quiz-5-04-08-10-answers

quiz-5-04-08-10-answers - Quiz 5 for Physics 176: Answers...

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Unformatted text preview: Quiz 5 for Physics 176: Answers Professor Greenside The following equations may be useful: E =- ln( Z ) , Z N = Z N 1 N ! , Z 1 = V Z int V Q , V Q = h 2 mkT 3 . (1) E = X s E ( s ) p s , p s = e- [ E ( s )- A N A ( s )- B N B ( s )] Z , Z = X s e- [ E ( s )- A N A ( s )- B N B ( s )] . (2) 1. (10 points) A certain olfactory receptor in your nose has three slots such that one slot can bind a single molecular odorant of type A with energy E A < 0 while each of the other two slots can independently bind a single molecular odorant of type B , with binding energy E B < 0. (The energy of the receptor is zero if nothing binds to it.) The receptor is in contact with a reservoir (the air inside your nose) whose temperature is T and whose chemical potentials for molecules A and B are A and B respectively, which you can assume to be constant. Write down an expression for the probability for an A odorant and a single B odorant to be bound to this receptor. Answer: The probability is given by 2 e- ( E A + E B- A- B ) 1 + e- ( E A- A ) + 2 e- ( E B- B ) + 2 e- ( E A + E B- A- B ) + e- (2 E B- 2 B ) + e- ( E A +2 E B- A- 2 B ) . (3) The first step in calculating any partition function is to identify the states s of the system. Here there are eight: (a) No molecules are bound to the receptor so E = 0 and N A = N B = 0. (b) An A molecule is bound to the receptor so E = E A , N A = 1, and N B = 0. (c) A B molecule is bound to the receptor so E = E B , N A = 0, and N B = 1. The degeneracy for this state is 2 since there are two separate slots that each can bind a B molecule. (d) An A molecule and a B molecule are bound to the receptor so E = E A + E B and N A = N B = 1. This state has a degeneracy of 2 since the B molecule can bind to either of the two B sites. (e) Two B molecules are bound to the receptor so E = 2 E B , N A = 0, and N B = 2. (f) Finally, an A molecule and two B molecules can bind to the receptor so E = E A + 2 E B , N A = 1, and N B = 2. Once the states s have been enumerated and the corresponding energies and particle numbers identified, the answer Eq. (3) is found by direct evaluation of the Gibbs sum in Eq. (2) by evaluating successive Gibbs factors in turn, using the known values of E s , N A ( s ), and N B ( s ). A few clever students realized that, because the three binding sites are independent and distinguishable, the partition function for the three binding sites factors like this: Z ABB = Z A Z 2 B (4) = 1 + e- ( E a- A ) 1 + e- ( E B- B ) 2 (5) = 1 + e- ( E A- A ) + 2 e- ( E B- B ) + 2 e- ( E A + E B- A- B ) + e- (2 E B- 2 B ) + e- ( E A- A +2 E B- 2 B ) , (6) which is the same as the grand partition function Z in the denominator of Eq. (3)....
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This note was uploaded on 10/20/2011 for the course PHYSICS 176 taught by Professor Behringer during the Spring '08 term at Duke.

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quiz-5-04-08-10-answers - Quiz 5 for Physics 176: Answers...

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