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Consider an isolated system of $N$identical fermions, inside a container where the allowed energy levels are nondegenerate and evenly spaced.* For instance, the fermions could be trapped in a one-dimensional harmonic oscillator potential. For simplicity, neglect the fact that fermions can have multiple spin orientations (or assume that they are all forced to have the same spin orientation). Then each energy level is either occupied or unoccupied, and any allowed system state can be represented by a column of dots, with a filled dot representing an occupied level and a hollow dot representing an unoccupied level. The lowest-energy system state has all levels below a certain point occupied, and all levels above that point unoccupied. Let $畏$be the spacing between energy levels, and let be the number of energy units (each of size $11$) in excess of the ground-state energy. Assume that$q<N$. Figure 7 .8 shows all system states up to $\mathrm{q}=3$.

(a) Draw dot diagrams, as in the figure, for all allowed system states with $q=4,q=5,\text{and}q=6$. (b) According to the fundamental assumption, all allowed system states with a given value of q are equally probable. Compute the probability of each energy level being occupied, for $q=6$. Draw a graph of this probability as a function of the energy of the level. ( c) In the thermodynamic limit where $q$is large, the probability of a level being occupied should be given by the Fermi-Dirac distribution. Even though 6 is not a large number, estimate the values of $渭$and T that you would have to plug into the Fermi-Dirac distribution to best fit the graph you drew in part (b).

A representation of the system states of a fermionic sytern with evenly spaced, nondegen erate energy levels. A filled dot rep- resents an occupied single-particle state, while a hollow dot represents an unoccupied single-particle state . {d) Calculate the entropy of this system for each value of q from $0to6$, and draw a graph of entropy vs. energy. Make a rough estimate of the slope of this graph near $q=6$, to obtain another estimate of the temperature of this system at that point. Check that it is in rough agreement with your answer to part ( c).

Step by step solution

01

Part(a) Step 1: Given information

We have been given that $q=4,5,6$

02

Part(a) Step 2:Simplify

Using dot diagram

For $q=4$

For $q=5$

For localid="1651066047791" $q=6$

03

Part(b) Step 1: Given information

We have been given that $\stackrel{炉}{n}=0/11$

04

Part(b) Step 2: Simplify

Probabilities are equal to the number of black dots in each row over the total states, therefore:$\begin{array}{llllllllllll}\frac{10}{11}& \frac{10}{11}& \frac{9}{11}& \frac{8}{11}& \frac{7}{11}& \frac{6}{11}& \frac{5}{11}& \frac{4}{11}& \frac{3}{11}& \frac{2}{11}& \frac{1}{11}& \frac{1}{11}\end{array}$

Therefore, the energy of the level equals the spacing between levels$畏$.

05

part(c) Step 1: Given information

we have been given that${\stackrel{炉}{n}}_{\mathrm{FD}}=\frac{1}{e^{(系-渭)/kT}+1}$

06

part(c) Step 2: Simplify

Plotting the function and the points on the graph will be:

07

part(d) Step 1: Given information

we have been given that$\frac{S}{k}=\ln \left(惟\right)$

08

part(d) Step 2: Simplify

A plot between $qand\raisebox{1ex}{$S$}\!\left/ \!\raisebox{-1ex}{$k$}\right.$

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