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PS2_solutions

# PS2_solutions - Problem Set 2 Solutions Chem 120B Question...

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Problem Set 2 Solutions Chem 120B Question 1. (i) Want to find Δ X 2 in terms of t t and Δ X 2 j Variable descriptions: Δ X j is the change in separation between the two molecules on taking one single step. Δ X is the overall change in separation between the two molecules after n steps. X = x 2 - x 1 , the distance between the two molecules. Hence: Δ X j = X final,j - X initial,j (1) Δ X j = ( x 2 ,final,j - x 1 ,final,j ) - ( x 2 ,initial,j - x 1 ,initial,j ) (2) Δ X j = ( x 2 ,final,j - x 2 ,initial,j ) - ( x 1 ,final,j - x 1 ,initial,j ) (3) Δ X j = Δ x 2 ,j - Δ x 1 ,j (4) So, as one would expect, the overall change per step is nothing but the difference in the changes the individual molecules undergo. The total change in seperation between the two molecules, Δ X , which is a function of time, t , can be expressed as the sum of the individual changes 1

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in seperations for each step j , i.e. Δ X ( t ) = Δ X ( n Δ t ) = n j =1 Δ X j (5) So in order to find Δ X 2 ( t ) , one will have to square equation (5) and take its average, i.e. Δ X 2 ( t ) = n i =1 Δ X i n j =1 Δ X j (6) Now insert the expression for Δ X j , c.f. equation (4) Δ X 2 ( t ) = n i =1 x 2 ,i - Δ x 1 ,i ) n j =1 x 2 ,j - Δ x 1 ,j ) (7) Δ X 2 ( t ) = n i =1 n j =1 x 2 ,i - Δ x 1 ,i )(Δ x 2 ,j - Δ x 1 ,j ) Now evaluate the sums Δ X 2 ( t ) = Δ x 2 , 1 Δ x 2 , 1 - Δ x 2 , 1 Δ x 1 , 1 - Δ x 1 , 1 Δ x 2 , 1 + ... (9) Δ x 2 , 1 Δ x 1 , 1 = Δ x 2 , 1 Δ x 1 , 1 = 0 as the particles are uncorrelated and thus Δ x i = 0. Similarly, Δ x 2 , 1 Δ x 2 , 2 = Δ x 2 , 1 Δ x 2 , 2 = 0 as the steps are uncorrelated. Thus the only remaining terms in the double sum will be: Δ X 2 ( t ) = n j =1 x 1 ,j ) 2 + (Δ x 2 ,j ) 2 (10) Now realize that: Δ X 2 j ( t ) = x 2 - Δ x 1 ) 2 (11) Δ X 2 j ( t ) = x 2 2 - x 1 Δ x 2 + Δ x 2 1 ) (12) 2
Again, 2Δ x 1 Δ x 2 = 0 as the particles are uncorrelated. Thus equation (9) can be rewritten as: Δ X 2 ( t ) = n j =1 Δ X 2 j = n j =1 Δ X 2 j = n Δ X 2 j (13) Recall: t = n Δ t n = t/ Δ t Thus: Δ X 2 ( t ) = t Δ t Δ X 2 j (14) (ii) Δ x 1 ,j P(Δ x 1 ,j ) Δ x 2 ,j P (Δ x 2 ,j ) Δ X j P(Δ X j ) + 1/2 + m 1/2 m - = ( m - 1) 1/4 + 1/2 - m 1/2 - m - = ( - m - 1) 1/4 - 1/2 + m 1/2 m + = ( m + 1) 1/4 - 1/2 - m 1/2 - m + = ( - m + 1) 1/4 (iii) Δ X 2 j ( t ) = 1 / 4[ 2 ( m - 1) 2 + 2 ( - m - 1) 2 + 2 ( m + 1) 2 + 2 ( - m + 1) 2 ] = 1 / 4 2 [ m 2 - 2 m + 1 + m 2 + 2 m + 1 + m 2 + 2 m + 1 + m 2 - 2 m + 1] = 1 / 4 2 [4 m 2 + 4] = 2 ( m 2 + 1) = ( ± m ) 2 + ( ± ) 2 Δ X 2 ( t ) = 2 1 + 2 2 (15) Question 2. (i) The multiplicity, W dist , of distinguishable particles is W = N N T , i.e. the way N T objects can be arranged in N slots. As described in the text, the first tail can be placed in N different slots, the second tail in N - 1 and so on. The last tail has N - ( N T - 1) slots available. So: W dist = N × ( N - 1) × ( N - 2) ... × ( N - ( N T - 1)) (16) 3

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PS2_solutions - Problem Set 2 Solutions Chem 120B Question...

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