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`        Ten identical particles are to be divided up into two containers (a) How many microstates belong to the configuration of three particles in one container and seven in the other(A) 120 (B) 30240(C) 3628800 (D) 6.3 × 109 (b) How many different configurations are possible? (A) 1 (B) 11(C) 120(D) 1024 (E) 3628800(c) what is the total number of microstates for the ten particle system? (A) 1 (B) 11 (C) 120 (D) 1024(E) 3628800(d)which configuration has the largest number of microstates?(A) 0, 10 (B) 3, 7 (C) 4, 6 (D) 5, 5`
4 years ago 396 Points
```							(a) The correct option is (A) 120.The number of microstates that lead to a given configuration the multiplicity w of that configuration is defined as,w = N!/N1! N2!Here N = 10, N1 = 3 and N2 = 7.So, w = N!/N1! N2! = 10!/3! 7! = 120From the above observation we conclude that, the microstates belong to the configuration of three particles in one container and seven in other will be 120. Therefore option (A) is correct. (b) The correct option is (E).Ten identical particles are to be divided up into two containers. The total number of possible different configuration will be equal to N!.Here N is the total number of particles.As N = 10, N! = 10!= 3628800From the above observation we conclude that, 3628800 different configuration are possible. Therefore option (E) is correct.(c) The correct option is (D).Ten identical particles are to be divided up into two containers. The total number of microstates for the ten particle system will be equal to 2N.Here N is the total number of particles.As N = 10, 2N = 210= 1024From the above observation we conclude that, the total number of microstates for the ten particle system will be 1024. Therefore option (D) is correct.(d) The correct option is (D).The number of microstates that lead to a given configuration the multiplicity w of that configuration is defined as,w = N!/N1! N2!In the configuration (0,10) , N = 10, N1 = 0 and N2 = 10.So, w = N!/N1! N2! = 10!/0! 10! = 1In the configuration (3,7) , N = 10, N1 = 3 and N2 = 7.So, w = N!/N1! N2! = 10!/3! 7! = 120In the configuration (4,6) , N = 10, N1 = 4 and N2 = 6.So, w = N!/N1! N2! = 10!/4! 6! = 210In the configuration (5,5) , N = 10, N1 = 5 and N2 = 5.So, w = N!/N1! N2! = 10!/5! 5! = 252From the above observation we conclude that, the configuration which has the largest number of microstates will be (5,5). Therefore option (D) is correct.
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4 years ago
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