From b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b Mon Sep 17 00:00:00 2001
From: priyanka
Date: Wed, 24 Jun 2015 15:03:17 +0530
Subject: initial commit / add all books

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 2912/CH5/EX5.6/Ex5_6.sce | 29 +++++++++++++++++++++++++++++
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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
-- 
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