Statistical Mechanics Questions and Answers

Answer: C

The counts are (i) 64 and (ii) 20.

(i) Distinguishable particles: each of the 3 particles independently chooses one of 4 levels, so there are \(4^{3} = 64\) arrangements.

(ii) Identical bosons: only the occupation numbers matter, and any number may share a level. The number of ways to put 3 identical objects into 4 boxes is

\[ \binom{3 + 4 - 1}{3} = \binom{6}{3} = 20 \]

24 is \(4 \times 3 \times 2\), the count for distinguishable particles that may not share a level. 4 is \(\binom{4}{3}\), the count for fermions (at most one per level). 16 has no basis.

Answer: A

The entropy is k ln(10).

Let \(n_{+}\) moments point along +m and \(n_{-}\) along -m. A total magnetization of m requires

\[ \begin{aligned} n_{+} + n_{-} &= 5, \quad n_{+} - n_{-} = 1 \\ n_{+} &= 3, \quad n_{-} = 2 \end{aligned} \]

The number of microstates is the number of ways to choose which 2 of the 5 moments point down:

\[ \begin{aligned} \Omega &= \binom{5}{2} = 10 \\ S &= k\ln\Omega = k\ln 10 \end{aligned} \]

Zero would require a unique arrangement, as for total magnetization 5m. The value k ln 5 counts arrangements with only one moment reversed (total 3m), and k ln 20 double counts the arrangements.

Answer: D

The answer is (i) \(7^{4}\) and (ii) 210.

For distinguishable particles each of the 4 particles can go into any of the 7 levels independently, giving \(7^{4} = 2401\) arrangements.

For indistinguishable bosons only the occupation numbers matter, and any number may share a level. The count is the number of ways to place 4 identical balls in 7 boxes:

\[ \binom{4 + 7 - 1}{4} = \binom{10}{4} = 210 \]

The option \(4^{7}\) reverses the roles of particles and levels. \(7!/(4!\,3!) = 35\) is the count for fermions, where no two particles may share a level.