Phase Space, Micro- and Macro-States MCQ Quiz - Objective Question with Answer for Phase Space, Micro- and Macro-States - Download Free PDF
Last updated on Jun 28, 2025
Latest Phase Space, Micro- and Macro-States MCQ Objective Questions
Phase Space, Micro- and Macro-States Question 1:
An isolated box of volume 𝑉 contains 5 identical, but distinguishable and noninteracting particles. The particles can either be in the ground state of zero energy or in an excited state of energy 𝜀. The ground state is non-degenerate while the excited state is doubly degenerate. There is no restriction on the number of particles that can be put in a given state. The number of accessible microstates corresponding to the macrostate of the system with energy 𝐸=2𝜀 are
Answer (Detailed Solution Below)
Phase Space, Micro- and Macro-States Question 1 Detailed Solution
Calculation:
Let the five identical distinguishable particles be A, B, C, D, E.
|
Particles in 0 |
Particles in ϵ |
Particles in ϵ |
Total Energy |
Number of Ways (Ω) |
|---|---|---|---|---|
|
A B C |
D E |
0 |
2ϵ |
5! / (3! 2! 0!) = 10 |
|
A B C |
D |
E |
2ϵ |
5! / (3! 1! 1!) = 20 |
|
A B C |
0 |
D E |
2ϵ |
5! / (3! 0! 2!) = 10 |
Hence, the total number of ways = 10 + 20 + 10 = 40.
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Phase Space, Micro- and Macro-States Question 2:
An isolated box of volume 𝑉 contains 5 identical, but distinguishable and noninteracting particles. The particles can either be in the ground state of zero energy or in an excited state of energy 𝜀. The ground state is non-degenerate while the excited state is doubly degenerate. There is no restriction on the number of particles that can be put in a given state. The number of accessible microstates corresponding to the macrostate of the system with energy 𝐸=2𝜀 are
Answer (Detailed Solution Below)
Phase Space, Micro- and Macro-States Question 2 Detailed Solution
Calculation:
Let the five identical distinguishable particles be A, B, C, D, E.
|
Particles in 0 |
Particles in ϵ |
Particles in ϵ |
Total Energy |
Number of Ways (Ω) |
|---|---|---|---|---|
|
A B C |
D E |
0 |
2ϵ |
5! / (3! 2! 0!) = 10 |
|
A B C |
D |
E |
2ϵ |
5! / (3! 1! 1!) = 20 |
|
A B C |
0 |
D E |
2ϵ |
5! / (3! 0! 2!) = 10 |
Hence, the total number of ways = 10 + 20 + 10 = 40.