Six distinguishable particles are distributed over 3 non‐degenerate levels, of energies 0, ε and 2ε. The most probable value for the total energy is
6ε
Particles and Energy Levels
We are given a system containing six distinguishable particles. These particles can be distributed among three non-degenerate energy levels with energies 0, ε, and 2ε. We need to find the most probable value for the total energy of the system.
Microstates and Total Energy
For distinguishable particles, the number of microstates ($W$) corresponding to a specific distribution $(n_0, n_1, n_2)$, where $n_0$, $n_1$, and $n_2$ are the number of particles in the energy levels 0, ε, and 2ε respectively, is given by the multinomial coefficient formula:
$\text{W}(n_0, n_1, n_2) = \frac{\text{N}!}{\text{n}_0! \text{n}_1! \text{n}_2!}$
Here, the total number of particles is $N = 6$, and the sum of particles in each level must be $n_0 + n_1 + n_2 = 6$.
The total energy ($E$) for a given distribution $(n_0, n_1, n_2)$ is the sum of the energies of the particles in each level:
$\text{E}(n_0, n_1, n_2) = n_0 \times 0 + n_1 \times \epsilon + n_2 \times 2\epsilon = (n_1 + 2n_2)\epsilon$
Possible Distributions and Calculations
We list all possible combinations of $(n_0, n_1, n_2)$ that sum to 6 and calculate the number of microstates ($W$) and the total energy ($E$) for each distribution:
| Distribution (n0, n1, n2) | Number of Microstates (W) = $\frac{6!}{\text{n}_0! \text{n}_1! \text{n}_2!}$ | Total Energy (E) = (n1 + 2n2)ε |
|---|---|---|
| (6, 0, 0) | $\frac{720}{720 \times 1 \times 1} = 1$ | (0 + 0)ε = 0 |
| (5, 1, 0) | $\frac{720}{120 \times 1 \times 1} = 6$ | (1 + 0)ε = ε |
| (5, 0, 1) | $\frac{720}{120 \times 1 \times 1} = 6$ | (0 + 2)ε = 2ε |
| (4, 2, 0) | $\frac{720}{24 \times 2 \times 1} = 15$ | (2 + 0)ε = 2ε |
| (4, 1, 1) | $\frac{720}{24 \times 1 \times 1} = 30$ | (1 + 2)ε = 3ε |
| (4, 0, 2) | $\frac{720}{24 \times 1 \times 2} = 15$ | (0 + 4)ε = 4ε |
| (3, 3, 0) | $\frac{720}{6 \times 6 \times 1} = 20$ | (3 + 0)ε = 3ε |
| (3, 2, 1) | $\frac{720}{6 \times 2 \times 1} = 60$ | (2 + 2)ε = 4ε |
| (3, 1, 2) | $\frac{720}{6 \times 1 \times 2} = 60$ | (1 + 4)ε = 5ε |
| (3, 0, 3) | $\frac{720}{6 \times 1 \times 6} = 20$ | (0 + 6)ε = 6ε |
| (2, 4, 0) | $\frac{720}{2 \times 24 \times 1} = 15$ | (4 + 0)ε = 4ε |
| (2, 3, 1) | $\frac{720}{2 \times 6 \times 1} = 60$ | (3 + 2)ε = 5ε |
| (2, 2, 2) | $\frac{720}{2 \times 2 \times 2} = 90$ | (2 + 4)ε = 6ε |
| (2, 1, 3) | $\frac{720}{2 \times 1 \times 6} = 60$ | (1 + 6)ε = 7ε |
| (2, 0, 4) | $\frac{720}{2 \times 1 \times 24} = 15$ | (0 + 8)ε = 8ε |
| (1, 5, 0) | $\frac{720}{1 \times 120 \times 1} = 6$ | (5 + 0)ε = 5ε |
| (1, 4, 1) | $\frac{720}{1 \times 24 \times 1} = 30$ | (4 + 2)ε = 6ε |
| (1, 3, 2) | $\frac{720}{1 \times 6 \times 2} = 60$ | (3 + 4)ε = 7ε |
| (1, 2, 3) | $\frac{720}{1 \times 2 \times 6} = 60$ | (2 + 6)ε = 8ε |
| (1, 1, 4) | $\frac{720}{1 \times 1 \times 24} = 30$ | (1 + 8)ε = 9ε |
| (1, 0, 5) | $\frac{720}{1 \times 1 \times 120} = 6$ | (0 + 10)ε = 10ε |
| (0, 6, 0) | $\frac{720}{1 \times 720 \times 1} = 1$ | (6 + 0)ε = 6ε |
| (0, 5, 1) | $\frac{720}{1 \times 120 \times 1} = 6$ | (5 + 2)ε = 7ε |
| (0, 4, 2) | $\frac{720}{1 \times 24 \times 2} = 15$ | (4 + 4)ε = 8ε |
| (0, 3, 3) | $\frac{720}{1 \times 6 \times 6} = 20$ | (3 + 6)ε = 9ε |
| (0, 2, 4) | $\frac{720}{1 \times 2 \times 24} = 15$ | (2 + 8)ε = 10ε |
| (0, 1, 5) | $\frac{720}{1 \times 1 \times 120} = 6$ | (1 + 10)ε = 11ε |
| (0, 0, 6) | $\frac{720}{1 \times 1 \times 720} = 1$ | (0 + 12)ε = 12ε |
Density of States ($\Omega(E)$)
The most probable value for the total energy corresponds to the energy value that has the largest number of microstates associated with it. This is found by summing the $W$ values for all distributions that result in the same total energy $E$. This sum is called the density of states, $\Omega(E)$.
| Total Energy (E) | Distributions (n0, n1, n2) | Sum of Microstates ($\Omega(E) = \sum W$) |
|---|---|---|
| 0 | (6, 0, 0) | 1 |
| ε | (5, 1, 0) | 6 |
| 2ε | (5, 0, 1), (4, 2, 0) | 6 + 15 = 21 |
| 3ε | (4, 1, 1), (3, 3, 0) | 30 + 20 = 50 |
| 4ε | (4, 0, 2), (3, 2, 1), (2, 4, 0) | 15 + 60 + 15 = 90 |
| 5ε | (3, 1, 2), (2, 3, 1), (1, 5, 0) | 60 + 60 + 6 = 126 |
| 6ε | (3, 0, 3), (2, 2, 2), (1, 4, 1), (0, 6, 0) | 20 + 90 + 30 + 1 = 141 |
| 7ε | (2, 1, 3), (1, 3, 2), (0, 5, 1) | 60 + 60 + 6 = 126 |
| 8ε | (2, 0, 4), (1, 2, 3), (0, 4, 2) | 15 + 60 + 15 = 90 |
| 9ε | (1, 1, 4), (0, 3, 3) | 30 + 20 = 50 |
| 10ε | (1, 0, 5), (0, 2, 4) | 6 + 15 = 21 |
| 11ε | (0, 1, 5) | 6 |
| 12ε | (0, 0, 6) | 1 |
Most Probable Energy Value
From the density of states table, the maximum value of $\Omega(E)$ is 141. This maximum density of states occurs at a total energy of 6ε.
Therefore, the most probable value for the total energy of the system is 6ε.
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