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| author | priyanka | 2015-06-24 15:03:17 +0530 |
|---|---|---|
| committer | priyanka | 2015-06-24 15:03:17 +0530 |
| commit | b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b (patch) | |
| tree | ab291cffc65280e58ac82470ba63fbcca7805165 /2912 | |
| download | Scilab-TBC-Uploads-b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b.tar.gz Scilab-TBC-Uploads-b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b.tar.bz2 Scilab-TBC-Uploads-b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b.zip | |
initial commit / add all books
Diffstat (limited to '2912')
109 files changed, 2087 insertions, 0 deletions
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);
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