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Unformatted text preview: Problem Set 2 Solutions Chem 120B Question 1. (i) Want to find h X 2 i in terms of t , t and D X 2 j E Variable descriptions: X j is the change in separation between the two molecules on taking one single step. 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: X j = X final,j X initial,j (1) X j = ( x 2 ,final,j x 1 ,final,j ) ( x 2 ,initial,j x 1 ,initial,j ) (2) X j = ( x 2 ,final,j x 2 ,initial,j ) ( x 1 ,final,j x 1 ,initial,j ) (3) X j = x 2 ,j 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, X , which is a function of time, t , can be expressed as the sum of the individual changes 1 in seperations for each step j , i.e. X ( t ) = X ( n t ) = n X j =1 X j (5) So in order to find h X 2 ( t ) i , one will have to square equation (5) and take its average, i.e. D X 2 ( t ) E = * n X i =1 X i n X j =1 X j + (6) Now insert the expression for X j , c.f. equation (4) D X 2 ( t ) E = * n X i =1 ( x 2 ,i x 1 ,i ) n X j =1 ( x 2 ,j x 1 ,j ) + (7) D X 2 ( t ) E = n X i =1 n X j =1 h ( x 2 ,i x 1 ,i )( x 2 ,j x 1 ,j ) i Now evaluate the sums D X 2 ( t ) E = h x 2 , 1 x 2 , 1 i  h x 2 , 1 x 1 , 1 i  h x 1 , 1 x 2 , 1 i + ... (9) h x 2 , 1 x 1 , 1 i = h x 2 , 1 ih x 1 , 1 i = 0 as the particles are uncorrelated and thus x i i = 0. Similarly, h x 2 , 1 x 2 , 2 i = h x 2 , 1 ih x 2 , 2 i = 0 as the steps are uncorrelated. Thus the only remaining terms in the double sum will be: D X 2 ( t ) E = n X j =1 D ( x 1 ,j ) 2 + ( x 2 ,j ) 2 E (10) Now realize that: D X 2 j ( t ) E = D ( x 2 x 1 ) 2 E (11) D X 2 j ( t ) E = D ( x 2 2 2 x 1 x 2 + x 2 1 ) E (12) 2 Again, 2 x 1 x 2 = 0 as the particles are uncorrelated. Thus equation (9) can be rewritten as: D X 2 ( t ) E = * n X j =1 X 2 j + = n X j =1 D X 2 j E = n D X 2 j E (13) Recall: t = n t n = t/ t Thus: D X 2 ( t ) E = t t D X 2 j E (14) (ii) x 1 ,j P( x 1 ,j ) x 2 ,j P ( x 2 ,j ) X j P( 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) D X 2 j ( t ) E = 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 D X 2 ( t ) E = ` 2 1...
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This note was uploaded on 09/08/2010 for the course CHEM 120B taught by Professor Geissler during the Fall '08 term at University of California, Berkeley.
 Fall '08
 Geissler
 Mole

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