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authorprashantsinalkar2017-10-10 12:27:19 +0530
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+// Example no.3.10
+// To calculate the effective mass of the electron in the valence band.
+// Page no.134
+
+clc;
+clear;
+
+// Given data
+Eg=1.18; // Band gap in eV
+Eg=1.18*1.602*10^-19; // Band gap in J
+hk1=9*10^-26; // The crystal momentum in Kg.m/s
+h=1.054*10^(-34); // The distance between two levels
+f=3.94*10^14; // Light wave of frequency
+m=9.109*10^(-31); // The electron rest mass in kg
+
+mr=(hk1)^2/(2*(h*2*%pi*f-Eg)); // The reduced mass in kg
+meff1=0.07*m; // The effective mass of an electron in the conduction band
+
+// The effective mass of the electron in the valence band.
+meff2=(mr*meff1)/(meff1-mr); // The effective mass of the electron in the valence band.
+
+// Displaying the result in command window
+printf('\n The effective mass of the electron in the valence band = %0.2f X 10^-31 kg',meff2*10^31);
+// The answer is varrying due to round-off error