diff options
Diffstat (limited to '2912/CH4')
| -rwxr-xr-x | 2912/CH4/EX4.1/Ex4_1.sce | 18 | ||||
| -rwxr-xr-x | 2912/CH4/EX4.2/Ex4_2.sce | 18 | ||||
| -rwxr-xr-x | 2912/CH4/EX4.3/Ex4_3.sce | 18 | ||||
| -rwxr-xr-x | 2912/CH4/EX4.4/Ex4_4.sce | 23 | ||||
| -rwxr-xr-x | 2912/CH4/EX4.5/Ex4_5.sce | 23 | ||||
| -rwxr-xr-x | 2912/CH4/EX4.6/Ex4_6.sce | 26 | ||||
| -rwxr-xr-x | 2912/CH4/EX4.7/Ex4_7.sce | 27 | ||||
| -rwxr-xr-x | 2912/CH4/EX4.8/Ex4_8.sce | 29 | ||||
| -rwxr-xr-x | 2912/CH4/EX4.9/Ex4_9.sce | 33 |
9 files changed, 215 insertions, 0 deletions
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);
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