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+//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