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Unformatted text preview: 44. (a) We denote the conﬁguration with n heads out of N trials as (n; N ). We use Eq. 2118:
W (25; 50) = 50!
= 1.26 × 1014 .
(25!)(50 − 25)! (b) We use the result of problem #43: Ntotal = 250 = 1.13 × 1015 .
(c) The percentage of time in question is equal to the probability for the system to be in the central
conﬁguration:
W (25; 50)
1.26 × 1014
p(25; 50) =
=
= 11.1% .
250
1.13 × 1015
(d) We use W (N/2, N ) = N !/[(N/2)!]2 , Ntotal = 2N and p(N/2; N ) = W (N/2, N )/Ntotal . The results
are as follows: For N = 100, W (N/2, N ) = 1.01 × 1029 , Ntotal = 1.27 × 1030 , and p(N/2; N ) = 8.0%.
(e) Similarly, for N = 250, we obtain W (N/2, N ) = 9.25 × 1058 , Ntotal = 1.61 × 1060 , and p(N/2; N ) =
5.7%.
(f) As N increases the number of available microscopic states increase as 2N , so there are more states
to be occupied, leaving the probability less for the system to remain in its central conﬁguration. ...
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 Fall '08
 SPRUNGER
 Physics

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