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**Unformatted text preview: **Homework #8 Solutions Question 1) K+K, Chapter 5, Problem 6 Gibbs sum for a 2-level system has the following list of states: 1) Un-occupied N = 0; = 0 2) Occupied with energy 0; N = 1, = 0 3) Occupied with energy ; N = 1, = a) The Gibbs sum can be written explicitely using the above list of all states. Z g ( T, ) = X N =0 X S ( N ) e ( N- s ) / (1) = e (0 - 0) / + e (1 - 0) + e (1 - ) / (2) = 1 + + e- / (3) where we have defined e / and Z g denotes the grand canonical partition function. b) Find the thermal average occupancy. < N > = d d log Z g (4) = 1 Z g d d Z g (5) = + e- / Z g ( T, ) (6) c) Average thermal occupancy of state of energy . P (1 , ) = e- / Z g This is the probability of occupancy of the state with N = 1 and energy . d) The thermal average of the system is given by a weighted sum of energies and corresponding probabilities of occupation. In this case there is only one term U = e- / Z g e) Now suppose that there can be an N = 2 state in which energy levels 0 and can be occupied simultaneously. This will add an N = 2 term to the sum: e (2 - (0+ )) / = 2 e- / And the total Gibbs function becomes Z g = 1 + + e- / + 2 e- / 1 Question 2) K+K, Chapter 5, Problem 7 As usual we start by forming the partition function. Z g = 2 X N =0 X s ( N ) e ( N- s ) / (7) = e (0 + / 2) + e ( + / 2) / + e ( - / 2) / + e (2 - / 2) / (8) = e / 2 + 2 e / cosh 2 + e (2 - / 2) / (9) We can use eq(5) for < N > . Z g = e / 2 + 2 cosh 2 + 2 e- / 2 (10) Z g = 2cosh 2 + 2 e- / 2 (11) < N > = 1 Z g 2cosh 2 + 2 e- / 2 (12) Now enforce < N > = 1. Using the explicit form of Z g we get Z g = 2 cosh 2 + 2 2 e- / 2 = e / 2 + 2 cosh 2 + 2 e- / 2 (13) 2 e- / 2 = e / 2 (14) 2 = e / (15) Since = e / we get = / 2 . Question 3a) K+K, Chapter 5, Problem 8 First consider the case with O 2 only. In this case this is a two state system. The heme site can be occupied or unoccupied. The Gibbs sum has only two terms: Z g = 1 + e ( - A ) / = 1 + e- A / (16) We want to find A such that < N > = 0 . 9 N . Again using eq(5): < N > = log Z g = Z g e- A / (17) = e- A / 1 + e- A / = 1 1 + - 1 e A / = 0 . 9 (18) Note that in the last equality we used 0.9 instead of 0Note that in the last equality we used 0....

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