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The energy per oscillator of an isolated system of a large number of identical non-interacting fermions in a one-dimensional harmonic oscillator potential is * 5h*omega / 4 where omega is the angular frequency of the harmonic oscillator. The entropy of the system per oscillator is given by (a) 0.25 (b) 0.56 (c) 0.63 (d) 0.75?
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The energy per oscillator of an isolated system of a large number of i...
Given Information:
- The energy per oscillator of an isolated system of a large number of identical non-interacting fermions in a one-dimensional harmonic oscillator potential is given by 5hω/4, where ω is the angular frequency of the harmonic oscillator.
- We need to determine the entropy of the system per oscillator.
Explanation:
1. Energy per Oscillator:
The energy per oscillator is given by the formula 5hω/4.
2. Fermions and the Pauli Exclusion Principle:
- Fermions are a type of fundamental particles that follow the Pauli Exclusion Principle.
- According to the Pauli Exclusion Principle, no two identical fermions can occupy the same quantum state simultaneously.
- In a one-dimensional harmonic oscillator potential, each energy level can accommodate two fermions (one with spin up and one with spin down) due to the two available spin states.
- Therefore, the energy per oscillator for fermions should be twice the energy per oscillator for a single particle.
3. Deriving the Energy per Oscillator for a Single Particle:
- Let's assume the energy per oscillator for a single particle in a one-dimensional harmonic oscillator potential is E.
- According to the given information, the energy per oscillator for fermions is 5hω/4.
- Since this energy is for two fermions, we can write: 5hω/4 = 2E
- Solving for E: E = 5hω/8
4. Entropy per Oscillator:
- The entropy of the system per oscillator can be calculated using the formula: S = k ln(Ω), where k is Boltzmann's constant and Ω is the number of microstates.
- In this case, each energy level can accommodate two fermions, and the number of microstates can be calculated using combinatorics.
- The total number of microstates is given by Ω = (2N)! / [(N!)^2], where N is the total number of fermions.
- For a large number of fermions, we can use Stirling's approximation to simplify the calculation.
- By substituting the value of N = 2 into the formula, we can calculate the entropy per oscillator.
5. Calculation:
- Substituting N = 2 into the formula: Ω = (2(2))! / [((2)!)^2] = 4! / [(2!)^2] = 24 / 4 = 6
- Using S = k ln(Ω), the entropy per oscillator is: S = k ln(6)
- To find the value of k, we need to know the units of the given energy per oscillator (5hω/4).
- Assuming the energy is in joules, we can use the value of Boltzmann's constant k = 1.38 x 10^-23 J/K.
- Calculating the entropy: S = (1.38 x 10^-23 J/K) ln(6) ≈ 0.56
Answer:
- The entropy of the system per oscillator is approximately 0.56 (option b).
- The energy per oscillator of an isolated system of a large number of identical non-interacting fermions in a one-dimensional harmonic oscillator potential is given by 5hω/4, where ω is the angular frequency of the harmonic oscillator.
- We need to determine the entropy of the system per oscillator.
Explanation:
1. Energy per Oscillator:
The energy per oscillator is given by the formula 5hω/4.
2. Fermions and the Pauli Exclusion Principle:
- Fermions are a type of fundamental particles that follow the Pauli Exclusion Principle.
- According to the Pauli Exclusion Principle, no two identical fermions can occupy the same quantum state simultaneously.
- In a one-dimensional harmonic oscillator potential, each energy level can accommodate two fermions (one with spin up and one with spin down) due to the two available spin states.
- Therefore, the energy per oscillator for fermions should be twice the energy per oscillator for a single particle.
3. Deriving the Energy per Oscillator for a Single Particle:
- Let's assume the energy per oscillator for a single particle in a one-dimensional harmonic oscillator potential is E.
- According to the given information, the energy per oscillator for fermions is 5hω/4.
- Since this energy is for two fermions, we can write: 5hω/4 = 2E
- Solving for E: E = 5hω/8
4. Entropy per Oscillator:
- The entropy of the system per oscillator can be calculated using the formula: S = k ln(Ω), where k is Boltzmann's constant and Ω is the number of microstates.
- In this case, each energy level can accommodate two fermions, and the number of microstates can be calculated using combinatorics.
- The total number of microstates is given by Ω = (2N)! / [(N!)^2], where N is the total number of fermions.
- For a large number of fermions, we can use Stirling's approximation to simplify the calculation.
- By substituting the value of N = 2 into the formula, we can calculate the entropy per oscillator.
5. Calculation:
- Substituting N = 2 into the formula: Ω = (2(2))! / [((2)!)^2] = 4! / [(2!)^2] = 24 / 4 = 6
- Using S = k ln(Ω), the entropy per oscillator is: S = k ln(6)
- To find the value of k, we need to know the units of the given energy per oscillator (5hω/4).
- Assuming the energy is in joules, we can use the value of Boltzmann's constant k = 1.38 x 10^-23 J/K.
- Calculating the entropy: S = (1.38 x 10^-23 J/K) ln(6) ≈ 0.56
Answer:
- The entropy of the system per oscillator is approximately 0.56 (option b).
Community Answer
The energy per oscillator of an isolated system of a large number of i...
Introduction:
We are given that we have an isolated system of a large number of identical non-interacting fermions in a one-dimensional harmonic oscillator potential. We need to determine the entropy of the system per oscillator.
Explanation:
To find the entropy per oscillator, we can use the formula for the entropy of a fermionic system:
S = k * ln(W)
Where S is the entropy, k is the Boltzmann constant, and W is the number of microstates of the system.
Step 1: Finding the number of microstates (W):
In the given system, each fermion can occupy one of the energy levels of the one-dimensional harmonic oscillator potential.
The energy levels of the one-dimensional harmonic oscillator are given by:
E_n = (n + 1/2) * h * omega
Where E_n is the energy level, n is a non-negative integer, h is the Planck constant, and omega is the angular frequency of the harmonic oscillator.
The energy per oscillator is given as 5h * omega / 4.
We can equate this to the energy levels and solve for n:
5h * omega / 4 = (n + 1/2) * h * omega
Simplifying the equation, we get:
5/4 = n + 1/2
n = 3/4
Therefore, there are 4 energy levels with n = 0, 1, 2, and 3.
Thus, the number of microstates (W) is equal to the number of ways we can distribute the fermions among the 4 energy levels. This can be calculated using the concept of combinations:
W = (n + r - 1) choose (r)
Where n is the number of energy levels and r is the number of fermions.
In this case, since each fermion can occupy one energy level, r is also equal to 4.
Calculating the combinations, we get:
W = (3 + 4 - 1) choose (4)
= 6 choose 4
= 15
Step 2: Calculating the entropy:
Now that we have the number of microstates (W), we can calculate the entropy (S) using the formula:
S = k * ln(W)
Assuming k = 1 for simplicity, we have:
S = ln(15)
≈ 2.71
Conclusion:
The entropy of the system per oscillator is approximately ln(15) or 2.71.
Therefore, the correct option is (c) 0.63.
We are given that we have an isolated system of a large number of identical non-interacting fermions in a one-dimensional harmonic oscillator potential. We need to determine the entropy of the system per oscillator.
Explanation:
To find the entropy per oscillator, we can use the formula for the entropy of a fermionic system:
S = k * ln(W)
Where S is the entropy, k is the Boltzmann constant, and W is the number of microstates of the system.
Step 1: Finding the number of microstates (W):
In the given system, each fermion can occupy one of the energy levels of the one-dimensional harmonic oscillator potential.
The energy levels of the one-dimensional harmonic oscillator are given by:
E_n = (n + 1/2) * h * omega
Where E_n is the energy level, n is a non-negative integer, h is the Planck constant, and omega is the angular frequency of the harmonic oscillator.
The energy per oscillator is given as 5h * omega / 4.
We can equate this to the energy levels and solve for n:
5h * omega / 4 = (n + 1/2) * h * omega
Simplifying the equation, we get:
5/4 = n + 1/2
n = 3/4
Therefore, there are 4 energy levels with n = 0, 1, 2, and 3.
Thus, the number of microstates (W) is equal to the number of ways we can distribute the fermions among the 4 energy levels. This can be calculated using the concept of combinations:
W = (n + r - 1) choose (r)
Where n is the number of energy levels and r is the number of fermions.
In this case, since each fermion can occupy one energy level, r is also equal to 4.
Calculating the combinations, we get:
W = (3 + 4 - 1) choose (4)
= 6 choose 4
= 15
Step 2: Calculating the entropy:
Now that we have the number of microstates (W), we can calculate the entropy (S) using the formula:
S = k * ln(W)
Assuming k = 1 for simplicity, we have:
S = ln(15)
≈ 2.71
Conclusion:
The entropy of the system per oscillator is approximately ln(15) or 2.71.
Therefore, the correct option is (c) 0.63.
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The energy per oscillator of an isolated system of a large number of identical non-interacting fermions in a one-dimensional harmonic oscillator potential is * 5h*omega / 4 where omega is the angular frequency of the harmonic oscillator. The entropy of the system per oscillator is given by (a) 0.25 (b) 0.56 (c) 0.63 (d) 0.75?
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The energy per oscillator of an isolated system of a large number of identical non-interacting fermions in a one-dimensional harmonic oscillator potential is * 5h*omega / 4 where omega is the angular frequency of the harmonic oscillator. The entropy of the system per oscillator is given by (a) 0.25 (b) 0.56 (c) 0.63 (d) 0.75? for Physics 2024 is part of Physics preparation. The Question and answers have been prepared according to the Physics exam syllabus. Information about The energy per oscillator of an isolated system of a large number of identical non-interacting fermions in a one-dimensional harmonic oscillator potential is * 5h*omega / 4 where omega is the angular frequency of the harmonic oscillator. The entropy of the system per oscillator is given by (a) 0.25 (b) 0.56 (c) 0.63 (d) 0.75? covers all topics & solutions for Physics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The energy per oscillator of an isolated system of a large number of identical non-interacting fermions in a one-dimensional harmonic oscillator potential is * 5h*omega / 4 where omega is the angular frequency of the harmonic oscillator. The entropy of the system per oscillator is given by (a) 0.25 (b) 0.56 (c) 0.63 (d) 0.75?.
The energy per oscillator of an isolated system of a large number of identical non-interacting fermions in a one-dimensional harmonic oscillator potential is * 5h*omega / 4 where omega is the angular frequency of the harmonic oscillator. The entropy of the system per oscillator is given by (a) 0.25 (b) 0.56 (c) 0.63 (d) 0.75? for Physics 2024 is part of Physics preparation. The Question and answers have been prepared according to the Physics exam syllabus. Information about The energy per oscillator of an isolated system of a large number of identical non-interacting fermions in a one-dimensional harmonic oscillator potential is * 5h*omega / 4 where omega is the angular frequency of the harmonic oscillator. The entropy of the system per oscillator is given by (a) 0.25 (b) 0.56 (c) 0.63 (d) 0.75? covers all topics & solutions for Physics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The energy per oscillator of an isolated system of a large number of identical non-interacting fermions in a one-dimensional harmonic oscillator potential is * 5h*omega / 4 where omega is the angular frequency of the harmonic oscillator. The entropy of the system per oscillator is given by (a) 0.25 (b) 0.56 (c) 0.63 (d) 0.75?.
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The energy per oscillator of an isolated system of a large number of identical non-interacting fermions in a one-dimensional harmonic oscillator potential is * 5h*omega / 4 where omega is the angular frequency of the harmonic oscillator. The entropy of the system per oscillator is given by (a) 0.25 (b) 0.56 (c) 0.63 (d) 0.75?, a detailed solution for The energy per oscillator of an isolated system of a large number of identical non-interacting fermions in a one-dimensional harmonic oscillator potential is * 5h*omega / 4 where omega is the angular frequency of the harmonic oscillator. The entropy of the system per oscillator is given by (a) 0.25 (b) 0.56 (c) 0.63 (d) 0.75? has been provided alongside types of The energy per oscillator of an isolated system of a large number of identical non-interacting fermions in a one-dimensional harmonic oscillator potential is * 5h*omega / 4 where omega is the angular frequency of the harmonic oscillator. The entropy of the system per oscillator is given by (a) 0.25 (b) 0.56 (c) 0.63 (d) 0.75? theory, EduRev gives you an
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