hw8solv5

# hw8solv5 - Homework#8 Solutions Question 1 K K Chapter 5...

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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; ε = 0 2) Occupied with energy 0; N = 1, ε = 0 3) Occupied with energy ε ; N = 1, ε = ε a) The Gibbs sum can be written explicitely using the above list of all states. Z g ( T,μ ) = ∞ X N =0 X S ( N ) e ( Nμ- ε s ) /τ (1) = e (0 μ- 0) /τ + e (1 μ- 0) τ + e (1 μ- ε ) /τ (2) = 1 + λ + λe- ε/τ (3) where we have defined λ ≡ e μ/τ and Z g denotes the grand canonical partition function. b) Find the thermal average occupancy. < N > = λ d dλ log Z g (4) = λ 1 Z g d dλ Z g (5) = λ + λe- ε/τ Z g ( T,μ ) (6) c) Average thermal occupancy of state of energy ε . P (1 ,ε ) = λe- ε/τ Z g This is the probability of occupancy of the state with N = 1 and energy ε . 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 = ελe- ε/τ Z g e) Now suppose that there can be an N = 2 state in which energy levels 0 and ε can be occupied simultaneously. This will add an N = 2 term to the sum: e (2 μ- (0+ ε )) /τ = λ 2 e- ε/τ And the total Gibbs function becomes Z g = 1 + λ + λe- ε/τ + λ 2 e- ε/τ 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 ε s ( N ) e ( Nμ- ε s ) /τ (7) = e (0 μ + δ/ 2) τ + e ( μ +Δ / 2) /τ + e ( μ- Δ / 2) /τ + e (2 μ- δ/ 2) /τ (8) = e δ/ 2 τ + 2 e μ/τ cosh Δ 2 τ + e (2 μ- δ/ 2) /τ (9) We can use eq(5) for < N > . Z g = e δ/ 2 τ + 2 λ cosh Δ 2 τ + λ 2 e- δ/ 2 τ (10) ∂ Z g ∂λ = 2cosh Δ 2 τ + 2 λe- δ/ 2 τ (11) < N > = λ 1 Z g 2cosh Δ 2 τ + 2 λe- δ/ 2 τ (12) Now enforce < N > = 1. Using the explicit form of Z g we get Z g = 2 λ cosh Δ 2 τ + 2 λ 2 e- δ/ 2 τ = e δ/ 2 τ + 2 λ cosh Δ 2 τ + λ 2 e- δ/ 2 τ (13) λ 2 e- δ/ 2 τ = e δ/ 2 τ (14) ⇒ λ 2 = e δ/τ (15) Since λ = e μ/τ we get μ = δ/ 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 ( μ- ε A ) /τ = 1 + λe- ε A /τ (16) We want to find ε A such that < N > = 0 . 9 N . Again using eq(5): < N > = λ ∂ ∂λ log Z g = λ Z g e- ε A /τ (17) = λe- ε A /τ 1 + λe- ε A /τ = 1 1 + λ- 1 e ε A /τ = 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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hw8solv5 - Homework#8 Solutions Question 1 K K Chapter 5...

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