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

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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. ...
View Full Document

Ask a homework question - tutors are online