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authorpriyanka2015-06-24 15:03:17 +0530
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+//example 7.2
+
+clear;
+clc;
+
+//Given:
+T=298;//Temperature [K]
+v=6.5*10^13;//Frequency in [sec-1]
+//Consider zero point energy = 0.
+h=6.627*10^-34;//planck's constant[J.s]
+k=1.381*10^-23;//Boltzmann constant
+N=1;//Since N=summation(gj*exp(-Ej/kT))
+
+//To find the fraction of molecules present in vibrational level
+E1=h*v;//for energy level 1[J]
+E2=2*h*v;//for energy level 2[J]
+x=k*T;
+g1=1;
+g2=1;
+N1=[g1*exp(-E1/x)];//molecules present in energy level 1
+N2=[g2*exp(-E2/x)];//molecules present in energy level 2
+n1=N1/N;//fraction of molecules present in energy level 1
+n2=N2/N;//fraction of molecules present in energy level 2
+printf("The fraction of molecule s present in energy level 1 is %f",n1);
+
+
+disp(n2,'The fraction of molecules present in energy level 2 is ');
+
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