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Chem390_solution_2

# Chem390_solution_2 - 2 Chem390_solution_2.nb Chem3900...

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Chem39 0 0 : Solution to Problem Set 2 Problem 1 : 1 The eigenvaluefor the energy of a harmonic oscillaotor is E v v 1 2 h . The probability of finding an oscillator in the v th state is given by P v 1 Q Exp E v k B T . Q T Simplify v 0 Exp v 1 2 h k B T ; P v Exp v 1 2 h k B T Q T ; AVGv Simplify v 0 v P v 1 1 h T k B 2 We need to solve v 1 for the temperature T. Solve 1 1 1 h T k B . constJSI . 2.75 10 13 s 1 , T T 1904.05 K 3 Let' s first plot v for the given T range. 1 1 h T k B .constJSI . 2.75 10 13 s 1 .T T K; Plot , T, 0.001, 2000 , PlotRange 0, 1.2 , AxesLabel "T K ", " v " 500 1000 1500 2000 T K 0.2 0.4 0.6 0.8 1 v The above plot shows that the average quantum number v stays near zero below about 500 K and then it increases almost linearly at higher temperatures. Since the average energy is proportional to v and C V is the rate of the energy increase with T, this plot is consistent with the Figure 17.4, where the C V slowly increases near 500 K and then stays constant at higher temperatures. Problem 2 : Let' s first define p j s and the partition function Q T . g 5, 3, 1, 5 ; E Oxy 0.00 eV, 0.02 eV, 0.03 eV, 1.97 eV ; Q T j 1 4 g j Exp E Oxy j k B T . constJSI; p g Exp E Oxy k B T Q T . constJSI 5 0. K T 5 22 860.8 K

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