initial commit / add all books

17 files changed, 290 insertions, 0 deletions

diff --git a/635/CH8/EX8.1/Ch08Ex1.sci b/635/CH8/EX8.1/Ch08Ex1.sci
new file mode 100755
index 000000000..0094a2333
--- /dev/null
+++ b/635/CH8/EX8.1/Ch08Ex1.sci
@@ -0,0 +1,10 @@
+// Scilab Code Ex08.1 Determination of shortest wavelength and frequency of X-rays from accelerating potential Page-250 (2010)
+V = 50e+03;    // Accelerating potential, volt
+c = 3e+08;    // Speed of light in free space
+Lambda_min = 1.24e-06/V;    // Minimum wavelength, metre
+F_max = c/Lambda_min;    // Maximum frequency, Hz
+printf("\nThe shortest wavelength present in X-rays = %4.2f angstrom", Lambda_min/1D-10);
+printf("\nThe maximum frequency present in X-rays = %3.1e Hz", F_max);
+// Result
+// The shortest wavelength present in X-rays = 0.25 angstrom
+// The maximum frequency present in X-rays = 1.2e+19 Hz 
diff --git a/635/CH8/EX8.10/Ch08Ex10.sci b/635/CH8/EX8.10/Ch08Ex10.sci
new file mode 100755
index 000000000..7b70868d6
--- /dev/null
+++ b/635/CH8/EX8.10/Ch08Ex10.sci
@@ -0,0 +1,22 @@
+// Scilab code Ex8.10: Determining angle of reflection by using wavelength of X-ray Page 261 (2010)

+lambda = 0.440e-010;    // Wavelength of X-rays, m

+d = 2.814e-010;    // Interplanar spacing of rocksalt crystal, m

+// 2*d*sin(theta) = n*lambda    **Bragg's law, n is the order of diffraction

+// Solving for theta, we have

+// theta = asin(n*lambda/(2*d)) 

+// Declare a function for converting angle into degrees and minutes

+function [d,m] = degree_minute(n)    

+         d = int(n);

+         m = (n-int(n))*60;

+endfunction  

+for n = 1:1:5    // For diffraction order from 1 to 5

+    theta = asind(n*lambda/(2*d));    // Bragg's angle

+    [deg, mint] = degree_minute(theta);    // Call conversion function

+    printf("\nTheta%d = %2d degree(s), %2d minute(s)", n, deg, mint);

+end

+// Result

+// Theta1 =  4 degree(s), 29 minute(s)

+// Theta2 =  8 degree(s), 59 minute(s)

+// Theta3 = 13 degree(s), 33 minute(s)

+// Theta4 = 18 degree(s), 13 minute(s)

+// Theta5 = 23 degree(s),  0 minute(s)
\ No newline at end of file
diff --git a/635/CH8/EX8.11/Ch08Ex11.sci b/635/CH8/EX8.11/Ch08Ex11.sci
new file mode 100755
index 000000000..bba9877e6
--- /dev/null
+++ b/635/CH8/EX8.11/Ch08Ex11.sci
@@ -0,0 +1,19 @@
+// Scilab code Ex8.11: Determining the wavelength of diffracted X-rays Page 262 (2010)

+d = 2.814e-010;    // Interplanar spacing of rocksalt crystal, m

+theta = 9;    // Bragg's angle, degree

+// 2*d*sin(theta) = n*lambda    **Bragg's law, n is the order of diffraction

+// Solving for lambda, we have

+// lambda = 2*d*sin(theta)/n;

+printf("\nThe first four wavelengths of diffracted beam are:");

+for n = 1:1:5    // For diffraction order from 1 to 5

+    lambda = 2*d*sind(theta)/n;     // Wavelength of X-rays, m

+    if lambda >= 0.2e-010 & lambda <= 1.0e-010 then

+        printf("\nLambda%d = %6.4e angstrom", n, lambda/1D-10);

+    end

+end

+// Result

+// The first four wavelengths of diffracted beam are:

+// Lambda1 = 8.8041e-001 angstrom

+// Lambda2 = 4.4021e-001 angstrom

+// Lambda3 = 2.9347e-001 angstrom

+// Lambda4 = 2.2010e-001 angstrom
\ No newline at end of file
diff --git a/635/CH8/EX8.12/Ch08Ex12.sci b/635/CH8/EX8.12/Ch08Ex12.sci
new file mode 100755
index 000000000..783ae3008
--- /dev/null
+++ b/635/CH8/EX8.12/Ch08Ex12.sci
@@ -0,0 +1,14 @@
+// Scilab code Ex8.12: Reciprocal lattice parameters from 2-D direct lattice parameters Page 277 (2010)

+a = 3e-010;    // First lattice parameter of direct lattice

+b = 5e-010;    // Second lattice parameter of direct lattice

+theta = 60;    // Angle between two lattice vectors of the direct lattice

+// if a_prime and b_prime are the lattice vectors for the reciprocal lattice, then

+// a_prime*a = 2*%pi and a_prime*b = 0 

+// Similarly, b_prime*b = 2*%pi and b_prime*a = 0

+// Solving for a_prime and b_prime, we have

+a_prime = 2*%pi/(a*cosd(90-theta)); // Lattice vector for reciprocal lattice, per metre

+b_prime = 2*%pi/(b*cosd(90-theta)); // Lattice vector for reciprocal lattice, per metre

+printf("\nThe reciprocal lattice vectors are:\n a_prime = %5.2f per angstrom and b_prime = %5.2f per angstrom", a_prime*1e-010, b_prime*1e-010);

+// Result

+// The reciprocal lattice vectors are:

+// a_prime =  2.42 per angstrom and b_prime =  1.45 per angstrom
\ No newline at end of file
diff --git a/635/CH8/EX8.13/Ch08Ex13.sci b/635/CH8/EX8.13/Ch08Ex13.sci
new file mode 100755
index 000000000..c79943c5f
--- /dev/null
+++ b/635/CH8/EX8.13/Ch08Ex13.sci
@@ -0,0 +1,31 @@
+// Scilab code Ex8.13: Bragg angle and the indices of diffraction of Powder Lines Page 285 (2010)

+n = 1;    // Cosider first order diffraction

+a = 6e-010;    // First lattice parameter of direct lattice, m

+lambda = 1.54e-010;    // Wavelength used in diffraction of X-rays by Powder Method, m

+// Declare a function for converting angle into degrees and minutes

+function [d,m] = degree_minute(n)    

+         d = int(n);

+         m = (n-int(n))*60;

+endfunction  

+// Calculate the hkl and hence interpalnar spacing 'd' for three lowest powder lines

+printf("\nThe Bragg angles and the indices of diffraction for the three lowest powder lines are:");

+for h = 0:1:2

+    for k = 0:1:2

+        for l = 0:1:1

+            if (modulo(h,2) == 1 & modulo(k,2) == 1 & modulo (l,2) == 1) | (modulo(h,2) == 0 & modulo(k,2) == 0 & modulo (l,2) == 0) then

+                if (h <> 0) then

+                    N = h^2+k^2+l^2;

+                    d = a/sqrt(N);    // Interplanar spacing, metre

+                    theta = asind(n*lambda/(2*d));

+                    [deg, mint] = degree_minute(theta);    // Call conversion function

+                    printf("\nd[%d%d%d] = %4.2e and theta[%d%d%d] = %d deg %d min", h, k, l, d, h, k, l, deg, mint);

+                end

+            end

+         end      

+    end

+end    

+// Result

+// The Bragg angles and the indices of diffraction for the three lowest powder lines are:

+// d[111] = 3.46e-010 and theta[111] = 12 deg 50 min

+// d[200] = 3.00e-010 and theta[200] = 14 deg 52 min

+// d[220] = 2.12e-010 and theta[220] = 21 deg 17 min
\ No newline at end of file
diff --git a/635/CH8/EX8.14/Ch08Ex14.sci b/635/CH8/EX8.14/Ch08Ex14.sci
new file mode 100755
index 000000000..72ec59da5
--- /dev/null
+++ b/635/CH8/EX8.14/Ch08Ex14.sci
@@ -0,0 +1,14 @@
+// Scilab code Ex8.14: Minimum distance from the centre of the Laue pattern  of an fcc crystal Page 289 (2010)

+n = 1;    // Consider the first order diffraction

+a = 4.5e-010;    // Lattice parameter for fcc lattice, m

+V = 50e+03;    // Potential difference across the X-ray tube, volt

+D = 5;    // Crystal to film distance, cm 

+h = 1, k = 1, l = 1;    // Incides for the planes of maximum spacing

+lambda_min = 1.24e-06/V;    // The cut-off wavelength of X-rays, m

+d_111 = a/sqrt(h^1+k^2+l^2);

+theta_111 = asind(n*lambda_min/(2*d_111));

+// As tan(2*theta_111) = x/D, solving for x

+x = D*tand(2*theta_111);    // // Minimum distance from the centre of Laue pattern

+printf("\nThe minimum distance from the centre of the Laue pattern at which reflections can occur from the planes of maximum spacing = %4.2f cm", x);

+// Result

+// The minimum distance from the centre of the Laue pattern at which reflections can occur from the planes of maximum spacing = 0.48 cm
\ No newline at end of file
diff --git a/635/CH8/EX8.15/Ch08Ex15.sci b/635/CH8/EX8.15/Ch08Ex15.sci
new file mode 100755
index 000000000..401f062a7
--- /dev/null
+++ b/635/CH8/EX8.15/Ch08Ex15.sci
@@ -0,0 +1,15 @@
+// Scilab code Ex8.15: Calculating unit cell height along the axis of a rotation photograph Page 291 (2010)

+n = 1;    // Consider the first order diffraction of X-rays

+S = [0.29,0.59,0.91,1.25,1.65,2.12];    // An array of heights of first six layers above(below) the zero layer, cm

+R = 3;        // Radius of the camera, cm

+lambda = 1.54e-08;    // Wavelength of the X-rays, cm

+// For an a-axis rotation photograph, the unit cell parameter is given by

+// a = n*lambda/S(n)*(R^2 + S(n)^2)^(1/2) 

+// Calculate 'a' for six different values of n from 1 to 6

+for n = 1:1:6

+    a = (n*lambda/S(n))*(R^2 + S(n)^2)^(1/2);

+end 

+printf("\nThe unit cell height of the crystal = %2.0f angstrom", a/1D-8);

+

+// Result

+// The unit cell height of the crystal = 16 angstrom
\ No newline at end of file
diff --git a/635/CH8/EX8.16/Ch08Ex16.sci b/635/CH8/EX8.16/Ch08Ex16.sci
new file mode 100755
index 000000000..892e3a6d0
--- /dev/null
+++ b/635/CH8/EX8.16/Ch08Ex16.sci
@@ -0,0 +1,19 @@
+// Scilab code Ex8.16: Diffraction of thermal neutrons from planes of Ni crystal  Page 294 (2010)

+k = 1.38e-023;    // Boltzmann constant, J/mol/K

+h = 6.626e-034;    // Planck's constant, Js

+theta = 28.5;    // Bragg's angle, degree

+a = 3.52e-010;    // Lattice parameter of fcc structure of nickel, m

+m_n = 1.67e-027;    // Rest mass of neutron, kg

+// For fcc lattice, the interplanar spacing is given by

+d = a/sqrt(3);    // Interplanar spacing of Ni, m

+// Bragg's equation for first order diffraction (n = 1) is

+lambda = 2*d*sind(theta);    // Bragg's law, m

+// From kinetic interpretaion of temperature, we have

+// (1/2)*m*v^2 = (3/2)*k*T    -- (a)

+// Further from de-Broglie relation

+// lambda = h/(m*v)           -- (b)

+// From (a) and (b), solving for T, we have

+T = h^2/(3*m_n*k*lambda^2);    // Effective temperature of the neutrons, K

+printf("\nThe effective temperature of neutrons = %d K", T);

+// Result

+// The effective temperature of neutrons = 168 K 
\ No newline at end of file
diff --git a/635/CH8/EX8.17/Ch08Ex17.sci b/635/CH8/EX8.17/Ch08Ex17.sci
new file mode 100755
index 000000000..775f9adfd
--- /dev/null
+++ b/635/CH8/EX8.17/Ch08Ex17.sci
@@ -0,0 +1,20 @@
+// Scilab code Ex8.17: Diffraction of electrons from fcc crystal planes Page 295 (2010)

+// Declare a function for converting angle into degrees and minutes

+function [d,m] = degree_minute(n)    

+         d = int(n);

+         m = (n-int(n))*60;

+endfunction

+h = 6.626e-034;    // Planck's constant, Js

+m = 9.1e-031;    // Rest mass of electron, kg

+e = 1.602e-019;    // charge on an electron, coulomb

+a = 3.5e-010;    // Lattice parameter of fcc crystal, m

+V = 80;    // Accelerating potential for electrons, volt

+lambda = h/sqrt(2*m*e*V);    // de-Broglie wavelength of electrons, m

+d_111 = a/sqrt(3);    // Interplanar spacing for (111) planes of fcc crystal, m

+// Bragg's equation for first order diffraction (n = 1) is

+// lambda = 2*d_111*sind(theta_111);    // Bragg's law, m

+theta_111 = asind(lambda/(2*d_111));    // Bragg's angle, degree

+[deg, mint] = degree_minute(theta_111);    // Call conversion function

+printf("\nThe Bragg angle for electron diffraction = %d deg %d min", deg, mint);

+// Result

+// The Bragg angle for electron diffraction = 19 deg 50 min

diff --git a/635/CH8/EX8.2/Ch08Ex2.sci b/635/CH8/EX8.2/Ch08Ex2.sci
new file mode 100755
index 000000000..2fb5acbad
--- /dev/null
+++ b/635/CH8/EX8.2/Ch08Ex2.sci
@@ -0,0 +1,19 @@
+// Scilab Code Ex8.2 Calculation of impinging electrons on the target and characteristics of X-rays Page-253 (2010)
+I = 2.5e-03;    // Current through X-ray tube, ampere
+V = 6e+03;    // Potential across the X-ray  tube, volt
+e = 1.6e-19;    // Charge on an electron, coulomb
+m = 9.1e-031;    // mass of an electron, kg
+t = 1;    // Transit time, second
+Q = I*t;    // Total charge flowing per second through the x-ray tube, coulomb
+n = Q/e;    // Number of electrons striking the target per second
+// We have eV = 1/2*m*v^2 (stopping potential = maximum Kinetic energy)
+// Solving for v
+v = sqrt(2*e*V/m);    // speed of electrons striking the target, m/s
+Lambda_min = 1.24e-06/V;    // Minimum wavelength of X-rays produced, metre
+printf("\nThe number of electrons striking the target = %4.2e",n);
+printf("\nThe velocity of electrons striking the target = %4.2e m/s",v);
+printf("\nThe shortest wavelength present in X-rays = %4.2e m", Lambda_min);
+// Result
+// The number of electrons striking the target = 1.56e+016
+// The velocity of electrons striking the target = 4.59e+007 m/s
+// The shortest wavelength present in X-rays = 2.07e-010 m 
\ No newline at end of file
diff --git a/635/CH8/EX8.3/Ch08Ex3.sci b/635/CH8/EX8.3/Ch08Ex3.sci
new file mode 100755
index 000000000..4a42cf639
--- /dev/null
+++ b/635/CH8/EX8.3/Ch08Ex3.sci
@@ -0,0 +1,19 @@
+// Scilab Code Ex8.3 Calculation of wavelength of characteristic X-rays Page-253 (2010)
+h = 6.626e-034;    // Planck's constant, Js
+c = 3e+08;    // Speed of light in free space, m/s
+e = 1.602e-019;    // Charge on an electron, coulomb
+E_K = -78;    // Energy of K shell for platinum, keV
+E_L = -12;    // Energy of L shell for platinum, keV
+E_M = -3 ;    // Energy of M shell for platinum, keV
+E_K_alpha = E_L - E_K;    // Energy of K_alpha line, keV
+E_K_beta = E_M - E_K;    // Energy of K_beta line, keV
+// We have E = h*f, where f = c/Lambda this implies E = h*c/lambda
+// Solving for Lambda
+// Lambda = h*c/E
+lambda_K_alpha = h*c/(E_K_alpha*e*1e+03);    // Wavelength of K_alpha line, metre
+lambda_K_beta = h*c/(E_K_beta*e*1e+03);    // Wavelength of K_beta line, metre
+printf("\nThe wavelength of K_alpha line = %4.2f angstrom", lambda_K_alpha/1D-10);
+printf("\nThe wavelength of K_beta line  = %4.2f angstrom", lambda_K_beta/1D-10);
+// Result
+// The wavelength of K_alpha line = 0.19 angstrom
+// The wavelength of K_beta line  = 0.17 angstrom
diff --git a/635/CH8/EX8.4/Ch08Ex4.sci b/635/CH8/EX8.4/Ch08Ex4.sci
new file mode 100755
index 000000000..6de56cf9d
--- /dev/null
+++ b/635/CH8/EX8.4/Ch08Ex4.sci
@@ -0,0 +1,13 @@
+// Scilab Code Ex8.4 Calculation of atomic number of an unknown element Page-255 (2010)
+lambda_Pt = 1.321e-010;    // Wavelength of L_alpha line of Pt, m
+Z_Pt = 78;    // Atomic number of platinum
+b = 7.4;    // Constant
+lambda_x = 4.174e-010;    // Wavelength of unknown element, m
+// We have f = [a*(Z-b)]^2        (Moseley's law)
+// As f_Pt = c/lambda_Pt = [a*(Z_Pt-b)]^2 
+// Similarly f_x = c/lambda_x = [a*(Z_x-b)]^2 
+// Dividing f_Pt by f_x and solving for x
+Z_x = b + sqrt(lambda_Pt/lambda_x)*(Z_Pt-b);    // Atomic number of unknown element 
+printf("\nThe atomic number of unknown element = %4.1f", Z_x);
+// Result
+// The atomic number of unknown element = 47.1  
\ No newline at end of file
diff --git a/635/CH8/EX8.5/Ch08Ex5.sci b/635/CH8/EX8.5/Ch08Ex5.sci
new file mode 100755
index 000000000..3f907925a
--- /dev/null
+++ b/635/CH8/EX8.5/Ch08Ex5.sci
@@ -0,0 +1,13 @@
+// Scilab Code Ex8.5 Calculation of wavelength of copper using Moseley's law Page-256 (2010)
+c = 3.0e+08;    // Speed of light, m/s
+lambda_W = 210e-010;    // Wavelength of K_alpha line of W, m
+Z_W = 74;    // Atomic number of tungsten
+Z_Cu = 29;    // Atomic number of copper
+b = 1;    // Constant for K-series
+// f_W = c/lambda_W = (a*73)^2, The frequency K_alpha line for tungsten, Hz
+// f_Cu = c/lambda_Cu = (a*28)^2, The frequency K_alpha line for copper, Hz
+// Dividing f_W by f_Cu and solving for lambda_Cu
+lambda_Cu = ((Z_W-b)/(Z_Cu-b))^2*lambda_W; // Wavelength of K_alpha line of Cu, m
+printf("\nThe wavelength of K_alpha line of copper = %4.0f angstrom", lambda_Cu/1D-10);
+// Result
+// The wavelength of K_alpha line of copper = 1427 angstrom
\ No newline at end of file
diff --git a/635/CH8/EX8.6/Ch08Ex6.sci b/635/CH8/EX8.6/Ch08Ex6.sci
new file mode 100755
index 000000000..8f1e6f5b9
--- /dev/null
+++ b/635/CH8/EX8.6/Ch08Ex6.sci
@@ -0,0 +1,16 @@
+// Scilab Code Ex8.6 Calculation of atomic number from wavelength using Moseley's law Page-256 (2010)
+c = 3.0e+08;    // Speed of light, m/s
+h = 6.626e-034;    // Planck's constant, Js
+epsilon_0 = 8.85e-012;    // Absolute electrical permittivity of free space, coulomb square per newton per metre square
+m = 9.1e-031;    // Mass of an electron, kg
+e = 1.6e-019;    // Charge on an electron, C
+lambda = 0.7185e-010;    // Wavelength of K_alpha line of unknown element
+b = 1;    // Mosley's constant for K-series
+n_1 = 1; n_2 = 2;    // Lower and upper energy levels
+// We know that f = c/lambda = m*e^4*(Z-b)^2/(8*epsilon_0^2*h^3)*(1/n_2^2-1/n_1^2)
+// This implies that lambda = (8*epsilon_0^2*c*h^3/(m*e^4*(Z-b)^2*(1/n_2^2-1/n_1^2))
+// Solving for Z
+Z = sqrt(8*epsilon_0^2*c*h^3/(m*e^4*lambda*(1/n_1^2-1/n_2^2)))+b; // Atomic number unknown element
+printf("\nThe atomic number unknown element = %2d", Z); 
+// Result
+// The atomic number unknown element = 42 
\ No newline at end of file
diff --git a/635/CH8/EX8.7/Ch08Ex7.sci b/635/CH8/EX8.7/Ch08Ex7.sci
new file mode 100755
index 000000000..5f87c26da
--- /dev/null
+++ b/635/CH8/EX8.7/Ch08Ex7.sci
@@ -0,0 +1,17 @@
+// Scilab Code Ex8.7 Calculation of wavelengths of tin and barium using Moseley's law Page-257 (2010)
+Z_Fe = 26;    // Atomic number of iron
+Z_Pt = 78;    // Atomic number of platinum
+Z_Sn = 50;    // Atomic number of tin
+Z_Ba = 56;    // Atomic number of barium
+b = 1;    // Mosley's constant for K-series
+lambda_Fe = 1.93e-010;    // Wavelength of K_alpha line of Fe
+lambda_Pt = 0.19e-010;    // Wavelength of K_alpha line of Pt
+// From Moseley's Law,
+// f = a*(Z-1)^2. This implies lambda  = C*1/(Z-1)^2
+// so that lambda_Fe  = C*1/(Z_Fe-1)^2 and lambda_Sn  = C*1/(Z_Sn-1)^2
+// Dividing lambda_Sn by lambda_Fe and solving for lambda_Sn
+lambda_Sn = (Z_Fe-1)^2/(Z_Sn-1)^2*lambda_Fe;    // Wavelength of K_alpha line for tin, m
+lambda_Ba = (Z_Pt-1)^2/(Z_Ba-1)^2*lambda_Pt;    // Wavelength of K_alpha line for barium, m
+printf("\nThe wavelengths of tin and barium = %3.1f angstrom and %4.2f angstrom respectively", lambda_Sn/1D-10, lambda_Ba/1D-10); 
+// Result
+// The wavelengths of tin and barium = 0.5 angstrom and 0.37 angstrom respectively
\ No newline at end of file
diff --git a/635/CH8/EX8.8/Ch08Ex8.sci b/635/CH8/EX8.8/Ch08Ex8.sci
new file mode 100755
index 000000000..ca053cb87
--- /dev/null
+++ b/635/CH8/EX8.8/Ch08Ex8.sci
@@ -0,0 +1,11 @@
+// Scilab Code Ex8.8 Percentage transmitted energy of X-rays: Page 259 (2010)

+mu = 139;    // Attenuation co-efficient of aluminium, per metre

+x = 0.005;    // Thickness of aluminium sheet, m

+// If X% is the intensity of the X-ray transmitted through the aluminium sheet then

+// X% = I/I_0

+// or X/100 = exp(-absorb_coeff*x)

+// Solving for X

+X = 100*exp(-mu*x);    // Transmitted percentage of X-rays

+printf("\nThe intensity of the X-ray transmitted through the aluminium sheet = %g percent", round(X));

+// Result

+// The intensity of the X-ray transmitted through the aluminium sheet = 50 percent
\ No newline at end of file
diff --git a/635/CH8/EX8.9/Ch08Ex9.sci b/635/CH8/EX8.9/Ch08Ex9.sci
new file mode 100755
index 000000000..f48083c8f
--- /dev/null
+++ b/635/CH8/EX8.9/Ch08Ex9.sci
@@ -0,0 +1,18 @@
+// Scilab code Ex8.9 : Determination of thickness of lead piece by using two equal intensity X-ray wavelengths : Page 259 (2010)

+lambda_1 = 0.064e-010;    // First wavelength of X-ray, metre

+lambda_2 = 0.098e-010;    // Second wavelength of X-ray, metre

+I1_ratio_I2 = 3;    // Ratio of attenuated beam intensity

+mu_m1 = 0.164;    // Mass absorption coefficient for first wavelength, metre square per kg

+mu_m2 = 0.35;    // Mass absorption coefficient for second wavelength, metre square per kg

+d = 0.164;    // Density of the lead, kg per metre cube

+mu1 = mu_m1*d;    // absorption co-efficient of the lead for first wavelength, per metre

+mu2 = mu_m2*d;    // absorption co-efficient of the lead for second wavelength, per metre

+x = poly(0,"x");    // Declare 'x' as the thickness variable

+// Now I = exp(-ac*x) thus

+// I1_ratio_I2 = exp(-ac_1*x)/exp(-ac_2*x)

+// or 3 = exp(2109.24)*x this implies

+// 2104.24*x = log(3) and assume

+p = 2104.24*x-log(3);

+printf("\nThe thickness of lead piece = %4.2e m", roots(p));

+// Result

+// The thickness of lead piece = 5.22e-004 m 
\ No newline at end of file