Question 1

a) A lattice is divided into two sub-lattices consisting of <math>N_1</math> and <math>N_2</math> sites respectively <math>(N_

Unknown macro: {tot}

=N_1+ N_2)</math>. Placing atoms of type <math>A</math> on sub-lattice <math>2</math> has an energy penalty of <math>\epsilon_A</math> but no penalty on sub-lattice <math>1</math>. Specie <math>B</math> can sit on both sub-lattices without any energy penalty. Atoms of both species are assumed to be noninteracting. Find the distribution of species <math>A</math> and <math>B</math> that maximizes the configurational entropy. Distribution: the number of <math>A</math> and <math>B</math> on each sub-lattice.

Entropy is maximum at <math>x=0.5</math> for both sub-lattices because this concetration generates the maximum number of configurations (note this is unconstrained maximization). Energy penalty does not contribute.

b) Are Fermi-Dirac particles distinguishable or indistinguishable? Briefly motivate your answer.

They can be either depending on the situation. If they can be labeled (for example localized in a solid or on a surface) they are distinguishable. If they cannot be labeled, they are indistingiushable (for example a fermion gas).

c) Sketch the Fermi-Dirac, Bose-Einstein, and Maxwell-Boltzmann average occupation numbers, <math>f(\epsilon)</math>, as a function of particle energy for <math>T>0</math>. Clearly mark the figure <math>f(\epsilon)=1</math>.

Unable to render embedded object: File (Occupation_probability_vs_energy.PNG) not found.

</center>

Question 2

A dilute ideal gas of <math>N</math> point-like dipole particles is placed in a constant electric field <math>E</math> at constant temperature <math>T</math> (<math>T</math> is not low) and volume <math>V</math>. Each particle has a polarization <math>p_0</math> which can align itself parallel or anti-parallel with the electric field. The particles do not interact with each other. As a reminder, the differential work performed while changing the polarization of a substance is given by <math>\delta W = E dP</math>, where <math>P</math> is the total polarization of the substance.

a) Find the proper Legendre transform of the energy and derive an expression for the characteristic potential of this system. Also give the total differential of the characteristic potential.

<math>dE = TdS - pdV + \mu dN + \epsilon dP</math>

<math>\wedge (N, V, T, \epsilon) = E(S, V, N, P) - TS - P \epsilon</math>

<math>d \wedge = -S dT - pdV + \mu dN - Pd\epsilon</math>

</center>

b) Write down an expression for the partition function for this system and clearly indicate which states you are summing over.

<math>-\beta \wedge = \frac

Unknown macro: {S}

Unknown macro: {k}

- \beta E + \beta P \epsilon</math>

<math>\gamma = \sum_

Unknown macro: {nu}

e^{-\beta \epsilon_

+ \beta p_

Unknown macro: {nu}

\epsilon}</math>

</center>

The term <math>\nu</math> is a sum over single particle states with all possible energies and polarizations. A system partition function is expressed below.

<math>\Gamma = \frac{ \left ( \sum_

e^{-\beta \epsilon_

Unknown macro: {nu}

+ \beta \epsilon p_{\nu}} \right )^N }

Unknown macro: {N!}

</math>

<math>p_

= \pm p_o</math>

<math>\epsilon_

Unknown macro: {nu}

= \frac

Unknown macro: {h^2}

Unknown macro: {8ma^2}

\left (n_x^2 +n_y^2 + n_z^2 \right )</math>

</center>

To a first approximation, translational energy eigenvalues are decoupled from the dipole's interaction with the electric field.

<math>\Gamma = \frac{ \left ( e^

Unknown macro: {beta epsilon p_o}

+ e^{-\beta \epsilon p_o} \right )^N}

Unknown macro: {N!}

q_t^N</math>

</center>

c) Write down an equation of state for the polarization expressed in terms of the partition function. There is no need to substitute the expression for the partition function as derived in part b).

<math>d \wedge = -S dT - pdV + \mu dN - P d \epsilon</math>

<math>\wedge = -kT \ln \gamma</math>

<math>P = 0 \left ( \frac

Unknown macro: {partial wedge}

Unknown macro: {partial epsilon}

\right )_

Unknown macro: {T, V, N}

</math>

<math>P = kT \left ( \frac

Unknown macro: {partial ln Gamma }

\right )_

Unknown macro: {T, V, N}

</math>

</center>

d) Derive a general expression for the fluctuations of the polarization of the gas (i.e. <math>\overline

Unknown macro: {P^2}

- \overline P^2 =</math> ?. Leave the answer in terms of general thermodynamic quantities.

<math>\frac

Unknown macro: {partial (overline P Gamma)}

Unknown macro: {partial epsilon}

= \frac

Unknown macro: {partial}

\left ( \sum_

P_

Unknown macro: {nu}

e^{-\beta (E_

-P_

Unknown macro: {nu}

\epsilon ) \right )</math>

<math>\frac

Unknown macro: {partial overline P}

Unknown macro: {partial epsilon}

+ \overline P \sum_

\beta P_

Unknown macro: {nu}

e^{-\beta (E_

- P_

Unknown macro: {nu}

\epsilon) } = \sum_

\beta P_

Unknown macro: {nu}

^{2 e}{-\beta (E_

- P_

Unknown macro: {nu}

\epsilon) }</math>

<math>\frac

Unknown macro: {partial overline P}

Unknown macro: {partial epsilon}

+ \overline P \frac{\sum_

\beta P_

Unknown macro: {nu}

e^{-\beta (E_

- P_

Unknown macro: {nu}

\epsilon) }}

Unknown macro: {Gamma}

= \frac{\sum_

\beta P_

Unknown macro: {nu}

^{2 e}{-\beta (E_

- P_

Unknown macro: {nu}

\epsilon) }}

Unknown macro: {Gamma}

</math>

<math>\overline

Unknown macro: {P^2}

- \overline P^2 = kT \frac

Unknown macro: {partial overline P}

Unknown macro: {partial epsilon}

</math>

</center>

Question 3

A gas with <math>N</math> non-interacting particles is enclosed in a chamber with volume <math>V</math> and kept at constant temperature <math>T</math> (<math>T</math> is not low). Each particle has two energy levels, separated by an energy <math>\epsilon</math>. The lower level is two-fold degenerate and the higher level is <math>\Omega</math> fold degenerate (<math>\Omega>>N</math>).

a) Calculate the internal energy of the system in terms of <math>\epsilon</math>,<math>N</math>,<math>\Omega</math>, and <math>T</math>.

Since absolute energy reference is arbitrary, set ground state level energy equal to zero.

<math>q_

Unknown macro: {single}

= \sum_

e^{-\beta E_{\nu}}</math>

<math>q_

Unknown macro: {single}

= 2 + \Omega e^{- \beta \epsilon}</math>

<math>Q_

Unknown macro: {system}

\frac{ \left (q_

\right )^N }

Unknown macro: {N!}

</math>

<math>\overline E = kT^2 \left ( \frac

Unknown macro: {partial ln Q}

Unknown macro: {partial T}

\right )_

Unknown macro: {N, V}

</math>

<math>\overline E = \frac{N \epsilon \Omega e^{\beta \epsilon}}{2 + \Omega e^{\beta \epsilon}}</math>

</center>

b) Calculate the heat capacity of the system in terms of the same quantities as part (a).

<math>C_V = \left ( \frac

Unknown macro: {partial overline E}

\right )_

Unknown macro: {N, V}

</math>

<math>C_V = \left ( \frac

Unknown macro: {epsilon}

Unknown macro: {kT}

\right )^{2 \frac{2Nk \Omega e}{\beta \epsilon}}{\left ( 2 + \Omega e^{\beta \epsilon} \right )^2</math>

</center>

c) What values does the heat capacity approach in the limit of high and low temperatures? (The limits can be studied even though the heat capacity is derived for a moderate/high temperature case.)

<math>x = \frac

Unknown macro: {kT}

</math>

<math>C_V \prop x^2 \frac{e^{-x}}{\left ( 2 + \Omega e^{-x} \right )^2</math>

<math>\lim_

Unknown macro: {T right 0}

C_v = \lim_

Unknown macro: {x right infty}

C_v</math>

<math>\lim_

C_v = 0</math>

<math>\lim_

Unknown macro: {T right infty}

C_v = \lim_

Unknown macro: {x right 0}

C_v</math>

<math>\lim_

C_v = 0</math>

</center>

This is a general result of any system with an upper bound on the available energy states.

Question 4

A system with <math>N</math> (non-interacting or interacting) particles is occupying a volume <math>V</math> at constant <math>T</math>.

NOTE: the energy eigenvalue spectrum is the set of energies <math>

Unknown macro: {e_1,e_2,..,e_i,...}

</math>. The index <math>i</math> extends over all eigenvectors of the <math>N</math>-particle hamiltonian.

a) Suppose we reversibly perturb the temperature of the system, keeping <math>N</math> and <math>V</math> fixed. Check all system properties that will change due to this perturbation.

____ Energy eigenvalue spectrum ____ State probabilities

</center>

The state probabilities change

b) Suppose we reversibly perturb the volume of the system, keeping <math>N</math> and <math>T</math> fixed. Check all system properties that will change due to this perturbation.

____ Energy eigenvalue spectrum ____ State probabilities

</center>

The energy eigenvalue spectrum changes

c) Suppose we add one particle to the system, keeping <math>T</math> and <math>V</math> fixed. Check all system properties that will change due to this perturbation.

____ Energy eigenvalue spectrum ____ State probabilites

</center>

Both the energy eigenvalue spectrum and state probabilities change. Energy levels are generally functions of <math>N</math> and <math>V</math>, which is why it would change when adding a particle. One way to thank about whether the state probabilities would change is to consider a wavefunctin of <math>N</math> particles to which we would add another particle. The wavefunction then needs to be renormalized and if you work it out, the probability for the particle to enter the ground state will change by just a factor of <math>N+1</math> (even if the particles are non-interacting...). When considering interacting particles, it is more obvious that the probabilities will change when a particle is added.

Question 5

Researchers @ MIT investigating materials for hydrogen storage have recently discovered a new element hydroamorium (<math>Hr</math>) showing promise as a hydrogen storage material. The available experimental evidence is summarized as follows:

Hydrogen is incorporated as an interstitial and for each <math>Hr</math> atom, there is one interstitial site available for hydrogen to occupy
when <math>H</math> is incorporated into the crystal structure of <math>Hr</math>, it bonds (with energy <math>e_0</math>) to a single <math>Hr</math> and vibrates as an independent harmonic oscillator (with frequency <math>\omega_H</math>)in one direction��
when no hydrogen is incorporated, each <math>Hr</math> atom vibrates as an independent harmonic oscillator with frequency <math>\omega_0</math> in each of three available directions
If bonded to a hydrogen, the vibrational frequency of <math>Hr</math> changes from <math>\omega_0</math> to <math>\omega_1</math>.

Assume that the <math>H</math> atoms do not interact with one another.

DATA:

mass of <math>Hr</math> atom <math>= m_
Unknown macro: {Hr}
</math>
- mass of <math>H</math> atom <math>=m_
 Unknown macro: {H}
 </math>
- <math>H</math> vibration frequency <math>= \omega_H</math>
- <math>Hr</math> vibration frequency <math>= \omega_o</math>
- <math>Hr</math> vibration frequency when bonded to <math>H</math> <math>=</math> <math>\omega_1</math>
- Absorption energy of <math>H</math> (per atom) <math>=e_0</math>
(a) for a crystal consisting of <math>N=</math>��number of incorporated <math>H</math> atoms��, and <math>M=</math>��number of <math>Hr</math> atoms�� <math>(N < M)</math> at constant <math>T</math> and <math>V</math> what are the available excitations/fluctuations?




The data given indicate configurational and vibrational excitations (although the actual solid will have other available excitations). The energy of the system in equilibrium with its environment will fluctuate.




(b) in the same conditions, what is the characteristic potential and appropriate ensemble ?




A characteristic potential with constant <math>N</math>, <math>M</math>, <math>V</math>, and <math>T</math> is expressed below. It is the Hemholtz free energy corresponding to a canonical ensemble.

<center>

 

<math>F = E - TS</math>

 

<math>dF = -S dT - p dV + \mu_H dN + \mu_
dM</math>

</center>

(c) What is the chemical potential of hydrogen, when <math>N < M</math> hydrogen atoms are incorporated?

A chemical potential is expressed below.

<math>dF = -SdT - pdV + \mu_H dN + \mu_

Unknown macro: {Hr}

dM</math>

<math>\left ( \frac

Unknown macro: { partial F}

Unknown macro: {partial N}

\right )_

Unknown macro: {T, V, M}

= \mu_H</math>

<math>Z = \sum_

Unknown macro: {nu}

\exp \beta E_

</math>

</center>

The energy of any microstate of the system decomposes into an additive sum of terms since individual vibrational states are decoupled from one another as well as different possible arrangements of <math>H</math> atoms. It is possible to write the total energy as below.

<math>E_

Unknown macro: {nu}

= \sum_

Unknown macro: {i=1}

^M \Gamma_i e_0 + \sum_

Unknown macro: {j=1}

^3 \sum_

^

Unknown macro: {M-N}

\hbar \omega_0 \left ( n_

Unknown macro: {ij}

+ \frac

Unknown macro: {1}

Unknown macro: {2}

\right ) + \sum_

Unknown macro: {j=1}

^3 \sum_

Unknown macro: {i=1}

^

Unknown macro: {N}

\hbar \omega_1 \left ( n'_

+ \frac

Unknown macro: {1}

Unknown macro: {2}

\right ) + \sum_

Unknown macro: {i=1}

^

\hbar \omega_H \left ( n_

Unknown macro: {i}

+ \frac

Unknown macro: {1}

Unknown macro: {2}

\right )</math>

</center>

The term <math>\Gamma_i</math> indicates the occupancy (by hydrogen) of the ith interstitial site and the <math>n_

Unknown macro: {ij}

</math>'s and <math>n_i</math>'s are quantum numbers associated with various available vibrational modes. Because of an additive separation, the partition function of the system factors.

<math>Z = z_

Unknown macro: {config}

z_

Unknown macro: {vib}

^

Unknown macro: {(Hr)}

z_

^

Unknown macro: {(Hr')}

z_

Unknown macro: {vib}

^

Unknown macro: {H}

</math>

<math>z_

= \sum_

Unknown macro: {arrangements}

\exp \beta e_o N</math>

<math>z_

Unknown macro: {config}

= \frac

Unknown macro: {M!}

Unknown macro: {N!(M-N)!}

\exp \beta e_o N</math>

</center>

As stated, <math>Hr</math> atoms vibrate in three directions, while <math>H</math> atoms oscillate in only one.

<math>x_

Unknown macro: {vib}

^H = \left [2 \sinh \left ( \frac

Unknown macro: {beta hbar omega_H}

\right ) \right]^N</math>

<math>z_

Unknown macro: {vib}

^H = (z_H)^N</math>

<math>x_

^

Unknown macro: {(Hr)}

= \left [2 \sinh \left ( \frac

Unknown macro: {beta hbar omega_o}

Unknown macro: {2}

\right ) \right]^

Unknown macro: {3(M-N)}

</math>

<math>z_

Unknown macro: {vib}

^{H = (z_o)}

</math>

<math>x_

Unknown macro: {vib}

^

Unknown macro: {(Hr')}

= \left [2 \sinh \left ( \frac

Unknown macro: {beta hbar omega_1}

\right ) \right]^

Unknown macro: {3N}

</math>

<math>z_

Unknown macro: {vib}

^{H = (z_1)}

</math>

</center>

Combine everything.

<math>Z = \frac

Unknown macro: {M!}

Unknown macro: {N!(M-N)!}

\exp [-\beta \epsilon_o N] (z_H)^{N (z_0)}

Unknown macro: {3(M-N)}

(z_1)^N</math>

</center>

Find the chemical potential of hydrogen.

<math>\mu_H = - kT \left ( \frac

Unknown macro: {partial ln(Z)}

\right )_

Unknown macro: {T, V, M}

</math>

<math>\mu_H = kT \left [ \beta e_o - \left ( \frac

Unknown macro: {partial}

Unknown macro: {partial N}

\ln [ N! (M-N)!] \right )_

Unknown macro: {T, M, V}

- 3 \ln(z_o) + 3 \ln(z_1) + 3 \ln(z_H) \right ]</math>

</center>

Use Stirling's approximation of \ln(N!) and express in terms of <math>x</math>.

<math>\mu_H = e_o + kT \left [\ln \left ( \frac

Unknown macro: {x}

Unknown macro: {1-x}

\right ) + 3 \ln \left ( \frac

Unknown macro: {z_o}

Unknown macro: {z_1}

\right ) - \ln (z_H) \right]</math>

<math>x=\frac

Unknown macro: {N}

Unknown macro: {M}

</math>

</center>

Child pages

3.20 Exam QuestionsExam22005

Question 1

Question 2

Question 3

Question 4

Question 5