Microstates and Macrostates Questions and Answers

Q: Three particles are to be distributed in four non-degenerate energy levels. The possible number of ways of distribution: (i) for distinguishable particles, and (ii) for identical Bosons, respectively, is

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.

Q: Consider 5 independent magnetic moments. Each moment has only two directions +m and -m. Hamiltonian of such system is given by \(x = \sum_{i=1}^{5} m_{i}\). If the total magnetization of the system is equal to m, the entropy is equal to :

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.

Q: The number of distinct ways in which 4 particles can be distributed in 7 energy levels if (i) they are distinguishable and (ii) if they are indistinguishable bosons respectively is :

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.

Q: A system of three indistinguishable particles has a total energy of \(4\varepsilon\). There are four single particle energy states with energies 0, \(\varepsilon\), \(2\varepsilon\) and \(3\varepsilon\). The number of microstates accessible to the system will be :

On the key, there are 3 accessible microstates.For indistinguishable particles a microstate is fixed by how many particles occupy each level, not by which particle is where. With no limit on the number per level (Bose counting), we need three energies from {0, 1, 2, 3} in units of \(\varepsilon\) adding to 4:\[ \begin{aligned} 4 &= 3 + 1 + 0 \\ 4 &= 2 + 2 + 0 \\ 4 &= 2 + 1 + 1 \end{aligned} \](4 + 0 + 0 is impossible because the highest level is \(3\varepsilon\).) That gives 3 microstates. One microstate would be the answer for fermions, which forbid the doubly occupied levels in the second and third sets. Four would come from counting an arrangement twice; distinguishable particles would give many more, since each set could be permuted.

Q: The volume of a perfect gas is doubled, the number of atoms N and the energy E being held constant, the change in entropy will be :

The entropy change is Nk ln 2.At constant N and energy E (free expansion of an ideal gas), only the volume term of the Sackur-Tetrode entropy changes: \(S = Nk\ln V + \) terms independent of V. So:\[ \Delta S = Nk\ln\frac{2V}{V} = Nk\ln 2 \]The answer depends only on the ratio of volumes, so options containing ln V or ln 2V alone cannot be a change in entropy.

Three particles are to be distributed in four non-degenerate energy levels. The possible number of ways of distribution: (i) for distinguishable particles, and (ii) for identical Bosons, respectively, is

A.(i) 24, (ii) 4
B.(i) 24, (ii) 20
C.(i) 64, (ii) 20
D.(i) 64, (ii) 16

Exam Year(s)

GATE Physics - 2018,

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.

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Consider 5 independent magnetic moments. Each moment has only two directions +m and -m. Hamiltonian of such system is given by \(x = \sum_{i=1}^{5} m_{i}\). If the total magnetization of the system is equal to m, the entropy is equal to :

A.k ln(10)
B.Zero
C.k ln(20)
D.k ln(5)

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

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.

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The number of distinct ways in which 4 particles can be distributed in 7 energy levels if (i) they are distinguishable and (ii) if they are indistinguishable bosons respectively is :

A.(i) \(\frac{7!}{4!}\) and (ii) \(\frac{7!}{4!\,3!}\)
B.(i) \(4^{7}\) and (ii) 210
C.(i) 7 ! 4 ! and (ii) \(\frac{7!}{4!\,3!}\)
D.(i) \(7^{4}\) and (ii) 210

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

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.

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A system of three indistinguishable particles has a total energy of \(4\varepsilon\). There are four single particle energy states with energies 0, \(\varepsilon\), \(2\varepsilon\) and \(3\varepsilon\). The number of microstates accessible to the system will be :

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: D

On the key, there are 3 accessible microstates.

For indistinguishable particles a microstate is fixed by how many particles occupy each level, not by which particle is where. With no limit on the number per level (Bose counting), we need three energies from {0, 1, 2, 3} in units of \(\varepsilon\) adding to 4:

\[ \begin{aligned} 4 &= 3 + 1 + 0 \\ 4 &= 2 + 2 + 0 \\ 4 &= 2 + 1 + 1 \end{aligned} \]

(4 + 0 + 0 is impossible because the highest level is \(3\varepsilon\).) That gives 3 microstates. One microstate would be the answer for fermions, which forbid the doubly occupied levels in the second and third sets. Four would come from counting an arrangement twice; distinguishable particles would give many more, since each set could be permuted.

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The volume of a perfect gas is doubled, the number of atoms N and the energy E being held constant, the change in entropy will be :

A.Nk ln V
B.2Nk ln V
C.Nk ln 2
D.½ Nk ln (2V)

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: C

The entropy change is Nk ln 2.

At constant N and energy E (free expansion of an ideal gas), only the volume term of the Sackur-Tetrode entropy changes: \(S = Nk\ln V + \) terms independent of V. So:

\[ \Delta S = Nk\ln\frac{2V}{V} = Nk\ln 2 \]

The answer depends only on the ratio of volumes, so options containing ln V or ln 2V alone cannot be a change in entropy.

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