P21_044

# P21_044 - 44. (a) We denote the conﬁguration with n heads...

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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. 21-18: 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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