Question

Consider a two-level system with energy states \( +\epsilon \) and \( -\epsilon \). The number of particles at \( +\epsilon \) level is \( N+ \) and the number of particles at \( -\epsilon \) level is \( N- \). The total energy of the system is \( E \) and the total number of particles is \( N = N+ + N- \). In the thermodynamic limit, the inverse of the absolute temperature of the system is: 
(Given: \( \ln N! \approx N \ln N - N \))

• A. \( \frac{k_B}{2 \epsilon} \ln \left[ \frac{N-}{E/\epsilon} \frac{N+}{E/\epsilon} \right] \)
• B. \( \frac{k_B}{\epsilon} \ln N \)
• C. \( \frac{k_B}{2 \epsilon} \ln N \)
• D. \( \frac{k_B}{\epsilon} \ln \left[ \frac{N-}{E/\epsilon} \frac{N+}{E/\epsilon} \right] \)

Hint:

For a two-level system, the inverse temperature can be determined by considering the relationship between the particle distribution, the energy, and the partition function.

Answer 1

The Correct Option is A

We are given a two-level system where the energies of the two states are \( +\epsilon \) and \( -\epsilon \). The number of particles at each energy level is \( N+ \) and \( N- \), and the total energy of the system is \( E \). We are asked to find the inverse of the absolute temperature. 
1. The partition function: The partition function for this system is given by: \[ Z = e^{-\beta \epsilon} + e^{\beta \epsilon} \] where \( \beta = \frac{1}{k_B T} \) is the inverse temperature. 
2. Average number of particles in each state: The average number of particles in the state \( +\epsilon \) is proportional to the Boltzmann factor \( e^{-\beta \epsilon} \), and the average number in the state \( -\epsilon \) is proportional to \( e^{\beta \epsilon} \). Therefore, we have: \[ N+ = \frac{N}{Z} e^{-\beta \epsilon} \] \[ N- = \frac{N}{Z} e^{\beta \epsilon} \] 
3. Using the thermodynamic relation: The total energy of the system is \( E = \epsilon N+ - \epsilon N- \). Substituting for \( N+ \) and \( N- \) from the above equations, we get: \[ E = \epsilon \left( \frac{N}{Z} e^{-\beta \epsilon} - \frac{N}{Z} e^{\beta \epsilon} \right) \] Simplifying this expression, we get the relation for the energy in terms of the temperature \( T \). 
4. Finding the inverse temperature: Using the above relation and the logarithmic approximation for large \( N \), the inverse temperature \( \beta = \frac{1}{k_B T} \) is given by: \[ \beta = \frac{1}{k_B} \ln \left[ \frac{N-}{E/\epsilon} \frac{N+}{E/\epsilon} \right] \] Therefore, the inverse temperature is: \[ \frac{1}{T} = \frac{k_B}{2 \epsilon} \ln \left[ \frac{N-}{E/\epsilon} \frac{N+}{E/\epsilon} \right] \] Thus, the correct answer is (A).