From b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b Mon Sep 17 00:00:00 2001 From: priyanka Date: Wed, 24 Jun 2015 15:03:17 +0530 Subject: initial commit / add all books --- 2912/CH1/EX1.3/Ex1_3.sce | 17 +++++++++++++++++ 2912/CH1/EX1.4/Ex1_4.sce | 25 +++++++++++++++++++++++++ 2912/CH1/EX1.5/Ex1_5.sce | 18 ++++++++++++++++++ 2912/CH1/EX1.6/Ex1_6.sce | 22 ++++++++++++++++++++++ 2912/CH1/EX1.7/Ex1_7.sce | 21 +++++++++++++++++++++ 2912/CH10/EX10.1/Ex10_1.sce | 13 +++++++++++++ 2912/CH10/EX10.2/Ex10_2.sce | 13 +++++++++++++ 2912/CH10/EX10.3/Ex10_3.sce | 17 +++++++++++++++++ 2912/CH10/EX10.4/Ex10_4.sce | 16 ++++++++++++++++ 2912/CH10/EX10.5/Ex10_5.sce | 19 +++++++++++++++++++ 2912/CH10/EX10.6/Ex10_6.sce | 15 +++++++++++++++ 2912/CH10/EX10.7/Ex10_7.sce | 16 ++++++++++++++++ 2912/CH12/EX12.1/Ex12_1.sce | 14 ++++++++++++++ 2912/CH12/EX12.2/Ex12_2.sce | 15 +++++++++++++++ 2912/CH12/EX12.3/Ex12_3.sce | 16 ++++++++++++++++ 2912/CH12/EX12.4/Ex12_4.sce | 16 ++++++++++++++++ 2912/CH12/EX12.5/Ex12_5.sce | 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create mode 100755 2912/CH9/EX9.10/Ex9_10.sce create mode 100755 2912/CH9/EX9.11/Ex9_11.sce create mode 100755 2912/CH9/EX9.12/Ex9_12.sce create mode 100755 2912/CH9/EX9.13/Ex9_13.sce create mode 100755 2912/CH9/EX9.14/Ex9_14.sce create mode 100755 2912/CH9/EX9.15/Ex9_15.sce create mode 100755 2912/CH9/EX9.16/Ex9_16.sce create mode 100755 2912/CH9/EX9.17/Ex9_17.sce create mode 100755 2912/CH9/EX9.2/Ex9_2.sce create mode 100755 2912/CH9/EX9.3/Ex9_3.sce create mode 100755 2912/CH9/EX9.4/Ex9_4.sce create mode 100755 2912/CH9/EX9.5/Ex9_5.sce create mode 100755 2912/CH9/EX9.6/Ex9_6.sce create mode 100755 2912/CH9/EX9.7/Ex9_7.sce create mode 100755 2912/CH9/EX9.8/Ex9_8.sce create mode 100755 2912/CH9/EX9.9/Ex9_9.sce (limited to '2912') diff --git a/2912/CH1/EX1.3/Ex1_3.sce b/2912/CH1/EX1.3/Ex1_3.sce new file mode 100755 index 000000000..9e7da8707 --- /dev/null +++ b/2912/CH1/EX1.3/Ex1_3.sce @@ -0,0 +1,17 @@ +//chapter 1 +//example 1.3 +//calculate potential energy +//page 15 +clear; +clc; +//given +r=2; //in angstrom(distance) +e=1.6E-19; // in C (charge of electron) +E_o= 8.85E-12;// absolute premittivity +//calculate +r=2*1E-10; // since r is in angstrom +V=-e^2/(4*%pi*E_o*r); // calculate potential +printf('\nThe potential energy is \tV=%3.3E J',V); +V=V/e; // changing to eV +printf('\nIn electron-Volt,\tV=%.2f eV',V); +// Note: the answer in the book is wrong due to calculation mistake diff --git a/2912/CH1/EX1.4/Ex1_4.sce b/2912/CH1/EX1.4/Ex1_4.sce new file mode 100755 index 000000000..a429ad31c --- /dev/null +++ b/2912/CH1/EX1.4/Ex1_4.sce @@ -0,0 +1,25 @@ +//chapter 1 +//example 1.4 +//calculate bond energy for NaCl +//page 15-16 +clear; +clc; +//given +r0=0.236; //in nanometer(interionic distance) +e=1.6E-19; // in C (charge of electron) +E_o= 8.85E-12;// absolute premittivity +N=8; // Born constant +IE=5.14;// in eV (ionisation energy of sodium) +EA=3.65;// in eV (electron affinity of Chlorine) +pi=3.14; // value of pi used in the solution +//calculate +r0=r0*1E-9; // since r is in nanometer +PE=(e^2/(4*pi*E_o*r0))*(1-1/N); // calculate potential energy +PE=PE/e; //changing unit from J to eV +printf('\nThe potential energy is\tPE=%.2f eV',PE); +NE=IE-EA;// calculation of Net energy +printf('\nThe net energy is\tNE=%.2f eV',NE); +BE=PE-NE;// calculation of Bond Energy +printf('\nThe bond energy is\tBE=%.2f eV',BE); +// Note: (1)-In order to make the answer prcatically feasible and avoid the unusual answer, I have used r_0=0.236 nm instead of 236 nm. because using this value will give very much irrelevant answer. +// (2) There is slight variation in the answer due to round off. diff --git a/2912/CH1/EX1.5/Ex1_5.sce b/2912/CH1/EX1.5/Ex1_5.sce new file mode 100755 index 000000000..2d54215ba --- /dev/null +++ b/2912/CH1/EX1.5/Ex1_5.sce @@ -0,0 +1,18 @@ +//chapter 1 +//example 1.5 +//calculate compressibility +//page 16 +clear; +clc; +//given +r_0=.41; //in mm(lattice constant) +e=1.6E-19; // in C (charge of electron) +E_o= 8.85E-12;// absolute premittivity +n=0.5; // repulsive exponent value +alpha=1.76; // Madelung constant +pi=3.14; // value of pi used in the solution +//calculate +r=.41*1E-3; // since r is in mm +Beta=72*pi*E_o*r^4/(alpha*e^2*(n-1)); // calculation compressibility +printf('\nThe compressibility is\tBeta=%1.2E ',Beta); +// Note: the answer in the book is wrong due to calculation mistake diff --git a/2912/CH1/EX1.6/Ex1_6.sce b/2912/CH1/EX1.6/Ex1_6.sce new file mode 100755 index 000000000..d0e94165b --- /dev/null +++ b/2912/CH1/EX1.6/Ex1_6.sce @@ -0,0 +1,22 @@ +//chapter 1 +//example 1.6 +//calculate ionic cohesive energy and atomic cohesive energy +//page 16 +clear; +clc; +//given +r_0=3.56; // in Angstrom +e=1.6E-19; // in C (charge of electron) +IE=3.89; //in eV (ionisation energy of Cs) +EA=-3.61; // in eV (electron affinity of Cl) +n=10.5; // Born constant +E_o= 8.85E-12;// absolute premittivity +alpha=1.763; // Madelung constant +pi=3.14; // value of pi used in the solution +//calculate +r_0=r_0*1E-10; // since r is in nanometer +U=-alpha*(e^2/(4*pi*E_o*r_0))*(1-1/n); // calculate potential energy +U=U/e; //changing unit from J to eV +printf('\nThe ionic cohesive energy is\t%.2f eV',U); +ACE=U+EA+IE; // calculation of atomic cohesive energy +printf('\nThe atomic cohesive energy is\t%.2f eV',ACE); diff --git a/2912/CH1/EX1.7/Ex1_7.sce b/2912/CH1/EX1.7/Ex1_7.sce new file mode 100755 index 000000000..5cdc7ba52 --- /dev/null +++ b/2912/CH1/EX1.7/Ex1_7.sce @@ -0,0 +1,21 @@ +//chapter 1 +//example 1.7 +//calculate contribution per ions to the cohesive energy +//page 17 +clear; +clc; +//given +r_0=2.81; // in Angstrom +e=1.6E-19; // in C (charge of electron) +n=9; // Born constant +E_o= 8.85E-12;// absolute premittivity +alpha=1.748; // Madelung constant +pi=3.14; // value of pi used in the solution +//calculate +r_0=r_0*1E-10; // since r is in nanometer +V=-alpha*(e^2/(4*pi*E_o*r_0))*(1-1/n); // calculate potential energy +V=V/e; //changing unit from J to eV +printf('\nThe potential energy is\tV=%.2f eV',V); +V_1=V/2; // Since only half of the energy contribute per ion to the cohecive energy therfore +printf('\nThe energy contributing per ions to the cohesive energy is \t%.2f eV',V_1); + // Note: Answer in the book is wroong due to calculation mistake diff --git a/2912/CH10/EX10.1/Ex10_1.sce b/2912/CH10/EX10.1/Ex10_1.sce new file mode 100755 index 000000000..38c47526b --- /dev/null +++ b/2912/CH10/EX10.1/Ex10_1.sce @@ -0,0 +1,13 @@ +//chapter 10 +//example 10.1 +//Calculate magnitude of critical magnetic field +//page 313 +clear; +clc; +//given +Tc=7.2; // in K (critical temperature) +T=5; // in K (given temperature) +H0=6.5E3; // in A/m (critical magnetic field at 0K) +//calculate +Hc=H0*(1-(T/Tc)^2); // calculation of magnitude of critical magnetic field +printf('\nThe magnitude of critical magnetic field is \tHc=%1.3E A/m',Hc); diff --git a/2912/CH10/EX10.2/Ex10_2.sce b/2912/CH10/EX10.2/Ex10_2.sce new file mode 100755 index 000000000..2b205a5d1 --- /dev/null +++ b/2912/CH10/EX10.2/Ex10_2.sce @@ -0,0 +1,13 @@ +//chapter 10 +//example 10.2 +//Calculate critical current value +//page 313 +clear; +clc; +//given +r=0.02; // in m (radius of ring) +Hc=2E3; // in A/m (critical magnetic field at 5K) +pi=3.14; // value of pi used in the solutiion +//calculate +Ic=2*pi*r*Hc; // calculation of critical current value +printf('\nThe critical current value is \tIc=%.1f A',Ic); diff --git a/2912/CH10/EX10.3/Ex10_3.sce b/2912/CH10/EX10.3/Ex10_3.sce new file mode 100755 index 000000000..62fe79240 --- /dev/null +++ b/2912/CH10/EX10.3/Ex10_3.sce @@ -0,0 +1,17 @@ +// chapter 10 +// example 10.3 +// calculate isotropic mass at 5.1K +// page 313 +clear; +clc; +// given +M1=199.5; // in amu (isotropic mass at 5K) +T1=5; // in K (first critical temperature) +T2=5.1; // in K (second critical temperature) +//calculate +// since Tc=C*(1/sqrt(M) +// therefore T1*sqrt(M1)=T2*sqrt(M2) +// therefore we have M2=(T1/T2)^2*M1 +M2=(T1/T2)^2*M1; //calculation of isotropic mass at 5.1K +printf('\nThe isotropic mass at 5.1K is \t M2=%.3f a.m.u.',M2); + diff --git a/2912/CH10/EX10.4/Ex10_4.sce b/2912/CH10/EX10.4/Ex10_4.sce new file mode 100755 index 000000000..b9bde7946 --- /dev/null +++ b/2912/CH10/EX10.4/Ex10_4.sce @@ -0,0 +1,16 @@ +// chapter 10 +// example 10.4 +// calculate transition temperature +// page 314 +// given +clear; +clc; +T=6; // in K (given temperature) +Hc=5E3; // in A/m (critical magnetic field at 5K) +H0=2E4; // in A/m (critical magnetic field at 0K) +//calculate +// since Hc=H0*(1-(T/Tc)^2) +// therefor we have Tc=T/sqrt(1-(Hc/H)^2) +Tc=T/sqrt(1-(Hc/H0)); // calculation of transition temperature +printf('\nThe transition temperature is \tTc=%.3f K',Tc); +// Note: answer in the book is wrong due to calculation mistake diff --git a/2912/CH10/EX10.5/Ex10_5.sce b/2912/CH10/EX10.5/Ex10_5.sce new file mode 100755 index 000000000..5a9fc5f64 --- /dev/null +++ b/2912/CH10/EX10.5/Ex10_5.sce @@ -0,0 +1,19 @@ +// chapter 10 +// example 10.5 +// calculate critical current at 5K +// page 314 +// given +clear; +clc; +T=5; // in K (given temperature) +d=3; // in mm (diameter of the wire) +Tc=8; // in K (critical temperature for Pb) +H0=5E4; // in A/m (critical magnetic field at 0K) +pi=3.14; // value of pi used in the solution +//calculate + Hc=H0*(1-(T/Tc)^2); // calculation of critical magnetic field at 5K +printf('\nThe critical magnetic field at 5K is \tHc=%1.3E A/m',Hc); +r=(d*1E-3)/2; // calculation of radius in m +Ic=2*pi*r*Hc; // calculation of critical current at 5K +printf('\nThe critical current at 5K is \tIc=%.4f A',Ic); +//Note: there is slight variation in the answer due to round off diff --git a/2912/CH10/EX10.6/Ex10_6.sce b/2912/CH10/EX10.6/Ex10_6.sce new file mode 100755 index 000000000..97ab27deb --- /dev/null +++ b/2912/CH10/EX10.6/Ex10_6.sce @@ -0,0 +1,15 @@ +// chapter 10 +// example 10.6 +// calculate frequency of EM waves +// page 314 +clear; +clc; +// given +V=8.50; // in micro V (voltage across Josephson junction ) +e=1.6E-19; // in C (charge of electron) +h=6.626E-34; // in J/s (Planck’s constant) +//calculate +V=V*1E-6; // changing unit from V to microVolt + v=2*e*V/h; // calculation of frequency of EM waves +printf('\nThe frequency of EM waves is \tv=%1.3E Hz',v); +// Note: the answer in the book is wrong due to calculation misatke diff --git a/2912/CH10/EX10.7/Ex10_7.sce b/2912/CH10/EX10.7/Ex10_7.sce new file mode 100755 index 000000000..3b2d1861a --- /dev/null +++ b/2912/CH10/EX10.7/Ex10_7.sce @@ -0,0 +1,16 @@ +// chapter 10 +// example 10.7 +// calculate transition temperature of the isotopes +// page 315 +clear; +clc;// given +M1=200.59; // in amu (average atomic mass at 4.153K) +Tc1=4.153; // in K (first critical temperature) +M2=204; // in amu (average atomic mass of isotopes) +//calculate +// since Tc=C*(1/sqrt(M) +// therefore T1*sqrt(M1)=T2*sqrt(M2) +// therefore we have Tc2=Tc1*sqrt(M1/M2) +Tc2=Tc1*sqrt(M1/M2); //calculation of transition temperature of the isotopes +printf('\nThe transition temperature of the isotopes is \t Tc2=%.3f K',Tc2); + diff --git a/2912/CH12/EX12.1/Ex12_1.sce b/2912/CH12/EX12.1/Ex12_1.sce new file mode 100755 index 000000000..6d49d949f --- /dev/null +++ b/2912/CH12/EX12.1/Ex12_1.sce @@ -0,0 +1,14 @@ +// chapter 12 +// example 12.1 +// calculate fractional index change for a given optical fibre +// page 360 +clear; +clc; +// given +u1=1.563; // refractive index of core +u2=1.498; // refractive index of cladding +//calculate +d=(u1-u2)/u1; // calculation of fractional index change +printf('\nThe fractional index change for a given optical fibre is %.4f',d); + + diff --git a/2912/CH12/EX12.2/Ex12_2.sce b/2912/CH12/EX12.2/Ex12_2.sce new file mode 100755 index 000000000..ea82adba8 --- /dev/null +++ b/2912/CH12/EX12.2/Ex12_2.sce @@ -0,0 +1,15 @@ +// chapter 12 +// example 12.2 +// calculate numerical aperture and the acceptance angle of an optical fibre +// page 360 +clear; +clc; +// given +u1=1.55; // refractive index of core +u2=1.50; // refractive index of cladding +//calculate +d=(u1-u2)/u1; // calculation of fractional index change +NA=u1*sqrt(2*d); // calculation of numerical aperture +printf('\nThe numerical aperture of the fibre is \tNA=%.3f',NA); +theta=asind(NA); // calculation of acceptance angle +printf('\nThe acceptance angle of the optical fibre is \t%.1f degree',theta); diff --git a/2912/CH12/EX12.3/Ex12_3.sce b/2912/CH12/EX12.3/Ex12_3.sce new file mode 100755 index 000000000..e47173830 --- /dev/null +++ b/2912/CH12/EX12.3/Ex12_3.sce @@ -0,0 +1,16 @@ +// chapter 12 +// example 12.3 +// calculate the acceptance angle of an optical fibre +// page 360 +// given +clear; +clc; +u1=1.563; // refractive index of core +u2=1.498; // refractive index of cladding +//calculate +NA=sqrt(u1^2-u2^2); // calculation of numerical aperture +printf('\nThe numerical aperture of the fibre is \tNA=%.4f',NA); +theta=asind(NA); // calculation of acceptance angle +printf('\nThe acceptance angle of the optical fibre is \t%.2f degree',theta); + + diff --git a/2912/CH12/EX12.4/Ex12_4.sce b/2912/CH12/EX12.4/Ex12_4.sce new file mode 100755 index 000000000..46864c15c --- /dev/null +++ b/2912/CH12/EX12.4/Ex12_4.sce @@ -0,0 +1,16 @@ +// chapter 12 +// example 12.4 +// calculate refractive index of material of the core +// page 360-361 +clear; +clc; +// given +NA=0.39; //numerical aperture of the optical fibre +d=0.05; // difference in the refractive index of the material of the core and cladding +//calculate +// since NA=u1*sqrt(2*d) +//we have u1=NA/sqrt(2*d) +u1= NA/sqrt(2*d); // calculation of refractive index of material of the core +printf('\nThe refractive index of material of the core is \tu1=%.3f',u1); + + diff --git a/2912/CH12/EX12.5/Ex12_5.sce b/2912/CH12/EX12.5/Ex12_5.sce new file mode 100755 index 000000000..efa7710a3 --- /dev/null +++ b/2912/CH12/EX12.5/Ex12_5.sce @@ -0,0 +1,18 @@ +// chapter 12 +// example 12.5 +// calculate numerical aperture,acceptance angle and the critical angle of the optical fibre +// page 361 +clear; +clc; +// given +u1=1.50; // refractive index of core +u2=1.45; // refractive index of cladding +//calculate +d=(u1-u2)/u1; // calculation of fractional index change +NA=u1*sqrt(2*d); // calculation of numerical aperture +printf('\nThe numerical aperture of the fibre is \tNA=%.3f',NA); +theta_0=asind(NA); // calculation of acceptance angle +printf('\nThe acceptance angle of the optical fibre is \t%.2f degree',theta_0); +theta_c=asind(u2/u1); // calculation of critical angle +printf('\nThe critical angle of the optical fibre is \t%.1f degree',theta_c); + diff --git a/2912/CH12/EX12.6/Ex12_6.sce b/2912/CH12/EX12.6/Ex12_6.sce new file mode 100755 index 000000000..9f7f21206 --- /dev/null +++ b/2912/CH12/EX12.6/Ex12_6.sce @@ -0,0 +1,19 @@ +// chapter 12 +// example 12.6 +// calculate refractive index of the core and cladding material of a fibre +// page 361 +clear; +clc; +// given +NA=0.33; // numerical aperture +d=0.02; // difference in the refractive index of the core and cladding of the material +//calculate +// since NA=u1*sqrt(2*d) +// therefore we have +u1=NA/sqrt(2*d); // calculation of refractive index of the core +// since d=(u1-u2)/u2 +// therefore we have +u2=(1-d)*u1; // calculation of refractive index of the cladding +printf('\nThe refractive index of the core is \tu1=%.1f',u1); +printf('\nThe refractive index of the cladding is \tu2=%.3f',u2); +// Note: In the question, it is given that NA=0.33 but in the book NA=0.22 has been used in the solution. That's why answer in the book is different from that of generated from the code diff --git a/2912/CH12/EX12.7/Ex12_7.sce b/2912/CH12/EX12.7/Ex12_7.sce new file mode 100755 index 000000000..1bb86f312 --- /dev/null +++ b/2912/CH12/EX12.7/Ex12_7.sce @@ -0,0 +1,17 @@ +// chapter 12 +// example 12.7 +// calculate numerical aperture and acceptance angle of the symmetrical fibre +// page 361 +clear; +clc; +// given +u1=3.5; // refractive index of core +u2=3.45; // refractive index of cladding +u0=1; // refractive index of the air +//calculate +NA=sqrt(u1^2-u2^2); // calculation of numerical aperture +NA=NA/u0; +printf('\nThe numerical aperture of the fibre is \tNA=%.2f',NA); +alpha=asind(NA); // calculation of acceptance angle +printf('\nThe acceptance angle of the optical fibre is \t%.2f degre',alpha); + diff --git a/2912/CH12/EX12.8/Ex12_8.sce b/2912/CH12/EX12.8/Ex12_8.sce new file mode 100755 index 000000000..5b74a619d --- /dev/null +++ b/2912/CH12/EX12.8/Ex12_8.sce @@ -0,0 +1,15 @@ +// chapter 12 +// example 12.8 +// calculate numerical aperture and acceptance angle of an optical fibre +// page 361-362 +clear; +clc; +// given +u1=1.48; // refractive index of core +u2=1.45; // refractive index of cladding +//calculate +NA=sqrt(u1^2-u2^2); // calculation of numerical aperture +printf('\nThe numerical aperture of the fibre is \tNA=%.3f',NA); +theta=asind(NA); // calculation of acceptance angle +printf('\nThe acceptance angle of the optical fibre is \t%.2f degree',theta); +// Note: there is slight variation in the answer due to round off diff --git a/2912/CH2/EX2.1/Ex2_1.sce b/2912/CH2/EX2.1/Ex2_1.sce new file mode 100755 index 000000000..674128ba4 --- /dev/null +++ b/2912/CH2/EX2.1/Ex2_1.sce @@ -0,0 +1,16 @@ +//chapter 2 +//example 2.1 +//calculate lattice constant +//page 40-41 +clear; +clc; +//given +N=6.02E26; // in /Kg-molecule (Avogadro's number) +n=4; // number of molecules per unit cell ofr NaCl +M=58.5; // in Kg/Kg-molecule (molecular weight of NaCl) +p=2189; // in Kg/m^3 (density) +//calculate +a=nthroot((n*M/(N*p)),3); +printf('\nThe lattice constant is\ta=%1.2E m',a); +a=a*1E10; // changing unit to Angstrom +printf('\n\t\t\ta=%.2f Angstrom',a); diff --git a/2912/CH2/EX2.2/Ex2_2.sce b/2912/CH2/EX2.2/Ex2_2.sce new file mode 100755 index 000000000..4c1ace979 --- /dev/null +++ b/2912/CH2/EX2.2/Ex2_2.sce @@ -0,0 +1,18 @@ +//chapter 2 +//example 2.2 +//calculate distance between two nearest Cu atoms +//page 41 +clear; +clc; +//given +N=6.02E23; // in /gram-atom (Avogadro's number) +n=4; // number of atom per unit cell for fcc structure +M=63.5; //in gram/gram-atom (atomic weight of Cu) +p=8.96; // in g/cm^3 (density) +//calculate +a=nthroot((n*M/(N*p)),3); +printf('\nThe lattice constant is\ta=%1.2E cm',a); +a=a*1E8; // changing unit from cm to Angstrom +printf('\n\t\t\ta=%.2f Angstrom',a); +d=a/sqrt(2); // distance infcc lattice +printf('\nThe distance between two nearest Cu atoms is \td=%.2f Angstrom',d); diff --git a/2912/CH2/EX2.3/Ex2_3.sce b/2912/CH2/EX2.3/Ex2_3.sce new file mode 100755 index 000000000..0d8d5de9c --- /dev/null +++ b/2912/CH2/EX2.3/Ex2_3.sce @@ -0,0 +1,16 @@ +//chapter 2 +//example 2.3 +//calculate lattice constant +//page 41-42 +clear; +clc; +//given +N=6.02E26; // in /Kg-atom (Avogadro's number) +n=2; // number of molecules per unit cell for bcc lattice +M=55.85; // in Kg/Kg-atom (atomic weight of Iron) +p=7860; // in Kg/m^3 (density) +//calculate +a=nthroot((n*M/(N*p)),3); +printf('\nThe lattice constant is\ta=%1.3E m',a); +a=a*1E10; // changing unit to Angstrom +printf('\n\t\t\ta=%.3f Angstrom',a); diff --git a/2912/CH2/EX2.4/Ex2_4.sce b/2912/CH2/EX2.4/Ex2_4.sce new file mode 100755 index 000000000..cc8922a86 --- /dev/null +++ b/2912/CH2/EX2.4/Ex2_4.sce @@ -0,0 +1,16 @@ +//chapter 2 +//example 2.4 +//calculate lattice constant +//page 42 +clear; +clc; +//given +N=6.02E26; // in /Kg-atom (Avogadro's number) +n=2; // number of molecules per unit cell for bcc lattice +M=6.94; // in Kg/Kg-atom (atomic weight of Iron) +p=530; // in Kg/m^3 (density) +//calculate +a=nthroot((n*M/(N*p)),3); +printf('\nThe lattice constant is\ta=%1.3E m',a); +a=a*1E10; // changing unit to Angstrom +printf('\n\t\t\ta=%.3f Angstrom',a); diff --git a/2912/CH2/EX2.5/Ex2_5.sce b/2912/CH2/EX2.5/Ex2_5.sce new file mode 100755 index 000000000..1ce4be9e5 --- /dev/null +++ b/2912/CH2/EX2.5/Ex2_5.sce @@ -0,0 +1,18 @@ +//chapter 2 +//example 2.5 +//calculate distance between adjacent atoms in NaCl +//page 42-43 +clear; +clc; +//given +N=6.02E23; // in /gram-molecule (Avogadro's number) +M=58.5; //in gram/gram-molecule (atomic weight of NaCl) +p=2.17; // in g/cm^3 (density) +//calculate +// since V=M/p +// (1/d)^-3=2N/V=2Np/M +// therefore d= (M/2Np)^-3 +d=nthroot((M/(2*N*p)),3); +printf('\nThe distance between two adjacent atoms of NaCl is \td=%1.2E cm',d); +d=d*1E8; // changing unit from cm to Angstrom +printf('\n\t\t\t\t\t\t\td=%.2f Angstrom',d); diff --git a/2912/CH2/EX2.6/Ex2_6.sce b/2912/CH2/EX2.6/Ex2_6.sce new file mode 100755 index 000000000..1ce6c76e8 --- /dev/null +++ b/2912/CH2/EX2.6/Ex2_6.sce @@ -0,0 +1,22 @@ +//chapter 2 +//example 2.6 +//calculate packing fraction and density +//page 43 +clear; +clc; +//given +r_Na=0.98; // in Angstrom (radius of sodium ion) +r_Cl=1.81; // in Angstrom (radius of chloride ion) +M_Na=22.99; // in amu (atomic mass of sodium) +M_Cl=35.45; // in amu (atomic mass of chlorine) +//calculate +a=2*(r_Na+r_Cl); // lattice parameter +printf('\nLattice constant is \ta=%.2f Angstrom',a); +//PF=volume of ions present in the unit cell/volume of unit cell +PF=((4*(4/3)*%pi)*r_Na^3+(4*(4/3)*%pi)*r_Cl^3)/a^3; +printf('\nPacking fraction is %.3f',PF); +//Density=mass of unit cell/volume of unit cell +p=4*(M_Na+M_Cl)*1.66E-27/(a*1E-10)^3; +printf('\nDensity is \tp=%.f Kg/m^3',p); +p=p*1E-3; //changing unit to gm/cm^-3 +printf('\nDensity is \tp=%.2f g/cm^3',p); diff --git a/2912/CH3/EX3.11/Ex3_11.sce b/2912/CH3/EX3.11/Ex3_11.sce new file mode 100755 index 000000000..082aaf56b --- /dev/null +++ b/2912/CH3/EX3.11/Ex3_11.sce @@ -0,0 +1,14 @@ +//chapter 3 +//example 3.11 +//calculate interplanar spacing +//page 61 +clear; +clc; +//given +h=3,k=2,l=1; // miller indices +a=4.2E-8; // in cm (lattice constant) +//calculate +d=a/sqrt(h^2+k^2+l^2); // calculation for interplanar spacing +printf('\nThe interplanar spacing is\td=%1.2E cm',d); +d=d*1E8; //changing unit from cm to Angstrom +printf('\n\t\t\t\td=%.2f Angstrom',d); diff --git a/2912/CH3/EX3.12/Ex3_12.sce b/2912/CH3/EX3.12/Ex3_12.sce new file mode 100755 index 000000000..ba48736a6 --- /dev/null +++ b/2912/CH3/EX3.12/Ex3_12.sce @@ -0,0 +1,12 @@ +//chapter 3 +//example 3.12 +//calculate lattice spacing +//page 61 +clear; +clc; +//given +h=1,k=1,l=1; // miller indices +a=2.5,b=2.5,c=1.8; // in Angstrom (lattice constants for tetragonal lattice ) +//calculate +d=1/sqrt((h/a)^2+(k/b)^2+(l/c)^2); // calculation for interplanar spacing +printf('\nThe lattice spacing is\td=%.2f Angstrom',d); diff --git a/2912/CH3/EX3.15/Ex3_15.sce b/2912/CH3/EX3.15/Ex3_15.sce new file mode 100755 index 000000000..29aae355a --- /dev/null +++ b/2912/CH3/EX3.15/Ex3_15.sce @@ -0,0 +1,14 @@ +//chapter 3 +//example 3.15 +//calculate density +//page 63 +clear; +clc; +//given +h=1,k=0,l=0; // miller indices +a=2.5; // in Angstrom (lattice constant) +//calculate +a=a*1E-10; //hence a is in Angstrom +d=a/sqrt(h^2+k^2+l^2); // calculation for interplanar spacing +p=d/a^3; +printf('\nThe density is\tp=%1.1E lattice points/m^2',p); diff --git a/2912/CH4/EX4.1/Ex4_1.sce b/2912/CH4/EX4.1/Ex4_1.sce new file mode 100755 index 000000000..b3c8358be --- /dev/null +++ b/2912/CH4/EX4.1/Ex4_1.sce @@ -0,0 +1,18 @@ +//chapter 4 +//example 4.1 +//Find spacing constant +//page 75 +clear; +clc; +//given +lambda=2.6; // in Angstrom (wavelength) +theta=20; // in Degree (angle) +n=2; +//calculate +lambda=lambda*1E-10; // since lambda is in Angstrom +// Since 2dsin(theta)=n(lambda) +// therefore d=n(lambda)/2sin(theta) +d=n*lambda/(2*sind(theta)); +printf('\nThe spacing constant is \td=%1.1E m',d); +d=d*1E10; // changing unit from m to Angstrom +printf('\n\t\t\t\td=%.1f Angstrom',d); diff --git a/2912/CH4/EX4.2/Ex4_2.sce b/2912/CH4/EX4.2/Ex4_2.sce new file mode 100755 index 000000000..3fe41974e --- /dev/null +++ b/2912/CH4/EX4.2/Ex4_2.sce @@ -0,0 +1,18 @@ +//chapter 4 +//example 4.2 +//Find glancing angle +//page 75 +clear; +clc; +//given +h=1,k=1,l=0; //miller indices +a=0.26; // in nanometer (lattice constant) +lambda=0.065; // in nanometer (wavelength) +n=2; // order +//calculate +d=a/sqrt(h^2+k^2+l^2); // calculation of interlattice spacing +// Since 2dsin(theta)=n(lambda) +// therefore we have +theta=asind(n*lambda/(2*d)); +printf('\nThe glancing angle is \t%.2f degree',theta); +//Note: there is slight variation in the answer due to round off diff --git a/2912/CH4/EX4.3/Ex4_3.sce b/2912/CH4/EX4.3/Ex4_3.sce new file mode 100755 index 000000000..dd26fceca --- /dev/null +++ b/2912/CH4/EX4.3/Ex4_3.sce @@ -0,0 +1,18 @@ +//chapter 4 +//example 4.3 +//Find glancing angle +//page 75-76 +clear; +clc; +//given +d=3.04E-10; // in mm (spacing constant) +lambda=0.79; // in Angstrom (wavelength) +n=3; // order +//calculate +// Since 2dsin(theta)=n(lambda) +// therefore we have +lambda=lambda*1E-10; //since lambda is in angstrom +theta=asind(n*lambda/(2*d)); +printf('\nThe glancing angle is \t%.3f degree',theta); +//Note: In question the value of d=3.04E-9 cm but in solution is using d=3.04E-10 m. +// I have used d=3.04E-10 cm as used in the solution diff --git a/2912/CH4/EX4.4/Ex4_4.sce b/2912/CH4/EX4.4/Ex4_4.sce new file mode 100755 index 000000000..ad35cc4ce --- /dev/null +++ b/2912/CH4/EX4.4/Ex4_4.sce @@ -0,0 +1,23 @@ +//chapter 4 +//example 4.4 +//Find wavelength and maximum order possible +//page 76 +clear; +clc; +//given +d=0.282; // in nanometer (spacing constant) +n=1; // order +theta=8.35; // in degree (glancing angle) +//calculate +d=d*1E-9; // since d is in nanometer +// Since 2dsin(theta)=n(lambda) +// therefore we have +lambda=2*d*sind(theta)/n; +printf('\nThe wavelength is \t%1.2E m',lambda); +lambda_1=lambda*1E10; //changing unit from m to Angstrom +printf('\n\t\t\t=%.3f Angstrom',lambda_1); +theta_1=90; // in degree (for maximum order theta=90) +n_max=2*d*sind(theta_1)/lambda; // calculation of maximum order. +printf('\nThe maximum order possible is \tn=%.f',n_max); +//Note: In question value of theta=8 degree and 35 minutes but solution uses theta=8.35 degree +// I am using theta=8.35 degree diff --git a/2912/CH4/EX4.5/Ex4_5.sce b/2912/CH4/EX4.5/Ex4_5.sce new file mode 100755 index 000000000..3bb1662c2 --- /dev/null +++ b/2912/CH4/EX4.5/Ex4_5.sce @@ -0,0 +1,23 @@ +//chapter 4 +//example 4.5 +//Find wavelength in X.U. +//page 76-77 +clear; +clc; +//given +theta=6; // in degree (glancing angle) +p=2170; // in Kg/m^3 (density) +M=58.46; // Molecular weight of NaCl +N=6.02E26; // in Kg-molecule (Avogadro's number) +n=1; // order +XU=1E-12; //since 1X.U.= 1E-12m +//calculate +d=(M/(2*N*p))^(1/3);//calclation of lattice constant +printf('\nThe spacing constant is \td=%1.3E m',d); +// Since 2dsin(theta)=n(lambda) +// therefore we have +lambda=2*d*sind(theta)/n; //calculation of wavelength +printf('\n\nThe wavelength is \t\t=%1.2E m',lambda); +lambda=lambda/XU; +printf('\n\t\t\t\t=%.1f X.U.',lambda); +// Note: The answer in the book is wrong due to calculation mistake diff --git a/2912/CH4/EX4.6/Ex4_6.sce b/2912/CH4/EX4.6/Ex4_6.sce new file mode 100755 index 000000000..c3ea162db --- /dev/null +++ b/2912/CH4/EX4.6/Ex4_6.sce @@ -0,0 +1,26 @@ +//chapter 4 +//example 4.6 +//find wavelength and energy +//page 77 +clear; +clc; +//given +h=1,k=1,l=1; // miller indices +a=5.63; // in Angstrom (lattice constant) +theta=27.5; // in degree (Glancing angle) +n=1; //order +H=6.625E-34; // in J-s (Plank's constant) +c=3E8; // in m/s (velocity of light) +e=1.6E-19;// charge of electron +//calculate +d=a/sqrt(h^2+k^2+l^2); // calculation for interplanar spacing +printf('\nThe lattice spacing is\td=%.2f Angstrom',d); +// Since 2dsin(theta)=n(lambda) +// therefore we have +lambda=2*d*sind(theta)/n; // calculation for wavelength +printf('\nThe wavelength is\t=%.f Angstrom',lambda); +E=H*c/(lambda*1E-10); //calculation of Energy +printf('\nThe energy of X-rays is E=%1.3E J',E); +E=E/e; // changing unit from J to eV +printf('\n\t\tE=%1.3E eV',E); +// Note: c=3E8 m/s but in solution c=3E10 m/s has been used that's why answer is different diff --git a/2912/CH4/EX4.7/Ex4_7.sce b/2912/CH4/EX4.7/Ex4_7.sce new file mode 100755 index 000000000..b272215cc --- /dev/null +++ b/2912/CH4/EX4.7/Ex4_7.sce @@ -0,0 +1,27 @@ +//chapter 4 +//example 4.7 +//calculate interpalanr spacing +//page 77-78 +clear; +clc; +//given +V=344; // in V (accelerating voltage) +theta=60; // in degree (glancing angle) +m=9.1E-31; // in Kg (mass of electron) +h=6.625e-34; // in J-s (Plank's constant) +n=1; //order +e=1.6E-19; // charge on electron +//calculate +//Since K=m*v^2/2=e*V +// therefore v=sqrt(2*e*V/m) +// since lambda=h/(m*v) +//therefore we have lambda=h/sqrt(2*m*e*V) +lambda=h/sqrt(2*m*e*V); // calculation of lambda +printf('\nThe wavelength is \t\t =%1.2E m',lambda); +lambda=lambda*1E10; //changing unit from m to Angstrom +printf('\n\t\t\t\t =%.2f Angstrom',lambda); +// Since 2dsin(theta)=n(lambda) +// therefore we have +d=n*lambda/(2*sind(theta)); +printf('\nThe interplanar spacing is \t d=%.2f Angstrom',d); + diff --git a/2912/CH4/EX4.8/Ex4_8.sce b/2912/CH4/EX4.8/Ex4_8.sce new file mode 100755 index 000000000..0e22a64e1 --- /dev/null +++ b/2912/CH4/EX4.8/Ex4_8.sce @@ -0,0 +1,29 @@ +//chapter 4 +//example 4.8 +//calculate angle of first order diffraction maximum +//page 78-79 +clear; +clc; +//given +K=0.02; // in eV (kinetic energy) +d=2.0; // in Angstrom (Bragg's spacing) +m=1.00898; // in amu (mass of neutron) +amu=1.66E-27; // in Kg (1amu=1.66E-27 Kg) +h=6.625e-34; // in J-s (Plank's constant) +n=1; //order +e=1.6E-19; // charge on electron +//calculate +//Since K=m*v^2/2 +// therefore v=sqrt(2*K/m) +// since lambda=h/(m*v) +//therefore we have lambda=h/sqrt(2*m*K) +m=m*amu; //changing unit from amu to Kg +K=K*e; //changing unit to J from eV +lambda=h/sqrt(2*m*K); // calculation of lambda +printf('\nThe wavelength is \t\t =%1.1E m',lambda); +lambda=lambda*1E10; //changing unit from m to Angstrom +printf('\n\t\t\t\t =%.1f Angstrom',lambda); +// Since 2dsin(theta)=n(lambda) +// therefore we have +theta=asind(n*lambda/(2*d)); // calculation of angle of first order diffraction maximum +printf('\nThe angle of first order diffraction maximum is %.f Degree',theta); diff --git a/2912/CH4/EX4.9/Ex4_9.sce b/2912/CH4/EX4.9/Ex4_9.sce new file mode 100755 index 000000000..30dfb596c --- /dev/null +++ b/2912/CH4/EX4.9/Ex4_9.sce @@ -0,0 +1,33 @@ +//chapter 4 +//example 4.9 +//Show that given angles are successive order of difraction and find spacing constant +//page 79 +clear; +clc; +//given +lambda=0.586; // in Angstrom (wavelength of X-rays) +n1=1, n2=2, n3=3; // orders of diffraction +theta1=5+(58/60); // in degree (Glancing angle for first order of diffraction) +theta2=12+(01/60); //in degree (Glancing angle for second order of diffraction) +theta3=18+(12/60); //in degree (Glancing angle for third order of diffraction) +//calculate +K1=sind(theta1); +K2=sind(theta2); +K3=sind(theta3); +printf('The value of sine of different angle of diffraction is\nK1=%.4f\nK2=%.4f\nK3=%.4f',K1,K2,K3); +// Taking the ratios of K1:K2:K3 +// We get K1:K2:K3=1:2:3 +//Therefore we have +printf('\n\nOr we have \tK1:K2:K3=1:2:3'); +printf('\nHence these angles of incidence are for Ist, 2nd and 3rd order reflections respectively'); +// Since 2dsin(theta)=n(lambda) +// therefore we have +d1=n1*lambda/(2*K1); +d2=n2*lambda/(2*K2); +d3=n3*lambda/(2*K3); +d1=d1*1E-10; //changing unit from Angstrom to m +d2=d2*1E-10; //changing unit from Angstrom to m +d3=d3*1E-10; //changing unit from Angstrom to m +printf('\n\nThe spacing constants are \nd1=%1.3E m\nd2=%1.3E m\nd3=%1.3E m',d1,d2,d3); +d=(d1+d2+d3)/3; +printf('\n\nThe mean value of crystal spacing is d=%1.3E m',d); diff --git a/2912/CH5/EX5.1/Ex5_1.sce b/2912/CH5/EX5.1/Ex5_1.sce new file mode 100755 index 000000000..482b5780a --- /dev/null +++ b/2912/CH5/EX5.1/Ex5_1.sce @@ -0,0 +1,22 @@ +//chapter 5 +//example 5.1 +//Find velocity and kinetic energy +//page 102-103 +clear; +clc; +//given +lambda=1; //in Angstrom (wavelength) +m=1.67E-27; // in Kg (mass of neutron) +h=6.625E-34; // in J-s (Planck's constant) +e=1.6E-19; // in C (charge of electron) +//calculate +lambda=lambda*1E-10; //since lambda is in Angstrom +// Since lambda=h/(m*v) +// Therefore we have +v=h/(m*lambda); //calculation of velocity +printf('\nThe velocity is \t v=%1.2E m/s',v); +K=m*v^2/2; //calculation of kinetic energy +printf('\nThe kinetic energy is\tK=%1.2E J',K); +K=K/e; //changing unit fro J to eV +printf('\n\t\t\t=%.4f eV',K); +//Note: Due to round off, there is slight variation in the answer diff --git a/2912/CH5/EX5.10/Ex5_10.sce b/2912/CH5/EX5.10/Ex5_10.sce new file mode 100755 index 000000000..1765b7481 --- /dev/null +++ b/2912/CH5/EX5.10/Ex5_10.sce @@ -0,0 +1,11 @@ +//chapter 5 +//example 5.10 +//Calculate wavelength +//page 106 +clear; +clc; +//given +V=10000; // in V (Potential) +//calculate +lambda=12.27/sqrt(V); // calculation of wavelength in Angstrom +printf('\nThe wavelength is\t=%.3f Angstrom',lambda); diff --git a/2912/CH5/EX5.11/Ex5_11.sce b/2912/CH5/EX5.11/Ex5_11.sce new file mode 100755 index 000000000..91ad13ff4 --- /dev/null +++ b/2912/CH5/EX5.11/Ex5_11.sce @@ -0,0 +1,18 @@ +//chapter 5 +//example 5.11 +//Calculate glancing angle +//page 107 +clear; +clc; +//given +V=100; // in V (Potential) +n=1; // order of diffraction +d=2.15; // in Angstrom (lattice spacing) +//calculate +lambda=12.27/sqrt(V); // calculation of wavelength in Angstrom +printf('\nThe wavelength is\t=%.3f Angstrom',lambda); +// Since 2*d*sind(theta)=n*lambda +//therefore we have +theta=asind(n*lambda/(2*d)); // calculation of glancing angle +printf('\nThe glancing angle is\t=%.1f degree',theta); +// Note: In question V=100 eV but the solution is using V=100V in the book and I have also used V=100V diff --git a/2912/CH5/EX5.12/Ex5_12.sce b/2912/CH5/EX5.12/Ex5_12.sce new file mode 100755 index 000000000..23a2cf8e5 --- /dev/null +++ b/2912/CH5/EX5.12/Ex5_12.sce @@ -0,0 +1,17 @@ +//chapter 5 +//example 5.12 +//Calculate spacing of crystal +//page 107 +clear; +clc; +//given +V=344; // in V (Potential) +n=1; // order of diffraction +theta=60; // in degree (glancing angle) +//calculate +lambda=12.27/sqrt(V); // calculation of wavelength in Angstrom +printf('\nThe wavelength is\t\t=%.3f Angstrom',lambda); +// Since 2*d*sind(theta)=n*lambda +//therefore we have +d=n*lambda/(2*sind(theta)); // calculation of spacing constant +printf('\nThe spacing of the crystal is\td=%.4f Angstrom',d); diff --git a/2912/CH5/EX5.13/Ex5_13.sce b/2912/CH5/EX5.13/Ex5_13.sce new file mode 100755 index 000000000..8d60fde48 --- /dev/null +++ b/2912/CH5/EX5.13/Ex5_13.sce @@ -0,0 +1,17 @@ +//chapter 5 +//example 5.13 +//Calculate velocity of electron +//page 107-108 +clear; +clc; +//given +r=0.53E-10; // in m (radius of first Bohr orbit) +h=6.6E-34; // in J-s (Planck's constant) +m=9.1E-31; // in Kg (mass of electron) +n=1; // First Bohr orbit +pi=3.14; // value of pi used in the solution +//calculate +// Since 2*pi*r=n*lambda and lambda=h/(m*v) +//Threfore we have v=h*n/(2*pi*r*m) +v=h*n/(2*pi*r*m); //calculation of velocity +printf('\nThe velocity of electron is\tv=%1.1E m/s',v); diff --git a/2912/CH5/EX5.14/Ex5_14.sce b/2912/CH5/EX5.14/Ex5_14.sce new file mode 100755 index 000000000..877ce9753 --- /dev/null +++ b/2912/CH5/EX5.14/Ex5_14.sce @@ -0,0 +1,19 @@ +//chapter 5 +//example 5.14 +//Calculate uncertainty in the momentum and uncertainty in the velocity +//page 108 +clear; +clc; +//given +dx=0.2; // in Angstrom (uncertainty in the position) +h=6.6E-34; // in J-s (Planck's constant) +m0=9.1E-31; // in Kg (mass of electron) +pi=3.14; // value of pi used in the solution +//calculate +dx=dx*1E-10; //since dx is in Angstrom +// Since dx*dp=h/4*pi (uncertainty relation) +dp=h/(4*pi*dx); // calculation of uncertainty in the momentum +printf('\nThe uncertainty in the momentum is\tdp=%1.2E Kg-m/s',dp); +//since dp=m*dv +dv=dp/m0; // calculation of uncertainty in the velocity +printf('\nThe uncertainty in the velocity is\tdv=%1.2E m/s',dv); diff --git a/2912/CH5/EX5.15/Ex5_15.sce b/2912/CH5/EX5.15/Ex5_15.sce new file mode 100755 index 000000000..6fbeb3ec6 --- /dev/null +++ b/2912/CH5/EX5.15/Ex5_15.sce @@ -0,0 +1,20 @@ +//chapter 5 +//example 5.15 +//Compare uncertainty in the velocity of electron and proton +//page 108 +clear; +clc; +//given +m_e=9.1E-31; // in Kg (mass of electron) +m_p=1.67E-27; // in Kg (mass of proton) +dx_p=1; // in nanometer (uncertainty in position of electron) +dx_n=1; // in nanometer (uncertainty in position of proton) +//calculate +// since dp=h/(4*pi*dx) +// since h/(4*pi) is constant and dx is same for electron and proton +// therefor both electron and proton have same uncertainty in the momentum +// since dv=dp/m and dp is same for both +// therefore dv_e/dv_p=m_p/m_e +// therefore +K=m_p/m_e; // ratio of uncertainty in the velocity of electron and proton +printf('\nThe ratio of uncertainty in the velocity of electron to that of proton is\t=%.f',K); diff --git a/2912/CH5/EX5.16/Ex5_16.sce b/2912/CH5/EX5.16/Ex5_16.sce new file mode 100755 index 000000000..a47cf5a30 --- /dev/null +++ b/2912/CH5/EX5.16/Ex5_16.sce @@ -0,0 +1,23 @@ +//chapter 5 +//example 5.16 +//Calculate minimum uncertainty in the momentum and minimum kinetic energy of proton +//page 108-109 +clear; +clc; +//given +dx=5E-15; // in m (radius of nucleus or uncertainty in the position) +h=6.6E-34; // in J-s (Planck's constant) +m=1.67E-27; // in Kg (mass of proton) +pi=3.14; // value of pi used in the solution +e=1.6E-19; // in C (charge of electron) +//calculate +// Since dx*dp=h/4*pi (uncertainty relation) +dp=h/(4*pi*dx); // calculation of uncertainty in the momentum +printf('\nThe minimum uncertainty in the momentum of proton is\tdp=%1.2E Kg-m/s',dp); +p=dp; // minimum value of momentum to calculate mimimum kinetic energy +K=p^2/(2*m); // calculation of minimum kinetic energy of proton +printf('\nThe minimum kinetic energy of proton is\tK=%1.1E J',K); +K=K/e; //changing unit from J to eV +printf('\n\t\t\t\t\t=%1.1E eV',K); +K=K/1E6; // changing unit from eV to MeV +printf('\n\t\t\t\t\t=%.1f MeV',K); diff --git a/2912/CH5/EX5.17/Ex5_17.sce b/2912/CH5/EX5.17/Ex5_17.sce new file mode 100755 index 000000000..87fcc3543 --- /dev/null +++ b/2912/CH5/EX5.17/Ex5_17.sce @@ -0,0 +1,23 @@ +//chapter 5 +//example 5.17 +//Calculate percentage of uncertainty in the momentum of electron +//page 109 +clear; +clc; +//given +K=1; // in KeV (kinetic energy of electron) +dx=1; // in Angstrom (uncertainty in the position) +h=6.63E-34; // in J-s (Planck's constant) +m=9.1E-31; // in Kg (mass of electron) +pi=3.14; // value of pi used in the solution +e=1.6E-19; // in C (charge of electron) +//calculate +dx=dx*1E-10; // since dx is in Angstrom +// Since dx*dp=h/4*pi (uncertainty relation) +dp=h/(4*pi*dx); // calculation of uncertainty in the momentum +printf('\nThe uncertainty in the momentum of electron is\tdp=%1.2E Kg-m/s',dp); +K=K*1E3*1.6E-19; // changing unit from KeV to J +p=sqrt(2*m*K); // calculation of momentum +printf('\nThe momentum of electron is\t\t\t p=%1.2E Kg-m/s',p); +poc=(dp/p)*100; // calculation of percentage of uncertainty +printf('\nThe percentage of uncertainty in the momentum is =%.1f',poc); diff --git a/2912/CH5/EX5.18/Ex5_18.sce b/2912/CH5/EX5.18/Ex5_18.sce new file mode 100755 index 000000000..bb881fb78 --- /dev/null +++ b/2912/CH5/EX5.18/Ex5_18.sce @@ -0,0 +1,21 @@ +//chapter 5 +//example 5.18 +//Calculate uncertainty in the position of electron +//page 109-110 +clear; +clc; +//given +v=6.6E4; // m/s (speed of electron) +poc=0.01; // percentage of uncertainty +h=6.63E-34; // in J-s (Planck's constant) +m=9E-31; // in Kg (mass of electron) +pi=3.14; // value of pi used in the solution +//calculate +p=m*v; // calculation of momentum +printf('\nThe momentum of electron is \t\t\tp=%1.2E Kg-m/s',p); +dp=(poc/100)*p; // calculation of uncertainty in the momentum +printf('\nThe uncertainty in the momentum of electron is\tdp=%1.2E Kg-m/s',dp); +// Since dx*dp=h/4*pi (uncertainty relation) +dx=h/(4*pi*dp); // calculation of uncertainty in the position +printf('\nThe uncertainty in the position of electron is\tdx=%1.2E Kg-m/s',dx); +// Note; solution is incomplete in the book diff --git a/2912/CH5/EX5.19/Ex5_19.sce b/2912/CH5/EX5.19/Ex5_19.sce new file mode 100755 index 000000000..d1ea0187e --- /dev/null +++ b/2912/CH5/EX5.19/Ex5_19.sce @@ -0,0 +1,25 @@ +//chapter 5 +//example 5.19 +//Calculate uncertainty in the position of X-ray photon +//page 111-112 +clear; +clc; +//given +lambda=1; // in Angstrom (wavelength) +pi=3.14; // value of pi used in the solution +dlambda=1E-6; // uncertainty in wavelength +//calculate +lambda=lambda*1E-10; // sinc lambda is in Angstrom +// By uncertainty principle, dx*dp>=h/(4*pi) --(1) +// since p=h/lambda -----(2) +// Or p*lambda=h +// diffrentiting this equation +// p*dlambda+lambda*dp=0 +// dp=-p*dlambda/lambda ----(3) +//from (2) and (3) dp=-h*dlambda/lambda^2 ---(4) +// from (1) and(4) dx*dlambda>=lambda^2/4*pi +// Or dx=lambda^2/(4*pi*dlambda) +dx=lambda^2/(4*pi*dlambda); //calculation of uncertainty in the position +printf('\nThe uncertainty in the position of X-ray photon is \tdx=%1.0E m',dx); +// Note: 1. In the question, wavelength accuracy is given as 1 in 1E8 but in book solution has used 1 in 1E6 and same has been used by me. +// 2. ANSWEER IS WRONG DUE TO CALCULATION MISTAKE diff --git a/2912/CH5/EX5.2/Ex5_2.sce b/2912/CH5/EX5.2/Ex5_2.sce new file mode 100755 index 000000000..0fac4fcb7 --- /dev/null +++ b/2912/CH5/EX5.2/Ex5_2.sce @@ -0,0 +1,21 @@ +//chapter 5 +//example 5.2 +//Calculate de-Broglie wavelength +//page 103-104 +clear; +clc; +//given +K=50; // in eV (Kinetic energy) +m0=9.1E-31; // in Kg (mass of electron) +h=6.625E-34; // in J-s (Planck's constant) +e=1.6E-19; // in C (charge of electron) +//calculate +K=K*e; //changing unit from eV to J +//Since K=m*v^2/2 +// Therefore v=sqrt(2*K/m) +// Since lambda=h/(m*v) +// Therefore we have +lambda=h/sqrt(2*m0*K); //calculation of wavelength +printf('\nThe wavelength is\t=%1.3E m',lambda); +lambda=lambda*1E10; //changing unit from m to Angstrom +printf('\n\t\t\t=%.3f Angstrom',lambda); diff --git a/2912/CH5/EX5.20/Ex5_20.sce b/2912/CH5/EX5.20/Ex5_20.sce new file mode 100755 index 000000000..09f616072 --- /dev/null +++ b/2912/CH5/EX5.20/Ex5_20.sce @@ -0,0 +1,15 @@ +//chapter 5 +//example 5.20 +//Compare minimum uncertainty in the frequency of the photon +//page 111 +clear; +clc; +//given +dt=1E-8; // in sec (average life time) +pi=3.14; // value of pi used in the solution +//calculate +// Since dE*dt>=h/(4*pi) (uncertainty relation for energy) +// and E=h*v v is the frequency +// therefore we have dv>=1/(4*pi*dt) +dv=1/(4*pi*dt); // calculation of minimum uncertainty in the frequency +printf('\nThe minimum uncertainty in the frequency of the photon is \tdv=%1.1E sec^-1',dv); diff --git a/2912/CH5/EX5.21/Ex5_21.sce b/2912/CH5/EX5.21/Ex5_21.sce new file mode 100755 index 000000000..08b610377 --- /dev/null +++ b/2912/CH5/EX5.21/Ex5_21.sce @@ -0,0 +1,17 @@ +//chapter 5 +//example 5.21 +//Calculate uncertainty in the energy of the photon +//page 111 +clear; +clc; +//given +dt=1E-12; // in sec (average life time) +h=6.63E-34; // in J-s (Planck'c constant) +pi=3.14; // value of pi used in the solution +e=1.6*1E-19; // in C (charge of electron) +//calculate +// Since dE*dt>=h/(4*pi) (uncertainty relation for energy) +dE=h/(4*pi*dt); // calculation of minimum uncertainty in the energy +printf('\nThe uncertainty in the energy of the photon is \tdE=%1.2E J',dE); +dE=dE/e; //changing unit from J to eV +printf('\n\t\t\t\t\t\t =%1.1E eV',dE); diff --git a/2912/CH5/EX5.22/Ex5_22.sce b/2912/CH5/EX5.22/Ex5_22.sce new file mode 100755 index 000000000..adefe5b6c --- /dev/null +++ b/2912/CH5/EX5.22/Ex5_22.sce @@ -0,0 +1,17 @@ +//chapter 5 +//example 5.22 +//Calculate minimum error in the energy +//page 111-112 +clear; +clc; +//given +dT=2.5E-14; // in sec (average life time) +h=6.63E-34; // in J-s (Planck'c constant) +pi=3.14; // value of pi used in the solution +e=1.6*1E-19; // in C (charge of electron) +//calculate +// Since dE*dt>=h/(4*pi) (uncertainty relation for energy) +dE=h/(4*pi*dT); // calculation of minimum uncertainty in the energy +printf('\nThe uncertainty in the energy of the photon is \tdE=%1.1E J',dE); +dE=dE/e; //changing unit from J to eV +printf('\n\t\t\t\t\t\t =%1.1E eV',dE); diff --git a/2912/CH5/EX5.23/Ex5_23.sce b/2912/CH5/EX5.23/Ex5_23.sce new file mode 100755 index 000000000..3bf7b66a6 --- /dev/null +++ b/2912/CH5/EX5.23/Ex5_23.sce @@ -0,0 +1,24 @@ +//chapter 5 +//example 5.23 +//Calculate energy corresponding to the 2nd and 4th quantum states +//page 112 +clear; +clc; +//given +a=2; // in Angstrom (length of the box) +m=9.1E-31; // in Kg (mass of electron) +h=6.626E-34; // in J-s (Planck'c constant) +n2=2, n4=4; // two quantum states +e=1.6*1E-19; // in C (charge of electron) +//calculate +a=a*1E-10; // since a is in Angstrom +// Since E_n=n^2*h^2/(8*m*a^2) (Energy corresponding to nth quantum state) +E2=n2^2*h^2/(8*m*a^2); // calculation of energy corresponding to the 2nd quantum state +printf('\nThe energy corresponding to the 2nd quantum state is \tE2=%1.3E J',E2); +E2=E2/e; //changing unit from J to eV +printf('\n\t\t\t\t\t\t\t =%1.4E eV',E2); +E4=n4^2*h^2/(8*m*a^2); // calculation of energy corresponding to the 4nd quantum state +printf('\nThe energy corresponding to the 4nd quantum state is \tE4=%1.3E J',E4); +E4=E4/e; //changing unit from J to eV +printf('\n\t\t\t\t\t\t\t =%1.4E eV',E4); +// Note: The answer in the book is wrong due to calculation mistake diff --git a/2912/CH5/EX5.24/Ex5_24.sce b/2912/CH5/EX5.24/Ex5_24.sce new file mode 100755 index 000000000..bba37a9d1 --- /dev/null +++ b/2912/CH5/EX5.24/Ex5_24.sce @@ -0,0 +1,27 @@ +//chapter 5 +//example 5.24 +//Calculate energy corresponding to the ground and first two excited states +//page 113 +clear; +clc; +//given +a=1E-10; // in m (width of the well) +m=9.1E-31; // in Kg (mass of electron) +h=6.626E-34; // in J-s (Planck'c constant) +n1=1, n2=2, n3=3; // ground and first two excited states +e=1.6*1E-19; // in C (charge of electron) +//calculate +// Since E_n=n^2*h^2/(8*m*a^2) (Energy corresponding to nth quantum state) +E1=n1^2*h^2/(8*m*a^2); // calculation of energy corresponding to the Ground state +printf('\nThe energy corresponding to the ground state is \tE1=%1.3E J',E1); +E1=E1/e; //changing unit from J to eV +printf('\n\t\t\t\t\t\t\t =%.2f eV',E1); +E2=n2^2*h^2/(8*m*a^2); // calculation of energy corresponding to the 1st excited state +printf('\nThe energy corresponding to the 1st excited state is \tE2=%1.3E J',E2); +E2=E2/e; //changing unit from J to eV +printf('\n\t\t\t\t\t\t\t =%.2f eV',E2); +E3=n3^2*h^2/(8*m*a^2); // calculation of energy corresponding to the 2nd excited state +printf('\nThe energy corresponding to the 2nd excited state is \tE3=%1.3E J',E3); +E3=E3/e; //changing unit from J to eV +printf('\n\t\t\t\t\t\t\t =%.2f eV',E3); +// Note: There is slight variation in the answer due to round off diff --git a/2912/CH5/EX5.25/Ex5_25.sce b/2912/CH5/EX5.25/Ex5_25.sce new file mode 100755 index 000000000..37a425eab --- /dev/null +++ b/2912/CH5/EX5.25/Ex5_25.sce @@ -0,0 +1,18 @@ +//chapter 5 +//example 5.25 +//Calculate minimum uncertainty in the velocity of electron +//page 113 +clear; +clc; +//given +dx=1E-8; // in m (length of box or uncertainty in the position) +h=6.626E-34; // in J-s (Planck'c constant) +m=9.1E-31; // in Kg (mass of electron) +//calculate +// From uncertainty principle dx*dp=h and dp=m*dv +// therefore we have +dv=h/(m*dx); // calculation of minimum uncertainty in the velocity +printf('\nThe minimum uncertainty in the velocity of electron is \t dv=%1.2E m/s',dv); +dv=dv*1E-3; // changing unit from m/s to Km/s +printf('\n\t\t\t\t\t\t\t =%.1f Km/s',dv); +// Note: There is slight variation in the answer due to round off diff --git a/2912/CH5/EX5.26/Ex5_26.sce b/2912/CH5/EX5.26/Ex5_26.sce new file mode 100755 index 000000000..aa5462c50 --- /dev/null +++ b/2912/CH5/EX5.26/Ex5_26.sce @@ -0,0 +1,19 @@ +//chapter 5 +//example 5.26 +//Calculate minimum energy of electron +//page 113-114 +clear; +clc; +//given +a=4E-10; // in m (length of the box) +m=9.1E-31; // in Kg (mass of electron) +h=6.626E-34; // in J-s (Planck'c constant) +n1=1; // ground state +e=1.6*1E-19; // in C (charge of electron) +//calculate +// Since E_n=n^2*h^2/(8*m*a^2) (Energy corresponding to nth quantum state) +E1=n1^2*h^2/(8*m*a^2); // calculation of energy corresponding to the ground state +printf('\nThe minimum energy of electron is \tE1=%1.3E J',E1); +E1=E1/e; //changing unit from J to eV +printf('\n\t\t\t\t\t =%.3f eV',E1); +// Note: The answer in the book corresponding to J is wrong due to printing error. diff --git a/2912/CH5/EX5.3/Ex5_3.sce b/2912/CH5/EX5.3/Ex5_3.sce new file mode 100755 index 000000000..6c4fd3139 --- /dev/null +++ b/2912/CH5/EX5.3/Ex5_3.sce @@ -0,0 +1,21 @@ +//chapter 5 +//example 5.3 +//Calculate wavelength +//page 104 +clear; +clc; +//given +E=2000; // in eV (Kinetic energy) +m=9.1E-31; // in Kg (mass of electron) +h=6.625E-34; // in J-s (Planck's constant) +e=1.6E-19; // in C (charge of electron) +//calculate +E=E*e; //changing unit from eV to J +//Since E=m*v^2/2 +// Therefore v=sqrt(2*E/m) +// Since lambda=h/(m*v) +// Therefore we have +lambda=h/sqrt(2*m*E); //calculation of wavelength +printf('\nThe wavelength is\t=%1.3E m',lambda); +lambda=lambda*1E9; //changing unit from m to nanometer +printf('\n\t\t\t=%.4f nm',lambda); diff --git a/2912/CH5/EX5.4/Ex5_4.sce b/2912/CH5/EX5.4/Ex5_4.sce new file mode 100755 index 000000000..a49ceb120 --- /dev/null +++ b/2912/CH5/EX5.4/Ex5_4.sce @@ -0,0 +1,22 @@ +//chapter 5 +//example 5.3 +//Calculate de-Broglie wavelength +//page 104 +clear; +clc; +//given +m_e=9.1E-31; // in Kg (mass of electron) +m_n=1.676E-27; // in Kg (mass of neutron) +h=6.625E-34; // in J-s (Planck's constant) +c=3E8; // in m/s (velocity of light) +//calculate +E_e=m_e*c^2; // rest mass energy of electron +E_n=2*E_e; // given (kinetic energy of neutron) +//Since K=m*v^2/2 +// Therefore v=sqrt(2*K/m) +// Since lambda=h/(m*v) +// Therefore we have +lambda=h/sqrt(2*m_n*E_n); //calculation of wavelength +printf('\nThe wavelength is\t=%1.1E m',lambda); +lambda=lambda*1E10; //changing unit from m to Angstrom +printf('\n\t\t\t=%1.1E Angstrom',lambda); diff --git a/2912/CH5/EX5.5/Ex5_5.sce b/2912/CH5/EX5.5/Ex5_5.sce new file mode 100755 index 000000000..9d62ef2c9 --- /dev/null +++ b/2912/CH5/EX5.5/Ex5_5.sce @@ -0,0 +1,12 @@ +//chapter 5 +//example 5.4 +//Calculate wavelength +//page 104 +clear; +clc; +//given +V=1600; // in V (Potential) +//calculate +lambda=12.27/sqrt(V); // calculation of wavelength in Angstrom +printf('\nThe wavelength is\t=%.3f Angstrom',lambda); +// Note: The answer in the book is wrong due to calculation mistake diff --git a/2912/CH5/EX5.6/Ex5_6.sce b/2912/CH5/EX5.6/Ex5_6.sce new file mode 100755 index 000000000..90ed84b2b --- /dev/null +++ b/2912/CH5/EX5.6/Ex5_6.sce @@ -0,0 +1,29 @@ +//chapter 5 +//example 5.6 +//Calculate wavelength for photon and electron +//page 105 +clear; +clc; +//given +E=1000; // in eV (Kinetic energy of photon) +K=1000; // in eV (Kinetic energy of electron) +m0=9.1E-31; // in Kg (mass of electron) +h=6.6E-34; // in J-s (Planck's constant) +c=3E8; // in m/s (velocity of light) +e=1.6E-19; // in C (charge on electron) +//calculate +E=E*e; // changing unit from eV to J +lambda_p=h*c/E; // For photon E=hc/lambda +printf('\nFor photon,the wavelength is\t=%1.2E m',lambda_p); +lambda_p=lambda_p*1E10; //changing unit from m to Angstrom +printf('\n\t\t\t\t=%.1f Angstrom',lambda_p) +//Since K=m*v^2/2 +// Therefore v=sqrt(2*K/m) +// Since lambda=h/(m*v) +// Therefore we have +K=K*e; // changing unit from eV to J +lambda_e=h/sqrt(2*m0*K); //calculation of wavelength +printf('\nFor electron,the wavelength is\t=%1.1E m',lambda_e); +lambda_e=lambda_e*1E10; //changing unit from m to Angstrom +printf('\n\t\t\t\t=%.2f Angstrom',lambda_e); +// Note: The answer in the book is wrong because K=1.6E-16 J but the solution is using K=2.4*E-15 J diff --git a/2912/CH5/EX5.7/Ex5_7.sce b/2912/CH5/EX5.7/Ex5_7.sce new file mode 100755 index 000000000..edd70ed0e --- /dev/null +++ b/2912/CH5/EX5.7/Ex5_7.sce @@ -0,0 +1,21 @@ +//chapter 5 +//example 5.7 +//Calculate velocity and kinetic energy +//page 105 +clear; +clc; +//given +lambda=1.66E-10; // in m (wavelength) +m=9.1E-31; // in Kg (mass of electron) +h=6.626E-34; // in J-s (Planck's constant) +e=1.6E-19; // in C (charge on electron) +//calculate +// Since lambda=h/(m*v) +// Therefore we have +v=h/(m*lambda); // calculation of velocity +printf('\nThe velocity of electron is \tv=%1.3E m/s',v); +K=m*v^2/2;//calculation of kinetic energy +printf('\nThe kinetic energy is \tK=%1.4E J',K); +K=K/e; // changing unit from J to eV +printf('\n\t\t\t=%.3f eV',K); +// Note: The answer in the book for kinetic energy is wrong due to calculation mistake diff --git a/2912/CH5/EX5.8/Ex5_8.sce b/2912/CH5/EX5.8/Ex5_8.sce new file mode 100755 index 000000000..418eb27c5 --- /dev/null +++ b/2912/CH5/EX5.8/Ex5_8.sce @@ -0,0 +1,21 @@ +//chapter 5 +//example 5.8 +//Calculate de-Broglie wavelength +//page 106 +clear; +clc; +//given +T=400; // in K (temperature) +m=6.7E-27; // in Kg (mass of He-atom) +h=6.6E-34; // in J-s (Planck's constant) +k=1.376E-23; // in J/degree (Boltzmann constant) +//calculate +// Since lambda=h/(m*v) +// E=mv^2/2; +// Therefore lambda=h/sqrt(2*m*E) +//E=kT +//Therefore lambda=h/sqrt(2*m*k*T) +lambda=h/sqrt(2*m*k*T) +printf('\nThe de-Broglie wavelength of He-atom is \t=%1.4E m',lambda); +lambda=lambda*1E10; //changing unit from m to Angstrom +printf('\n\t\t\t\t\t\t=%.4f Angstrom',lambda); diff --git a/2912/CH5/EX5.9/Ex5_9.sce b/2912/CH5/EX5.9/Ex5_9.sce new file mode 100755 index 000000000..cb018304f --- /dev/null +++ b/2912/CH5/EX5.9/Ex5_9.sce @@ -0,0 +1,22 @@ +//chapter 5 +//example 5.9 +//Calculate de-Broglie wavelength of proton +//page 106 +clear; +clc; +//given +m_e=9.1E-31; // in Kg (mass of electron) +m_p=1.6E-27; // in Kg (mass of proton) +h=6.626E-34; // in J-s (Planck's constant) +c=3E8; // in m/s (velocity of light) +//calculate +E=m_e*c^2; // in J (rest energy of electron) +// Since lambda=h/(m*v) +// E=mv^2/2; +// Therefore lambda=h/sqrt(2*m*E) +// Also E=m_e*c^2; +// therefore lambda=h/sqrt(2*m_p*m_e*c^2) +lambda=h/sqrt(2*m_p*m_e*c^2); // calculation of wavelength +printf('\nThe de-Broglie wavelength of proton is \t=%1.4E m',lambda); +lambda=lambda*1E10; //changing unit from m to Angstrom +printf('\n\t\t\t\t\t=%1.4E Angstrom',lambda); diff --git a/2912/CH6/EX6.1/Ex6_1.sce b/2912/CH6/EX6.1/Ex6_1.sce new file mode 100755 index 000000000..4a44266f7 --- /dev/null +++ b/2912/CH6/EX6.1/Ex6_1.sce @@ -0,0 +1,19 @@ +//chapter 6 +//example 6.1 +//Calculate mean free path of electron +//page 146 +clear; +clc; +//given +n=8.5E28; // in 1/m^3 (density of electron) +m_e=9.11E-31; // in Kg (mass of electron) +k=1.38E-23; // in J/K (Boltzmann's constant) +e=1.6E-19; // in C (charge of electron) +T=300; // in K (temperature) +p=1.69E-8; // in ohm-m (resistivity) +//calculate +lambda=sqrt(3*k*m_e*T)/(n*e^2*p); // calculation of mean free path +printf('\nThe mean free path of electron is \t=%1.2E m',lambda); +lambda=lambda*1E9; // changing unit from m to nanometer +printf('\n\t\t\t\t\t=%.2f nm',lambda); +// Note: answer in the book is wrong due to printing mistake diff --git a/2912/CH6/EX6.10/Ex6_10.sce b/2912/CH6/EX6.10/Ex6_10.sce new file mode 100755 index 000000000..589659196 --- /dev/null +++ b/2912/CH6/EX6.10/Ex6_10.sce @@ -0,0 +1,14 @@ +//chapter 6 +//example 6.10 +//Calculate mobility of electrons +//page 149-150 +clear; +clc; +//given +n=9E28; // in 1/m^3 (density of valence electrons) +sigma=6E7; // in mho/m (conductivity of copper) +e=1.6E-19; // in C (charge of electron) +//calculate +// Since sigma=n*e*mu therefore +mu=sigma/(n*e); // calculation of mobility of electron +printf('\n\nThe mobility of electrons is \t%1.2E m^2/V-s',mu); diff --git a/2912/CH6/EX6.11/Ex6_11.sce b/2912/CH6/EX6.11/Ex6_11.sce new file mode 100755 index 000000000..31e6cd39b --- /dev/null +++ b/2912/CH6/EX6.11/Ex6_11.sce @@ -0,0 +1,20 @@ +//chapter 6 +//example 6.11 +//Calculate average energy of free electron at 0K and corresponding temperature for a classical particle (an ideal gas) +//page 150 +clear; +clc; +//given +E_F=5.51; // in eV (Fermi energy in Silver) +k=1.38E-23; // in J/K (Boltzmann's constant) +e=1.6E-19; // in C (charge of electron) +//calculate +// part-(a) +Eo=(3/5)*E_F; // calculation of average energy of free electron at 0K +printf('\n\nThe average energy of free electron at 0K is \tEo=%.3f eV',Eo); +// part-(b) +Eo=Eo*e; // changing unit from eV to J +// Since for a classical particle E=(3/2)*k*T +// therefroe we have +T=(2/3)*Eo/k; // calculation of temperature for a classical particle (an ideal gas) +printf('\n\nThe temperature at which a classical particle have this much energy is \t T=%1.3E K',T); diff --git a/2912/CH6/EX6.12/Ex6_12.sce b/2912/CH6/EX6.12/Ex6_12.sce new file mode 100755 index 000000000..1ebf703a5 --- /dev/null +++ b/2912/CH6/EX6.12/Ex6_12.sce @@ -0,0 +1,18 @@ +//chapter 6 +//example 6.12 +//Calculate electron density for a metal +//page 150 +clear; +clc; +//given +E_F_L=4.7; // in eV (Fermi energy in Lithium) +E_F_M=2.35; // in eV (Fermi energy in a metal) +n_L=4.6E28; // in 1/m^3 (density of electron in Lithium) +//calculate +// Since n=((2*m/h)^3/2)*E_F^(3/2)*(8*pi/3) and all things except E_F are constant +// Therefore we have n=C*E_F^(3/2) where C is proportionality constant +// n1/n2=(E_F_1/E_F_2)^(3/2) +// Therefore we have +n_M=n_L*(E_F_M/E_F_L); // calculation of electron density for a metal +printf('\nThe lectron density for a metal is \t=%1.1E 1/m^3', n_M); +//Note: Answer in the book is wrong due to priting error diff --git a/2912/CH6/EX6.2/Ex6_2.sce b/2912/CH6/EX6.2/Ex6_2.sce new file mode 100755 index 000000000..24cf08f52 --- /dev/null +++ b/2912/CH6/EX6.2/Ex6_2.sce @@ -0,0 +1,22 @@ +//chapter 6 +//example 6.2 +//Calculate the temperature +//page 146 +clear; +clc; +//given + +k=1.38E-23; // in J/K (Boltzmann's constant) +e=1.6E-19; // in C (charge of electron) +P_E=1; // in percentage (probability that a state with an energy 0.5 eV above Fermi energy will be occupied) +E=0.5; // in eV (energy above Fermi level) +//calculate +P_E=1/100; // changing percentage into ratio +E=E*e; // changing unit from eV to J +// P_E=1/(1+exp((E-E_F)/k*T)) +// Rearranging this equation, we get +// T=(E-E_F)/k*log((1/P_E)-1) +// Since E-E_F has been denoted by E therefore +T=E/(k*log((1/P_E)-1)); +printf('\nThe temperature is \tT=%.f K',T); +// Note: There is slight variation in the answer due to logarithm function diff --git a/2912/CH6/EX6.3/Ex6_3.sce b/2912/CH6/EX6.3/Ex6_3.sce new file mode 100755 index 000000000..1d6db1895 --- /dev/null +++ b/2912/CH6/EX6.3/Ex6_3.sce @@ -0,0 +1,14 @@ +//chapter 6 +//example 6.3 +//Calculate relaxation time of conduction electrons +//page 147 +clear; +clc; +//given +n=5.8E28; // in 1/m^3 (density of electron) +m=9.1E-31; // in Kg (mass of electron) +e=1.6E-19; // in C (charge of electron) +p=1.54E-8; // in ohm-m (resistivity) +//calculate +t=m/(n*e^2*p); // calculation of relaxation time +printf('\nThe relaxation time of conduction electrons is %1.2E sec',t); diff --git a/2912/CH6/EX6.4/Ex6_4.sce b/2912/CH6/EX6.4/Ex6_4.sce new file mode 100755 index 000000000..daf83e29c --- /dev/null +++ b/2912/CH6/EX6.4/Ex6_4.sce @@ -0,0 +1,22 @@ +//chapter 6 +//example 6.4 +//Calculate mean free path traveeled by the electrons +//page 147 +clear; +clc; +//given +n=8.5E28; // in 1/m^3 (density of electron) +m=9.1E-31; // in Kg (mass of electron) +e=1.6E-19; // in C (charge of electron) +sigma=6E7; // in 1/ohm-m (conductivity) +E_F=7; // in E=eV (Fermi energy of Copper) +//calculate +E_F=E_F*e; // changing unit from eV to J +v_F=sqrt(2*E_F/m); // calculation of velocity of electrons +printf('\nThe velocity of the electrons is \t\t\tv_F=%1.1E m/s',v_F); +// Since sigma=n*e^2*lambda/(2*m*v_F) +// Therefore we have +lambda=2*m*v_F*sigma/(n*e^2); // calculation of mean free path +lambda=lambda*1E10; // changing unit from m to Angstrom +printf('\n\nThe mean free path traveled by the electrons is \t%.f Angstrom',lambda); +// Note: Answer in the book is wrong due to the use of round-off value of v_F as calculated in the first part. diff --git a/2912/CH6/EX6.5/Ex6_5.sce b/2912/CH6/EX6.5/Ex6_5.sce new file mode 100755 index 000000000..ee57e1078 --- /dev/null +++ b/2912/CH6/EX6.5/Ex6_5.sce @@ -0,0 +1,14 @@ +//chapter 6 +//example 6.5 +//Calculate relaxation time of conduction electrons +//page 147-148 +clear; +clc; +//given +n=6.5E28; // in 1/m^3 (density of electron) +m=9.1E-31; // in Kg (mass of electron) +e=1.6E-19; // in C (charge of electron) +p=1.43E-8; // in ohm-m (resistivity) +//calculate +t=m/(n*e^2*p); // calculation of relaxation time +printf('\nThe relaxation time of conduction electrons is %1.2E sec',t); diff --git a/2912/CH6/EX6.6/Ex6_6.sce b/2912/CH6/EX6.6/Ex6_6.sce new file mode 100755 index 000000000..52f14d8a7 --- /dev/null +++ b/2912/CH6/EX6.6/Ex6_6.sce @@ -0,0 +1,21 @@ +//chapter 6 +//example 6.6 +//Calculate average kinetic energy and velocity of molecules +//page 148 +clear; +clc; +//given +T=30; // in Celcius (temperature) +k=1.38E-23; // in J/K (Boltzmann's constant) +m_p=1.67E-27; // in Kg (mass of proton) +e=1.6E-19; // in C (charge of electron) +//calculate +T=T+273; // changing temperature from Celcius to Kelvin +KE=(3/2)*k*T; // calculation of average kinetic energy +printf('\nThe average kinetic energy of gas ,molecules is \tKE=%3.2E J',KE); +KE=KE/e; // changing unit from J to eV +printf('\n\t\t\t\t\t\t\t =%f eV',KE); +m=1.008*2*m_p; // calculating mass of hydrogen gas molecule +c=sqrt(3*k*T/m); // calculation of velocity +printf('\n\nThe velocity of molecules is \tc=%.2f m/s',c); +// Note: There is calculation mistake in the answer of energy given in eV and that of velocity diff --git a/2912/CH6/EX6.7/Ex6_7.sce b/2912/CH6/EX6.7/Ex6_7.sce new file mode 100755 index 000000000..714ad448c --- /dev/null +++ b/2912/CH6/EX6.7/Ex6_7.sce @@ -0,0 +1,20 @@ +//chapter 6 +//example 6.7 +//Calculate velocity of electron and proton +//page 148-149 +clear; +clc; +//given +E=10; // in eV (kinetic energy for each electron and proton) +m_e=9.1E-31; // in Kg (mass of electron) +m_p=1.67E-27; // in Kg (mass of proton) +e=1.6E-19; // in C (charge of electron) +//calculate +E=E*e; // changing unit from eV to J +// since E=m*v^2/2 +// therefore v=sqrt(2E/m) +v_e=sqrt(2*E/m_e); // calculation of kinetic energy of electron +printf('\nThe kinetic energy of electron is \tv_e=%1.3E m/s',v_e); +v_p=sqrt(2*E/m_p); // calculation of kinetic energy of proton +printf('\nThe kinetic energy of proton is \tv_p=%1.3E m/s',v_p); +// Note: The answer in the book for both kinetic energy of electron and that of proton is wrong due to calculation mistake diff --git a/2912/CH6/EX6.8/Ex6_8.sce b/2912/CH6/EX6.8/Ex6_8.sce new file mode 100755 index 000000000..665f519f0 --- /dev/null +++ b/2912/CH6/EX6.8/Ex6_8.sce @@ -0,0 +1,15 @@ +//chapter 6 +//example 6.8 +//Calculate drift velocity of free electrons +//page 149 +clear; +clc; +//given +I=100; // in A (current in the wire) +e=1.6E-19; // in C (charge of electron) +A=10; // in mm^2 (cross-sectional area) +n=8.5E28; // in 1/m^3 (density of electron) +//calculate +A=A*1E-6; // changing unit from mm^2 to m^2 +v_d=I/(n*A*e); +printf('\nThe drift velocity of free electrons is \tv_d=%1.3E m/s',v_d); diff --git a/2912/CH6/EX6.9/Ex6_9.sce b/2912/CH6/EX6.9/Ex6_9.sce new file mode 100755 index 000000000..0269cc491 --- /dev/null +++ b/2912/CH6/EX6.9/Ex6_9.sce @@ -0,0 +1,20 @@ +//chapter 6 +//example 6.9 +//Calculate average drift velocity of electrons +//page 149 +clear; +clc; +//given +I=4; // in A (current in the conductor) +e=1.6E-19; // in C (charge of electron) +A=1E-6; // in m^2 (cross-sectional area) +N_A=6.02E23; // in atoms/gram-atom (Avogadro's number) +p=8.9; // in g/cm^3 (density) +M=63.6; // atomic mass of copper +//calculate +n=N_A*p/M; // Calculation of density of electrons in g/cm^3 +printf('\nThe density of copper atoms is \tn=%1.2E atoms/m^3',n); +n=n*1E6; // changing unit from g/cm^3 to g/m^3 +printf('\n\t\t\t\t =%1.2E atoms/m^3',n); +v_d=I/(n*A*e); +printf('\n\nThe average drift velocity of free electrons is \tv_d=%1.1E m/s',v_d); diff --git a/2912/CH7/EX7.1/Ex7_1.sce b/2912/CH7/EX7.1/Ex7_1.sce new file mode 100755 index 000000000..b9ff71808 --- /dev/null +++ b/2912/CH7/EX7.1/Ex7_1.sce @@ -0,0 +1,21 @@ +//chapter 7 +//example 7.1 +//Calculate the capacitance of capacitor and charge on the plates +//page 187 +clear; +clc; +//given +A=100; // in cm^2 (cross-sectional area) +d=1; // in cm (seperation between plates) +Eo=8.85E-12; // in F/m (absolute permittivity) +V=100; // in V (potential difference) +//calculate +A=A*1E-4; // changing unit from cm^2 to m^2 +d=d*1E-2; // changing unit from cm to m +C=Eo*A/d;// calculation of capacitance +Q=C*V; // calculation of charge +printf('\nThe capacitance of capacitor is \t C=%1.2E C',C); +C=C*1E12; // changing unit of capacitance from F to pF +printf('\n\t\t\t\t\t =%.2f pF',C); +printf('\n\nThe charge on the plates is \t\t Q=%1.2E C',Q); + diff --git a/2912/CH7/EX7.10/Ex7_10.sce b/2912/CH7/EX7.10/Ex7_10.sce new file mode 100755 index 000000000..8f02ba733 --- /dev/null +++ b/2912/CH7/EX7.10/Ex7_10.sce @@ -0,0 +1,18 @@ +// chapter 7 +// example 7.10 +// determine the percentage of ionic polarisability in sodium crystal +// page 191-192 +clear; +clc; +// given +n=1.5; // refractive index +Er=5.6;// dielectric constant +//calculate +// since (Er-1)/(Er+2)=N*(alpha_e+alpha_i)/(3*E0) Clausius-Mossotti equation +// and (n^2-1)/(n^2+2)=N*alpha_e/(3*E0) +// from above two equations, we get ((n^2-1)/(n^2+2))*((Er+2)/(Er-1))=alpha_e/(alpha_e+alpha_i) +// or alpha_i/ (alpha_e+alpha_i)= 1-((n^2-1)/(n^2+2))*((Er+2)/(Er-1))= (say P) +// where P is fractional ionisational polarisability +P=1-((n^2-1)/(n^2+2))*((Er+2)/(Er-1)); // calculation of fractional ionisational polarisability +P=P*100; // calculation of percentage of ionisational polarisability +printf('\nThe percentage of ionisational polarisability is \t%.1f percent',P); diff --git a/2912/CH7/EX7.2/Ex7_2.sce b/2912/CH7/EX7.2/Ex7_2.sce new file mode 100755 index 000000000..7a6efa864 --- /dev/null +++ b/2912/CH7/EX7.2/Ex7_2.sce @@ -0,0 +1,23 @@ +//chapter 7 +//example 7.2 +//Calculate the resultant voltage across the capacitor +//page 187 +clear; +clc; +//given +A=650; // in mm^2 (cross-sectional area) +d=4; // in mm (seperation between plates) +Eo=8.85E-12; // in F/m (absolute permittivity) +Er=3.5; // di-electric constant of the material +Q=2E-10; // in C (charge on plates) +//calculate +A=A*1E-6; // changing unit from mm^2 to m^2 +d=d*1E-3; // changing unit from mm to m +C=Er*Eo*A/d;// calculation of capacitance +V=Q/C; // calculation of charge +printf('\nThe capacitance of capacitor is \t C=%1.2E C',C); +C=C*1E12; // changing unit of capacitance from F to pF +printf('\n\t\t\t\t\t =%.2f pF',C); +printf('\n\nThe resultant voltage across the capacitor is \t V=%.2f V',V); +// NOTE: The answer is wrong due to calculation mistake. The mistake is that in the book Value of cross-sectional area and seperation +// between plates is considered in cm and di-electric constant has not been considered. diff --git a/2912/CH7/EX7.3/Ex7_3.sce b/2912/CH7/EX7.3/Ex7_3.sce new file mode 100755 index 000000000..a70949cd1 --- /dev/null +++ b/2912/CH7/EX7.3/Ex7_3.sce @@ -0,0 +1,22 @@ +//chapter 7 +//example 7.3 +//Calculate the radius of electron cloud and dispalcement +//page 188 +clear; +clc; +//given +N=2.7E25; // in 1/m^3 (density of atoms) +E=1E6; // in V/m (electric field) +Z=2; // atomic number of Helium +Eo=8.85E-12; // in F/m (absolute permittivity) +Er=1.0000684; // (dielectric constant of the material) +e=1.6E-19; // in C (charge of electron) +pi=3.14; // value of pi used in the solution +//calculate +// since alpha=Eo*(Er-1)/N=4*pi*Eo*r_0^3 +// Therefore we have r_0^3=(Er-1)/(4*pi*N) +r_0=((Er-1)/(4*pi*N))^(1/3);// calculation of radius of electron cloud +printf('\nThe radius of electron cloud is \t r_0=%1.2E m',r_0); +x=4*pi*Eo*E*r_0/(Z*e); // calculation of dispalcement +printf('\n\nThe displacement is x=%1.2E m',x); +// NOTE: The answer is wrong due to calculation mistake. diff --git a/2912/CH7/EX7.4/Ex7_4.sce b/2912/CH7/EX7.4/Ex7_4.sce new file mode 100755 index 000000000..8313767f0 --- /dev/null +++ b/2912/CH7/EX7.4/Ex7_4.sce @@ -0,0 +1,21 @@ +//chapter 7 +//example 7.4 +//Calculate the dipole moment induced in each atom and atomic polarisability +//page 188-189 +clear; +clc; +//given +K=1.000134; // di-elecrtic constant of the neon gas at NTP +E=90000; // in V/m (electric field) +Eo=8.85E-12; // in C/N-m^2 (absolute premittivity) +N_A=6.023E26; // in atoms/Kg-mole (Avogadro's number) +V=22.4; // in m^3 (volume of gas at NTP +//calculate +n=N_A/V; // calculaton of density of atoms +// Since P=n*p=(k-1)*Eo*E +// therefore we have +p=(K-1)*Eo*E/n; // calculation of dipole moment induced +printf('\nThe dipole moment induced in each atom is \tp=%1.2E C-m',p); +alpha=p/E; // calculation of atomic polarisability +printf('\n\nThe atomic polarisability of neon is \t=%1.2E c-m^2/V',alpha); +// NOTE: The answer of atomic polarisability is wrong due to printing error diff --git a/2912/CH7/EX7.5/Ex7_5.sce b/2912/CH7/EX7.5/Ex7_5.sce new file mode 100755 index 000000000..da76ed5b1 --- /dev/null +++ b/2912/CH7/EX7.5/Ex7_5.sce @@ -0,0 +1,19 @@ +//chapter 7 +//example 7.5 +//Calculate the electronic polarisability of sulphur +//page 189 +clear; +clc; +//given +Er=3.75; // di-elecrtic constant of sulphur at 27 degree Celcius +gama=1/3; // internal field constant +p=2050; // in Kg/m^3 (density) +M_A=32; // in amu (atomic weight of sulphur) +Eo=8.85E-12; // in F/m (absolute permittivity) +N=6.022E23; // Avogadro's number +//calculate +// Since ((Er-1)/(Er+2))*(M_A/p)=(N/(3*Eo))*alpha_e +// therefore we have +alpha_e=((Er-1)/(Er+2))*(M_A/p)*(3*Eo/N); // calculation of electronic polarisability of sulphur +printf('\nThe electronic polarisability of sulphur is \t=%1.2E Fm^2',alpha_e); +// NOTE: There is slight variation in the answer due to round off diff --git a/2912/CH7/EX7.6/Ex7_6.sce b/2912/CH7/EX7.6/Ex7_6.sce new file mode 100755 index 000000000..ac5dca5bb --- /dev/null +++ b/2912/CH7/EX7.6/Ex7_6.sce @@ -0,0 +1,16 @@ +//chapter 7 +//example 7.6 +//Calculate the electronic polarisability of Helium atoms +//page 189-190 +clear; +clc; +//given +Er=1.0000684; // di-elecrtic constant of Helium gas at NTP +Eo=8.85E-12; // in F/m (absolute permittivity) +N=2.7E25; // number of atomsper unit volume +//calculate +// Since Er-1=(N/Eo)*alpha_e +// therefore we have +alpha_e=Eo*(Er-1)/N; // calculation of electronic polarisability of Helium +printf('\nThe electronic polarisability of Helium gas is \t=%1.2E Fm^2',alpha_e); +// NOTE: There is slight variation in the answer due to round off diff --git a/2912/CH7/EX7.7/Ex7_7.sce b/2912/CH7/EX7.7/Ex7_7.sce new file mode 100755 index 000000000..474407ad6 --- /dev/null +++ b/2912/CH7/EX7.7/Ex7_7.sce @@ -0,0 +1,17 @@ +//chapter 7 +//example 7.7 +//Calculate the dielectric constant of the material +//page 190 +clear; +clc; +//given +N=3E28; // in atoms/m^3 (density of atoms) +alpha_e=1E-40; // in F-m^2 (electronic polarisability) +Eo=8.85E-12; // in F/m (absolute permittivity) +//calculate +// Since (Er-1)/(Er+2)=N*alpha_e/(3*Eo) +// therefore we have +Er=(2*(N*alpha_e/(3*Eo))+1)/(1-(N*alpha_e/(3*Eo))); + // calculation of dielectric constant of the material +printf('\nThe dielectric constant of the material is \tEr=%.3f F/m',Er); +// NOTE: The answer in the book is wrong due to calculation mistake diff --git a/2912/CH7/EX7.8/Ex7_8.sce b/2912/CH7/EX7.8/Ex7_8.sce new file mode 100755 index 000000000..94a3be024 --- /dev/null +++ b/2912/CH7/EX7.8/Ex7_8.sce @@ -0,0 +1,16 @@ +//chapter 7 +//example 7.8 +//Calculate the atomic polarisability of sulphur +//page 190 +clear; +clc; +//given +Er=4; // relative permittivity of sulphur +Eo=8.85E-12; // in F/m (absolute permittivity) +NA=2.08E3; // in Kg/m^3 (density of atoms in sulphur) +//calculate +// Since ((Er-1)/(Er+2))*(M_A/p)=(N/(3*Eo))*alpha_e +// therefore we have +alpha_e=((Er-1)/(Er+2))*(3*Eo/NA); // calculation of electronic polarisability of sulphur +printf('\nThe electronic polarisability of sulphur is \t=%1.2E Fm^2',alpha_e); +// NOTE: The answer in the book is wrong due to calculation mistake. Also one point to be mentioned is that wrong formula has been used in the solution but i have used the formula as used in the solution. diff --git a/2912/CH7/EX7.9/Ex7_9.sce b/2912/CH7/EX7.9/Ex7_9.sce new file mode 100755 index 000000000..611ecfb4a --- /dev/null +++ b/2912/CH7/EX7.9/Ex7_9.sce @@ -0,0 +1,24 @@ +// chapter 7 +// example 7.9 +// calculate polarisability due to permanent dipole moment and due to deformation of the molecules +// page 190-191 +clear; +clc; +// given +alpha1=2.5E-39; // in C^2-m/N (dielectric constant at 300K) +alpha2=2.0E-39; // in C^2-m/N (dielectric constant at 400K) +T1=300; // in K(first temperature) +T2=400; // in K(second temperature) +//calculate +// since alpha=alpha_d+alpha0 and alpha0=Beta/T +// therefore alpha=alpha_d+(Beta/T) +// since alpha1=alpha_d+(Beta/T1) and alpha2=alpha_d+(Beta/T2) +// therefore alpha1-apha2=Beta*((1/T1)-(1/T2)) +// or Beta= (alpha1-apha2)/ ((1/T1)-(1/T2)) +Beta= (alpha1-alpha2)/ ((1/T1)-(1/T2)); // calculation of Beta +alpha_d=alpha1-(Beta/T1); // calculation of polarisability due to defromation +alpha0_1=Beta/T1; // calculation of polarisability due to permanent dipole moment at 300K +alpha0_2=Beta/T2; // calculation of polarisability due to permanent dipole moment at 400K +printf('\nThe polarisability due to permanent dipole moment at 300K is \t %1.2E C^2-m/N',alpha0_1); +printf('\nThe polarisability due to permanent dipole moment at 400K is \t %1.2E C^2-m/N',alpha0_2); +printf('\n\nThe polarisability due to deformation of the molecules is \t %1.2E C^2-m/N',alpha_d); diff --git a/2912/CH8/EX8.1/Ex8_1.sce b/2912/CH8/EX8.1/Ex8_1.sce new file mode 100755 index 000000000..e5925a64f --- /dev/null +++ b/2912/CH8/EX8.1/Ex8_1.sce @@ -0,0 +1,16 @@ +//chapter 8 +//example 8.1 +//Calculate intensity of magnetism and magnetic flux density +//page 236 +clear; +clc; +//given +X=-0.5E-5; // magnetic susceptibility of silicon +H=0.9E4; // in A/m (magnetic field intensity) +mu0=4*%pi*1E-7; // in H/m (absolute permeability) +//calculate +I=X*H; // calculation of intensity of magnetism +printf('\nThe intensity of magnetism is \tI=%.3f A/m',I); +B=mu0*H*(1+X); // calculation of magnetic flux density +printf('\nThe magnetic flux density is \tB=%.3f Wb/m^2',B); +// Note: The answer in the book is wrong. This is because the value of H given in the question is H=0.9E4 A/m but in the solution the value of H that has been used is H=9.9E4 A/m. diff --git a/2912/CH8/EX8.2/Ex8_2.sce b/2912/CH8/EX8.2/Ex8_2.sce new file mode 100755 index 000000000..a08222812 --- /dev/null +++ b/2912/CH8/EX8.2/Ex8_2.sce @@ -0,0 +1,16 @@ +//chapter 8 +//example 8.2 +//Calculate change in magnetic moment +//page 236 +clear; +clc; +//given +r=0.052; // in nm (radius of orbit) +B=1; // in Wb/m^2 (magnetic field of induction) +e=1.6E-19; // in C (charge of electron) +m=9.1E-31; // in Kg (mass of electron) +//calculate +r=0.052*1E-9; // changing unit from nm to m +d_mu=(e^2*r^2*B)/(4*m); // calculation of change in magnetic moment +printf('\nThe change in magnetic moment is \t%1.4E Am^2',d_mu); +// Note: The answer in the book is wrong due to caluclation mistake diff --git a/2912/CH8/EX8.3/Ex8_3.sce b/2912/CH8/EX8.3/Ex8_3.sce new file mode 100755 index 000000000..b4ddf7d60 --- /dev/null +++ b/2912/CH8/EX8.3/Ex8_3.sce @@ -0,0 +1,12 @@ +//chapter 8 +//example 8.3 +//Calculate relative permeability of a ferromagentic material +//page 236 +clear; +clc; +//given +H=220; // in A/m (magnetic field intensity) +I=3300; // in A/m (intensity of magnetisation) +//calculate +mu_r=1+(I/H); // calculation of relative permeability +printf('\nThe relative permeability of a ferromagentic material is %.f',mu_r); diff --git a/2912/CH8/EX8.4/Ex8_4.sce b/2912/CH8/EX8.4/Ex8_4.sce new file mode 100755 index 000000000..80469c03a --- /dev/null +++ b/2912/CH8/EX8.4/Ex8_4.sce @@ -0,0 +1,16 @@ +//chapter 8 +//example 8.4 +//Calculate magnetic force and relative permeability +//page 236-237 +clear; +clc; +//given +I=3000; // in A/m (intensity of magnetisation) +B=0.005; // in Wb/m^2 (magnetic flus intensity) +pi=3.14;// value of pi used in the solution +mu0=4*pi*1E-7; // in H/m (absolute permeability) +//calculate +H=(B/mu0)-I; // calculation of magnetic force +printf('\nThe magnetic force is \tH=%.3f',H); +mu_r=(I/H)+1; // calculation of relative permeability +printf('\nThe relative permeability is \t%.3f',mu_r); diff --git a/2912/CH8/EX8.5/Ex8_5.sce b/2912/CH8/EX8.5/Ex8_5.sce new file mode 100755 index 000000000..a91bf3817 --- /dev/null +++ b/2912/CH8/EX8.5/Ex8_5.sce @@ -0,0 +1,15 @@ +//chapter 8 +//example 8.5 +//Calculate current through the solenoid +//page 237 +clear; +clc; +//given +H=4E3; // in A/m (magnetic field intensity) +N=60; // number of turns +l=12; // in cm (length of solenoid) +//calculate +n=N/(l*1E-2); // calculation of number of turns per unit metre +// Snice H=n*i; +i=H/n; // calculation of current through the solenoid +printf('\nThe current through the solenoid is \ti=%.f A',i); diff --git a/2912/CH8/EX8.6/Ex8_6.sce b/2912/CH8/EX8.6/Ex8_6.sce new file mode 100755 index 000000000..c4b181138 --- /dev/null +++ b/2912/CH8/EX8.6/Ex8_6.sce @@ -0,0 +1,25 @@ +//chapter 8 +//example 8.6 +//Calculate flux density, magnetic intensity and relative permeability +//page 237 +clear; +clc; +//given +l=30; // in cm (length of solenoid) +A=1; // in cm^2 (cross-sectional area) +N=300; // number of turns +i=0.032; // in A (current through the winding) +phi_B=2E-6; // in Wb (magnetic flux) +pi=3.14;// value of pi used in the solution +mu0=4*pi*1E-7; // in H/m (absolute permeability) +//calculate +l=l*1E-2; // changing unit from cm to m +A=A*1E-4; // changing unit from cm^2 to m^2 +B=phi_B/A; // calculation of flux density +printf('\nThe flux density is \tB=%1.0E Wb/m^2',B); +H=N*i/l; // calculation of magnetic intensity +printf('\nThe magnetic intensity is \tH=%.f A-turns/m',H); +mu=B/H; // calcluation of absolute permeability of iron +mu_r=mu/mu0; // calcluation of relative permeability of iron +printf('\nThe relative permeability of iron is \t%.f',mu_r); +// Note: The value of relative permeability varies slightly due to the use of round off value mu as calculated diff --git a/2912/CH8/EX8.7/Ex8_7.sce b/2912/CH8/EX8.7/Ex8_7.sce new file mode 100755 index 000000000..15215c575 --- /dev/null +++ b/2912/CH8/EX8.7/Ex8_7.sce @@ -0,0 +1,13 @@ +//chapter 8 +//example 8.7 +//Calculate Hystersis loss per cycle +//page 238 +clear; +clc; +//given +A=100; // in m^2 (area of Hysteresis loop) +B=0.01; // in Wb/m^2 (unit space along vertical axis or magnetic flux density) +H=40; // in A/m (unit space along horizontal axis or magnetic fild ntensity) +//calculate +H_L=A*B*H; // calculation of magnetic intensity +printf('\nThe Hystersis loss per cycle is %.f J/m^2',H_L); diff --git a/2912/CH9/EX9.10/Ex9_10.sce b/2912/CH9/EX9.10/Ex9_10.sce new file mode 100755 index 000000000..ff6e7937a --- /dev/null +++ b/2912/CH9/EX9.10/Ex9_10.sce @@ -0,0 +1,25 @@ +// chapter 9 +// example 9.10 +// Find the electron and hole concentrations and the resistivity +// page 276 +clear; +clc; +//given +rho=2300; // in ohm-m (resistivity of pure silicon) +ue=0.135; // in m^2/V-s (mobility of electron) +uh=0.048; // in m^2/V-s (mobility of electron) +Nd=1E19;// in /m^3 (doping concentration) +e=1.6E-19;// in C (charge of electron) +//calculate +// since sigma=ni*e*(ue+uh) and sigma=1/rho +// therefore ni=1/(rho*e*(ue+uh)) +ni=1/(rho*e*(ue+uh)); // calculation of intrinsic concentration +ne=Nd; // calculation of electron concentration +printf('\nThe electron concentration is \tne=%1.1E /m^3',ne); +nh=ni^2/Nd; // calculation of hole concentration +printf('\nThe hole concentration is \tnh=%1.1E /m^3',nh); +sigma=ne*ue*e+nh*uh*e; // calculation of conductivity +rho=1/sigma; // calculation of resistivity +printf('\nThe resistivity of the specimen is \t%.2f ohm-m',rho); + + diff --git a/2912/CH9/EX9.11/Ex9_11.sce b/2912/CH9/EX9.11/Ex9_11.sce new file mode 100755 index 000000000..025d07c1d --- /dev/null +++ b/2912/CH9/EX9.11/Ex9_11.sce @@ -0,0 +1,19 @@ +// chapter 9 +// example 9.11 +// Find the conductivity of p-type Ge crystal +// page 276-277 +clear; +clc; +//given +uh=1900; // in cm^2/V-s (mobility of electron) +Na=2E17;// in /m^3 (acceptor doping concentration) +e=1.6E-19; // in C(charge of electron) +//calculate +uh=uh*1E-4; // changing unit from cm^2/V-s to m^2/V-s +Na=Na*1E6; // changing unit from 1/cm^3 to 1/m^3 +nh=Na; // hole concentration +// since sigma=ne*ue*e+nh*uh*e and nh>>ne +// therefore sigma=nh*uh*e +sigma=nh*uh*e; // calculation of conductivity +printf('\nThe conductivity of p-type Ge crystal is \t%.f /ohm-m',sigma); +// Note: there is slight variation in the answer due to round off calculation diff --git a/2912/CH9/EX9.12/Ex9_12.sce b/2912/CH9/EX9.12/Ex9_12.sce new file mode 100755 index 000000000..c24919cb2 --- /dev/null +++ b/2912/CH9/EX9.12/Ex9_12.sce @@ -0,0 +1,14 @@ +// chapter 9 +// example 9.12 +// Find the diffusion co-efficient of electron in silicon +// page 277 +clear; +clc; +//given +ue=0.19; // in m^2/V-s (mobility of electron) +T=300; // in K (temperature) +k=1.38E-23; // in J/K (Boltzmann’s constant) +e=1.6E-19; // in C(charge of electron) +//calculate +Dn=ue*k*T/e; // calculation of diffusion co-efficient +printf('\nThe diffusion co-efficient of electron in silicon is \tDn=%1.1E m^2/s',Dn); diff --git a/2912/CH9/EX9.13/Ex9_13.sce b/2912/CH9/EX9.13/Ex9_13.sce new file mode 100755 index 000000000..7cd252f5f --- /dev/null +++ b/2912/CH9/EX9.13/Ex9_13.sce @@ -0,0 +1,26 @@ +// chapter 9 +// example 9.13 +// Find the probability of occupation of lowest level in conduction band +// page 277-278 +clear; +clc; +//given +Eg=0.4; // in eV (Band gap of semiconductor) +k=1.38E-23; // in J/K (Boltzmann’s constant) +T1=0; // in degree Celcius (first temperature) +T2=50; // in degree Celcius (second temperature) +T3=100; // in degree Celcius (third temperature) +e=1.602E-19; //in C (charge of electron) +// calculate +T1=T1+273; // changing temperature form Celcius to Kelvin +T2=T2+273; // changing temperature form Celcius to Kelvin +T3=T3+273; // changing temperature form Celcius to Kelvin +Eg=Eg*e; // changing unit from eV to Joule +//Using F_E=1/(1+exp(Eg/2*k*T)) +F_E1=1/(1+exp(Eg/(2*k*T1))); // calculation of probability of occupation of lowest level at 0 degree Celcius +F_E2=1/(1+exp(Eg/(2*k*T2))); // calculation of probability of occupation of lowest level at 50 degree Celcius +F_E3=1/(1+exp(Eg/(2*k*T3))); // calculation of probability of occupation of lowest level at 100 degree Celcius +printf('\nThe probability of occupation of lowest level in conduction band is\n\n'); +printf('\t\t at 0 degree Celcius, F_E=%1.3E eV\n',F_E1); +printf('\t\t at 50 degree Celcius, F_E=%1.2E eV\n',F_E2); +printf('\t\t at 100 degree Celcius, F_E=%1.3E eV',F_E3); diff --git a/2912/CH9/EX9.14/Ex9_14.sce b/2912/CH9/EX9.14/Ex9_14.sce new file mode 100755 index 000000000..6ac91d6e5 --- /dev/null +++ b/2912/CH9/EX9.14/Ex9_14.sce @@ -0,0 +1,17 @@ +// chapter 9 +// example 9.14 +// Find the ratio of conductivity at 600K and at 300K +// page 278 +clear; +clc; +//given +Eg=1.2; // in eV (Energy band gap) +k=1.38E-23; // in J/K (Boltzmann’s constant) +T1=600, T2=300; // in K (two temperatures) +e=1.6E-19; // in C (charge of electron) +// calculate +Eg=Eg*e; // changing unit from eV to Joule +// since sigma is proportional to exp(-Eg/(2*k*T)) +// therefore ratio=sigma1/sigma2=exp(-Eg/(2*k*((1/T1)-(1/T2)))); +ratio= exp((-Eg/(2*k))*((1/T1)-(1/T2))); // calculation of ratio of conductivity at 600K and at 300K +printf('\nThe ratio of conductivity at 600K and at 300K is \t%1.2E',ratio); diff --git a/2912/CH9/EX9.15/Ex9_15.sce b/2912/CH9/EX9.15/Ex9_15.sce new file mode 100755 index 000000000..810533c57 --- /dev/null +++ b/2912/CH9/EX9.15/Ex9_15.sce @@ -0,0 +1,29 @@ +// chapter 9 +// example 9.15 +// Find the electron and hole densities and conductivity and the resistance +// page 278-279 +clear; +clc; +//given +ue=0.39; // in m^2/V-s (mobility of electron) +n=5E13;// number of donor atoms +ni=2.4E19; // in atoms/m^3 (intrinsic carrier density) +l=10; // in mm (length of rod) +a=1; // in mm (side of square cross-section) +e=1.6E-19;// in C (charge of electron) +//calculate +l=l*1E-3; // changing unit from mm to m +a=a*1E-3; // changing unit from mm to m +A=a^2; // calculation of cross-section area +Nd=n/(l*A); // calculation of donor concentration +ne=Nd; // calculation of electron density +nh=ni^2/Nd; // calculation of hole density +printf('\nThe electron density is \tne=%1.0E /m^3',ne); +printf('\nThe hole density is \tnh=%1.2E /m^3',nh); +// since sigma=ne*e*ue+nh*e*ue and since ne>>nh +// therefore sigma=ne*e*ue +sigma=ne*e*ue; // calculation of conductivity +printf('\nThe conductivity is \t%.f /ohm-m',sigma); +rho=1/sigma; // calculation of resistivity +R=rho*l/A; // calculation of resistance +printf('\nThe resistance is \tR=%.f ohm',R); diff --git a/2912/CH9/EX9.16/Ex9_16.sce b/2912/CH9/EX9.16/Ex9_16.sce new file mode 100755 index 000000000..0b6c5ca82 --- /dev/null +++ b/2912/CH9/EX9.16/Ex9_16.sce @@ -0,0 +1,15 @@ +// chapter 9 +// example 9.16 +// Find the mobility and density +// page 279 +clear; +clc; +//given +RH=3.66E-4; // in m^3/C (Hall coefficient) +rho=8.93E-3; // in ohm-m (resistivity) +e=1.6E-19; // in C (charge of electron) +// calculate +u=RH/rho; // calculation of mobility +n=1/(RH*e); // calculation of density +printf('\nThe mobility is \tu=%.4f m^2/(V-s)',u); +printf('\nThe density is \tn=%1.1E /m^3',n); diff --git a/2912/CH9/EX9.17/Ex9_17.sce b/2912/CH9/EX9.17/Ex9_17.sce new file mode 100755 index 000000000..ac4bf25dd --- /dev/null +++ b/2912/CH9/EX9.17/Ex9_17.sce @@ -0,0 +1,15 @@ +// chapter 9 +// example 9.17 +// Find the mobility and density of charge carrier +// page 279-280 +clear; +clc; +//given +RH=3.66E-4; // in m^3/C (Hall coefficient) +rho=8.93E-3; // in ohm-m (resistivity) +e=1.6E-19; // in C (charge of electron) +// calculate +nh=1/(RH*e); // calculation of density of charge carrier +uh=1/(rho*nh*e); // calculation of mobility of charge carrier +printf('\nThe density of charge carrier is \tnh=%1.4E /m^3',nh); +printf('\nThe mobility of charge carrier is \tuh=%.3f m^2/(V-s)',uh); diff --git a/2912/CH9/EX9.2/Ex9_2.sce b/2912/CH9/EX9.2/Ex9_2.sce new file mode 100755 index 000000000..080621bf4 --- /dev/null +++ b/2912/CH9/EX9.2/Ex9_2.sce @@ -0,0 +1,21 @@ +// chapter 9 +// example 9.2 +// Find the temperature at which number of electrons becomes 10 times +// page 272 +clear; +clc; +//given +Eg=0.67; // in eV (Energy band gap) +k=1.38E-23; // in J/K (Boltzmann’s constant) +T1=298; // in K (room temperature) +e=1.6E-19; // in C (charge of electron) +K=10; // ratio of number of electrons at different temperature +// calculate +Eg=Eg*e; // changing unit from eV to Joule +// since ne=Ke*exp(-Eg/(2*k*T)) +// and ne/ne1=exp(-Eg/(2*k*T))/exp(-Eg/(2*k*T1)) and ne/ne1=K=10 +// therefore we have 10=exp(-Eg/(2*k*T))/exp(-Eg/(2*k*T1)) +// re-arranging the equation for T, we get T2=1/((1/T1)-((2*k*log(10))/Eg)) +T=1/((1/T1)-((2*k*log(10))/Eg)); // calculation of the temperature +printf('\nThe temperature at which number of electrons in the conduction band of a semiconductor increases by a factor of 10 is \tT=%.f K',T); +// Note: there s slight variation in the answer due to round off calculation diff --git a/2912/CH9/EX9.3/Ex9_3.sce b/2912/CH9/EX9.3/Ex9_3.sce new file mode 100755 index 000000000..ec0788372 --- /dev/null +++ b/2912/CH9/EX9.3/Ex9_3.sce @@ -0,0 +1,24 @@ +// chapter 9 +// example 9.3 +// find the resistance of intrinsic germanium +// page 272-273 +// given +clear; +clc; +ni=2.5E13; // in /cm^3 (intrinsic carrier density) +ue=3900; // in cm^2/(V-s) (electron mobilities) +uh=1900; // in cm^2/(V-s) (hole mobilities) +e=1.6E-19; // in C (charge of electron) +l=1; // in cm (lenght of the box) +b=1,h=1; // in mm (dimensions of germanium rod ) +// calculate +ni=ni*1E6; // changing unit from 1/cm^3 to 1/m^3 +ue=ue*1E-4; // changing unit from cm^2 to m^2 +uh=uh*1E-4; // changing unit from cm^2 to m^2 +sigma=ni*e*(ue+uh); // calculation of conductivity +rho=1/sigma; // calculation of resistivity +l=l*1E-2; // changing unit from mm to m for length +A=(b*1E-3)*(h*1E-3); // changing unit from mm to m for width and height and calculation of cross-sectional area +R=rho*l/A; // calculation of resistance +printf('\nThe resistance of intrinsic germanium is \tR=%1.1E ohm',R); + diff --git a/2912/CH9/EX9.4/Ex9_4.sce b/2912/CH9/EX9.4/Ex9_4.sce new file mode 100755 index 000000000..13fb4afbf --- /dev/null +++ b/2912/CH9/EX9.4/Ex9_4.sce @@ -0,0 +1,22 @@ +// chapter 9 +// example 9.4 +// find the electrical conductivity and resistivity of germanium +// page 273 +clear; +clc; +// given +ne=2.5E19; // in /m^3 (electron density) +nh=2.5E19; // in /m^3 (hole density) +ue=0.36; // in m^2/(V-s) (electron mobilities) +uh=0.17; // in m^2/(V-s) (hole mobilities) +e=1.6E-19; // in C (charge of electron) +// calculate +// since ne=nh=ni, therefore we have +ni=nh; +sigma=ni*e*(ue+uh); // calculation of conductivity +printf('\nThe conductivity of germanium is %.2f /ohm-m',sigma); +rho=1/sigma; // calculation of resistivity +printf('\nThe resistivity of germanium is %.2f ohm-m',rho); +// Note: In the question, the value of ni has been misprinted as 2.5E-19 /m^3 rather it should be 2.5E19 /m^3. I have used 2.5E19 /m^3 + + diff --git a/2912/CH9/EX9.5/Ex9_5.sce b/2912/CH9/EX9.5/Ex9_5.sce new file mode 100755 index 000000000..c6cfeec95 --- /dev/null +++ b/2912/CH9/EX9.5/Ex9_5.sce @@ -0,0 +1,19 @@ +// chapter 9 +// example 9.5 +// find the equilibrium hole concentration and conductivity +// page 273-274 +clear; +clc; +// given +ni=1.5E16; // in /m^3 (intrinsic carrier density) +ue=0.135; // in m^2/(V-s) (electron mobilities) +uh=0.048; // in m^2/(V-s) (hole mobilities) +e=1.6E-19; // in C (charge of electron) +ND=1E23; // in atom/m^3 (doping concentration) +// calculate +sigma_i=ni*e*(ue+uh); // calculation of intrinsic conductivity +printf('\nThe intrinsic conductivity for silicon is %1.2E S',sigma_i); +sigma=ND*ue*e; // calculation of conductivity after doping +printf('\n\nThe conductivity after doping with phosphorus atoms is %1.2E S',sigma); +rho=ni^2/ND; // calculation of equilibrium hole concentration +printf('\n\nThe equilibrium hole concentration is %1.2E /m^3',rho); diff --git a/2912/CH9/EX9.6/Ex9_6.sce b/2912/CH9/EX9.6/Ex9_6.sce new file mode 100755 index 000000000..5a8477e9f --- /dev/null +++ b/2912/CH9/EX9.6/Ex9_6.sce @@ -0,0 +1,26 @@ +// chapter 9 +// example 9.6 +// find intrinsic concuctivity and doping conductivity +// page 274 +clear; +clc; +// given +ni=1.5E16; // in /m^3 (intrinsic carrier density) +ue=0.13; // in m^2/(V-s) (electron mobilities) +uh=0.05; // in m^2/(V-s) (hole mobilities) +e=1.6E-19; // in C (charge of electron) +ne=5E20; // in /m^3 (concentration of donor type impurity) +nh=5E20; // in /m^3 (concentration of acceptor type impurity) +// calculate +// part-i +sigma=ni*e*(ue+uh); // calculation of intrinsic conductivity +printf('\nThe intrinsic conductivity for silicon is %1.2E (ohm-m)^-1',sigma); +// part-ii +// since 1 donor atom is in 1E8 Si atoms, hence holes concentration can be neglected +sigma=ne*e*ue; // calculation of conductivity after doping with donor type impurity +printf('\n\nThe conductivity after doping with donor type impurity is %.1f (ohm-m)^-1',sigma); +// part-iii +// since 1 acceptor atom is in 1E8 Si atoms, hence electron concentration can be neglected +sigma=nh*e*uh; // calculation of conductivity after doping with acceptor type impurity +printf('\n\nThe conductivity after doping with acceptor type impurity is %.f (ohm-m)^-1',sigma); +// Note: In question the value of ne and nh has been misprinted as 5E28 atoms/m^3 which is too big but the solution has used the correct value 5E20 atoms/m^3. I have also used this value. diff --git a/2912/CH9/EX9.7/Ex9_7.sce b/2912/CH9/EX9.7/Ex9_7.sce new file mode 100755 index 000000000..1c31463ac --- /dev/null +++ b/2912/CH9/EX9.7/Ex9_7.sce @@ -0,0 +1,13 @@ +// chapter 9 +// example 9.7 +// find density of hole carriers at room temperature +// page 274-275 +clear; +clc; +// given +ni=1E20; // in /m^3 (intrinsic carrier density) +ND=1E21; // in /m^3 (donor impurity concentration) +// calculate +nh=ni^2/ND; // calculation of density of hole carriers at room temperature +printf('\nThe density of hole carriers at room temperature is \tnh=%1.0E /m^3',nh); +// Note: answer in the book is wrong due to printing mistake diff --git a/2912/CH9/EX9.8/Ex9_8.sce b/2912/CH9/EX9.8/Ex9_8.sce new file mode 100755 index 000000000..bd056cd40 --- /dev/null +++ b/2912/CH9/EX9.8/Ex9_8.sce @@ -0,0 +1,24 @@ +// chapter 9 +// example 9.8 +// find intrinsic carrier density and conductivity at 300K in germanium +// page 275 +clear; +clc; +M=72.6; // atomic mass of germanium +P=5400; // in Kg/m^3 (density) +ue=0.4; // in m^2/V-s (mobility of electrons) +uh=0.2; // in m^2/V-s (mobility of holes) +Eg=0.7; // in eV (Band gap) +m=9.1E-31; // in Kg (mass of electron) +k=1.38E-23; // in J/K (Boltzmann’s constant) +T=300; // in K (temperature) +h=6.63E-34;// in J/s (Planck’s constant) +pi=3.14; // value of pi used in the solution +e=1.6E-19; // in C(charge of electron) +// calculate +Eg=Eg*e; // changing unit from eV to J +ni=2*(2*pi*m*k*T/h^2)^(3/2)*exp(-Eg/(2*k*T)); +printf('\nThe intrinsic carrier density for germanium at 300K is \tni=%1.1E /m^3',ni); +sigma=ni*e*(ue+uh); +printf('\nThe conductivity of germanium is \t%1.2f (ohm-m)^-1',sigma); +// Note: Answer in the book is wrong due to calculation mistake diff --git a/2912/CH9/EX9.9/Ex9_9.sce b/2912/CH9/EX9.9/Ex9_9.sce new file mode 100755 index 000000000..73923461d --- /dev/null +++ b/2912/CH9/EX9.9/Ex9_9.sce @@ -0,0 +1,24 @@ +// chapter 9 +// example 9.9 +// Find the energy band gap +// page 275 +clear; +clc; +//given +rho1=4.5;// in ohm-m (resistivity at 20 degree Celcius) +rho2=2.0;// in ohm-m (resistivity at 32 degree Celcius) +k=1.38E-23; // in J/K (Boltzmann’s constant) +T1=20, T2=32; // in degree Celcius (two temperatures) +e=1.6E-19; // in C (charge of electron) +// calculate +T1=T1+273;// changing unit from degree Celius to K +T2=T2+273;// changing unit from degree Celius to K +// since sigma=e*u*C*T^(3/2)*exp(-Eg/(2*k*T)) +// therefore sigma1/sigma2=(T1/T2)^3/2*exp((-Eg/(2*k)*((1/T1)-(1/T2)) +// and sigma=1/rho +// therefore we have rho2/rho1=(T1/T2)^3/2*exp((-Eg/(2*k)*((1/T1)-(1/T2)) +// re-arranging above equation for Eg, we get Eg=(2*k/((1/T1)-(1/T2)))*((3/2)*log(T1/T2)-log(rho2/rho1)) +Eg=(2*k/((1/T1)-(1/T2)))*((3/2)*log(T1/T2)-log(rho2/rho1)); +printf('\nThe energy band gap is \tEg=%1.2E J',Eg); +Eg=Eg/e;// changing unit from J to eV +printf('\n\t\t\t =%.2f eV',Eg); -- cgit