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Q. 3.24

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Found in: Page 107

### An Introduction to Thermal Physics

Book edition 1st
Author(s) Daniel V. Schroeder
Pages 356 pages
ISBN 9780201380279

# Use a computer to study the entropy, temperature, and heat capacity of an Einstein solid, as follows. Let the solid contain 50 oscillators (initially), and from 0 to 100 units of energy. Make a table, analogous to Table 3.2, in which each row represents a different value for the energy. Use separate columns for the energy, multiplicity, entropy, temperature, and heat capacity. To calculate the temperature, evaluate $\Delta U/\Delta S$ for two nearby rows in the table. (Recall that $U=qϵ$ for some constant $ϵ$.) The heat capacity $\left(\Delta U/\Delta T\right)$ can be computed in a similar way. The first few rows of the table should look something like this:(In this table I have computed derivatives using a "centered-difference" approximation. For example, the temperature $.28$ is computed as $2/\left(7.15-0\right)$.) Make a graph of entropy vs. energy and a graph of heat capacity vs. temperature. Then change the number of oscillators to 5000 (to "dilute" the system and look at lower temperatures), and again make a graph of heat capacity vs. temperature. Discuss your prediction for the heat capacity, and compare it to the data for lead, aluminum, and diamond shown in Figure 1.14. Estimate the numerical value of $\epsilon$ in electron-volts, for each of those real solids.

The table can be prepared as:

The graphs can be made as:

For 50 oscillations

For 5000 oscillations

The values of $\epsilon$ are:

${\epsilon }_{\text{lead}}=6.875×{10}^{-3}\mathrm{eV}\phantom{\rule{0ex}{0ex}}{\epsilon }_{\text{alu}\mathrm{mi}nium}=2.587×{10}^{-2}\mathrm{eV}\phantom{\rule{0ex}{0ex}}{\epsilon }_{\text{diamond}}=0.19233\mathrm{eV}$

See the step by step solution

## Step 1: Given Information

The given Einstein solid contains 50 oscillators from 0 to 100 units of energy.

That is,

$N=50\phantom{\rule{0ex}{0ex}}q=1\text{to}100$

$U=qϵ$

## Step 2: Calculation for table values

The entropy of the system is given as:

$\frac{S}{k}=\mathrm{ln}\Omega \phantom{\rule{0ex}{0ex}}\frac{S}{k}=\mathrm{ln}\left(\frac{q+N-1}{⌊q⌊N-1}\right)$

The temperature is given as:

$\frac{1}{T}=\frac{\partial S}{\partial U}\phantom{\rule{0ex}{0ex}}T=\frac{\Delta U}{\Delta S}\phantom{\rule{0ex}{0ex}}T=\frac{\Delta \left(\epsilon q\right)}{\Delta \left(k\mathrm{ln}\Omega \right)}\phantom{\rule{0ex}{0ex}}\frac{kT}{ϵ}=\frac{\Delta \left(q\right)}{\Delta \left(\mathrm{ln}\Omega \right)}$

Heat capacity per oscillation is given as:

The table consisting of entropy, temperature, and heat capacity of an Einstein solid for 50 oscillations with 0 to 100 units of energy can be prepared as follows:

## Step 3: Graph

The graphs of entropy vs energy and heat capacity vs temperature can be sketched as follows:

For 50 oscillations:

The graph of entropy vs energy can be made as:

The graph of heat capacity vs temperature can be made as:

For 5000 oscillations:

The graph of entropy vs energy can be made as:

The graph of heat capacity vs temperature can be made as:

The heat capacity is substantially smaller than the values in figure 1.14, since, for 5000 oscillations with the same units of energy, the energy is lowered, lowering the system's temperature.

## Step 4: Calculation of ε

For $q=1,\Omega =1$

Solving for $\epsilon$, we get,

$U=ϵq\phantom{\rule{0ex}{0ex}}\epsilon =\frac{U}{q}\phantom{\rule{0ex}{0ex}}ϵ=\frac{Tk\left(\mathrm{ln}\Omega \right)}{q}\phantom{\rule{0ex}{0ex}}\epsilon =kT$

For lead, heat capacity is maximum at $T=80K$

${\epsilon }_{\text{lead}}=kT\phantom{\rule{0ex}{0ex}}{\epsilon }_{\text{lead}}=\left(1.38×{10}^{-23}\right)×80\phantom{\rule{0ex}{0ex}}{\epsilon }_{\text{lead}}=1.1×{10}^{-21}\mathrm{J}\phantom{\rule{0ex}{0ex}}{\epsilon }_{\text{lead}}=6.875×{10}^{-3}\mathrm{eV}$

For aluminum heat capacity is maximum at $T=300\mathrm{K}$

${\epsilon }_{alu\mathrm{min}um}=kT\phantom{\rule{0ex}{0ex}}{\epsilon }_{alu\mathrm{min}um}=\left(1.38×{10}^{-23}\right)×300\phantom{\rule{0ex}{0ex}}{\epsilon }_{alu\mathrm{min}um}=4.14×{10}^{-21}\mathrm{J}\phantom{\rule{0ex}{0ex}}{\epsilon }_{alu\mathrm{min}um}=2.587×{10}^{-2}\mathrm{eV}$

For diamond heat capacity is maximum at $T=2230\mathrm{K}$

${\epsilon }_{\text{diamond}}=kT\phantom{\rule{0ex}{0ex}}{\epsilon }_{\text{diamond}}=\left(1.38×{10}^{-23}\right)×2230\phantom{\rule{0ex}{0ex}}{\epsilon }_{\text{diamond}}=3.0774×{10}^{-20}\mathrm{J}\phantom{\rule{0ex}{0ex}}{\epsilon }_{\text{diamond}}=0.19233\mathrm{eV}$

The table of the values can be prepared as:

The graphs can be prepared as:

For 50 oscillations:

For 5000 oscillations:

The value of $\epsilon$ are:

${ϵ}_{\text{lead}}=6.875×{10}^{-3}\mathrm{eV}\phantom{\rule{0ex}{0ex}}{ϵ}_{\text{aluminium}}=2.587×{10}^{-2}\mathrm{eV}\phantom{\rule{0ex}{0ex}}{ϵ}_{\text{diamond}}=0.19233\mathrm{eV}$