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Q 6.1

Expert-verifiedFound in: Page 224

Book edition
1st

Author(s)
Daniel V. Schroeder

Pages
356 pages

ISBN
9780201380279

Consider a system of two Einstein solids, where the first "solid" contains just a single oscillator, while the second solid contains 100 oscillators. The total number of energy units in the combined system is fixed at 500. Use a computer to make a table of the multiplicity of the combined system, for each possible value of the energy of the first solid from 0 units to 20. Make a graph of the total multiplicity vs. the energy of the first solid, and discuss, in some detail, whether the shape of the graph is what you would expect. Also plot the logarithm of the total multiplicity, and discuss the shape of this graph.

Therefore, the first graph is exponential decaying of -q_{A. }In second graph the slope is $-\u03f5/kT$ where $\u03f5$ is the size of energy unit and T is the temperature.

A system of two Einstein solids, where the first "solid" contains just a single oscillator, while the second solid contains 100 oscillators. The total number of energy units in the combined system is fixed at 500.

Consider two Einstein solids, one of which has N_{A}=1 oscillators and the other of which includes N_{B}=100 oscillators. Because there are 500 energy units, the following formula is used:

${q}_{A}+{q}_{B}=500\left(1\right)$

We'll need to make a table of the system's overall multiplicity. The total multiplicity is calculated as follows:

${\Omega}_{\text{tot.}}=\Omega \left({N}_{A},{q}_{A}\right)\Omega \left({N}_{B},{q}_{B}\right)$

Where,

$\Omega \left({N}_{A},{q}_{A}\right)=\frac{\left({q}_{A}+{N}_{A}-1\right)!}{{q}_{A}!\left({N}_{A}-1\right)!}\Omega \left({N}_{B},{q}_{B}\right)=\frac{\left({q}_{B}+{N}_{B}-1\right)!}{{q}_{B}!\left({N}_{B}-1\right)!}$

Thus,

${\Omega}_{tot.}=\frac{\left({q}_{A}+{N}_{A}-1\right)!}{{q}_{A}!\left({N}_{A}-1\right)!}\frac{\left({q}_{B}+{N}_{B}-1\right)!}{{q}_{B}!\left({N}_{B}-1\right)!}$

Substitute from (1) with q_{B}

${\Omega}_{tot.}=\frac{\left({q}_{A}+{N}_{A}-1\right)!}{{q}_{A}!\left({N}_{A}-1\right)!}\frac{\left(499-{q}_{A}+{N}_{B}\right)!}{\left(500-{q}_{A}\right)!\left({N}_{B}-1\right)!}$

Substitute with N_{A} and N_{B }

${\Omega}_{\text{tot.}}=\frac{\left(599-{q}_{A}\right)!}{\left(500-{q}_{A}\right)!\left(99\right)!}$

Using python, table is generated, the code used is:

The table is:

$\begin{array}{cc}{q}_{A}& {\Omega}_{\text{tot}}\\ 1& 1.852\times {10}^{115}\\ 2& 1.546\times {10}^{115}\\ 3& 1.290\times {10}^{115}\\ 4& 1.076\times {10}^{115}\\ 5& 8.975\times {10}^{114}\\ 6& 7.481\times {10}^{114}\\ 7& 6.234\times {10}^{114}\\ 8& 5.193\times {10}^{114}\\ 9& 4.325\times {10}^{114}\\ 10& 3.600\times {10}^{114}\\ 11& 2.996\times {10}^{114}\\ 12& 2.492\times {10}^{114}\\ 13& 2.073\times {10}^{114}\\ 14& 1.723\times {10}^{114}\\ 15& 1.432\times {10}^{114}\\ 16& 1.190\times {10}^{114}\\ 17& 9.882\times {10}^{113}\\ 18& 8.204\times {10}^{113}\\ 19& 6.808\times {10}^{113}\\ 20& 5.648\times {10}^{113}\end{array}$

To plot the graph between ${q}_{A}\text{and}{\Omega}_{tot}$, the code is

The graph plotted is:

Now we need to plot a graph between ${q}_{A}\text{and}\mathrm{ln}\left(\Omega \right)$, the code used is:

The graph plotted is:

Therefore, the first graph is exponential decaying of -q_{A. }In second graph the slope is $-\u03f5/kT$ where $\u03f5$ is the size of energy unit and T is the temperature.

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