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sol5 - CHEM 444 HW#4 Answer key by Jiahao Chen due Grading...

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CHEM 444 HW#4 Answer key by Jiahao Chen due October 13, 2004 Grading policy: One point is given for each numbered equation shown in the solution, or for written statements and/or equations to the same effect. The total score for this problem set is 80 points. 1. A low-frequency breathing mode of a protein has frequency 69.5 cm - 1 . What is the vibrational tem- perature Θ of this mode? Round off your number to an integer. Assume the spectrum corresponding to this mode is harmonic and choose the zero of energy at the ground vibrational level, such that the energy eigenvalues are given by the expression E n = n ω (1) Express the Boltzmann factor e - En k B T in terms of the vibrational temperature. Evaluate the Boltz- mann factor at room temperature for several energy levels (up to quantum numbers for which the Boltzmann factor drops below 0.01). Notice that the Boltzmann factor decreases by e - Θ T each time the quantum number increments by 1. Sum your results to obtain an approximation to the partition function. Then normalize all the Boltzmann factors by the partition function to obtain occupation probabilities. Does the value of the partition function give you a reasonable estimate of the number of energy levels that are populated at this temperature? Now recall that the book (and lecture notes) placed the zero of energy at the minimum of the under- lying harmonic potential. Imagine shifting all the energy levels of this problem up by the zero-point energy, 1 2 ω . Determine the effect of this shift to the partition function. Does the rescaled partition function agree approximately with the value given by the result derived in lecture? Why is there a small difference in the two results? Imagine repeating the calculation at T = 3000 K. Calculate the Boltzmann factor for the 10 lowest vibrational states. Do you see a pattern? Explain why the value of the partition function gives you a very good estimate of the number of states that are populated when T Θ . Evaluate the partition function from the expression given in the book and lecture notes. Do you think the different choice of the zero of energy (where all eigenvalues are shifted by the zero point amount) makes a significant difference at this temperature? [ 42 points] Solution. Take the frequency to be ˜ ν = 69 . 5 cm - 1 . The vibrational temperature is then the energy spacing expressed in units of Boltzmann’s constant: Θ = ω k B = hc ˜ ν k B (2) = ( 69 . 5 cm - 1 ) 0 . 695 cm - 1 (3) = 100 K (4) Setting the zero of energy to the ground vibrational level, the Boltzmann factor for state n is e - En k B T = e - n ω k B T (5) = e - n Θ T (6) So the first fourteen energy levels have Boltzmann factors e - n Θ T [ 1 2 point per entry] : 1

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CHEM 444 HW#4 Solutions n e - n Θ T n e - n Θ T n e - n Θ T n e - n Θ T 0 1.0000 1 0.7151 2 0.5113 3 0.3656 4 0.2614 5 0.1869 6 0.1337 7 0.0956 8 0.0684 9 0.0489 10 0.0350 11 0.0250 12 0.0179 13 0.0128 So the partition function desired is (note calculation in fraction is to more significant figures than is expressed): Q 13 n =0 e - n Θ T = 1 - e - 13Θ T 1 - e - Θ T = 1 - 0 . 0128 1 - 0 . 715 = 3 . 47 (7) so the desired occupation probabilities p n = e - n Θ T Q (8) are [ 1 2 point each] : n p n n
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