273.0 #initial temperature, K\n",
+ "E1= -13.6 #energy of ground state, eV\n",
+ "E2= -3.4 #energy of first excited state, eV\n",
+ "dE= E2-E1 #difference in energy levels\n",
+ "g1=2.0 #number of energy states for E1\n",
+ "g2=8.0 #number of energy states for E2\n",
+ "\n",
+ "#Calculation\n",
+ "J= dE/(k*To) \n",
+ "Nratio1= (g2/g1)*math.exp(-J) #ratio of number of atoms in level 2 and level 1 at To\n",
+ "T1=10273.0 #K\n",
+ "J1= J*To/T1 \n",
+ "Nratio2= (g2/g1)*math.exp(-J1) #at T1\n",
+ "\n",
+ "#Result\n",
+ "print\"(a).The ratio at 273 K is:%.3g\"%Nratio1,\"(Approx)\"\n",
+ "print\"(b).The ratio at 10273 k is:%.g \"%Nratio2"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a).The ratio at 273 K is:1.97e-188 (Approx)\n",
+ "(b).The ratio at 10273 k is:4e-05 \n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:9.4,Page no:305"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration \n",
+ "Moxygen= 16.0 #atomic mass,u\n",
+ "Mo2= 32.0 #Molecular mass, u\n",
+ "u= 1.66*(10**(-27)) #atomic mass unit, kg\n",
+ "Moxygen= Mo2*u #mass, kg\n",
+ "t= 273 #temperature, K\n",
+ "k= 1.38*10**(-23) #Boltzmann constant, J/K\n",
+ "\n",
+ "#Calculation\n",
+ "Vrms= math.sqrt(3*k*t/Moxygen) # m/s\n",
+ "\n",
+ "#Result\n",
+ "print\"The rms velocity of oxygen is: \",round(Vrms),\"m/s\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The rms velocity of oxygen is: 461.0 m/s\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:9.5,Page no:314"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration \n",
+ "V= 1.00 #volume, cm**3\n",
+ "V= V*10**(-6) #converting to m**3\n",
+ "dI= 2.404 #standard value of definite Integral used\n",
+ "k= 8.617*10**(-5) #Boltzmann constant, eV/K\n",
+ "h= 4.13*(10**(-15)) #Planck's constant, eV.s\n",
+ "T= 1000 #temperature, K\n",
+ "c= 3 *(10**8) #speed of light, m/s\n",
+ "\n",
+ "#Calculation\n",
+ "#Part (a)\n",
+ "N= 8*(math.pi)*V*dI*((k*T/(h*c))**3)\n",
+ "#Part(b)\n",
+ "Sig= 5.670*10**(-8) #Stefan's constant, W/m**2.K**4 , refer to Page 317\n",
+ "Ephoton= Sig*(c**2)*(h**3)*T/(2.405*(2*(math.pi)*(k**3))) #J\n",
+ "e_to_J=6.23*10**18*Ephoton #Converting to eV \n",
+ "\n",
+ "#Result\n",
+ "print\"(a),The number of photons is:%.3g\"%N\n",
+ "print\"(b)The average energy of the photons is:%.3g\"%Ephoton,\"J=\",round(e_to_J,3),\"eV\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a),The number of photons is:2.03e+10\n",
+ "(b)The average energy of the photons is:3.72e-20 J= 0.232 eV\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:9.6,Page no:317"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration \n",
+ "T= 2.7 #blackbody temperature, K\n",
+ "Lambda= 2.898*10**(-3)/T #using wein's displacement law, Eqn 9.40, m\n",
+ "\n",
+ "#Calculation\n",
+ "Lambda= Lambda*10**(3) #converting to mm\n",
+ "\n",
+ "#Result\n",
+ "print\"The wavelength for maximum radiation is: \",Lambda,\"mm\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The wavelength for maximum radiation is: 1.07333333333 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:9.7,Page no:317"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration \n",
+ "Rearth= 1.5*10**(11) #radius of earth, m\n",
+ "r= 1.4 #rate of arrival of sunlight, kW/m**2\n",
+ "\n",
+ "#Calculation\n",
+ "P= (r*10**3)*4*(math.pi)*(Rearth**2) #total power reaching Earth\n",
+ "Rsun= 7*10**(8) #radius of Sun, m\n",
+ "r2= P/(4*(math.pi)*(Rsun**2)) #radiation rate of Sun, W/m**2\n",
+ "emissivity=1 #for blackbody\n",
+ "Sig= 5.670*10**(-8) #Stefan's constant, W/m**2.K**4\n",
+ "T= (r2/(emissivity*Sig))**(1.0/4.0) \n",
+ "\n",
+ "#Result\n",
+ "print\"The surface temperature of Sun is:%.2g\"%T,\"K\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The surface temperature of Sun is:5.8e+03 K\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:9.8,Page no:325"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "u= 1.66*(10**(-27)) #atomic mass unit, kg\n",
+ "density= 8.94*10**(3) # kg/m**3\n",
+ "M= 63.5 #atomic mass of copper, u\n",
+ "\n",
+ "#Calculation\n",
+ "Edensity= density/(M*u) #electron density, electrons/m**3\n",
+ "h= 6.63*(10**(-34)) #Planck's constant, J.s\n",
+ "Me= 9.1*(10**(-31)) #mass of electron, kg\n",
+ "Efermi= h**2/(2*Me)*((3*Edensity)/(8*(math.pi)))**(2.0/3.0) # J\n",
+ "e_to_J=6.23*10**18*Efermi #Converting to eV \n",
+ "\n",
+ "#Result\n",
+ "print\"The fermi energy is:%.3g\"%Efermi,\"J or\",round(e_to_J,2),\"eV\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The fermi energy is:1.13e-18 J or 7.04 eV\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Concepts_Of_Modern_Physics/README.txt b/Concepts_Of_Modern_Physics/README.txt new file mode 100755 index 00000000..230b914c --- /dev/null +++ b/Concepts_Of_Modern_Physics/README.txt @@ -0,0 +1,10 @@ +Contributed By: Anshul Chauhan +Course: btech +College/Institute/Organization: National institute of foundary and forge technology,Ranchi +Department/Designation: Manufacturing Engineering +Book Title: Concepts Of Modern Physics +Author: Arthur Beiser +Publisher: Tata McGraw - Hill Education +Year of publication: 2006 +Isbn: 0072448482 +Edition: 6
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