91Ó°ÊÓ

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

That is,

$$\begin{gathered} N=50 \\ q=1\text{to}100 \end{gathered}$$

$U=q\mathrm{Ï\mu }$

The entropy of the system is given as:

$$\begin{gathered} \frac{S}{k}=\ln \mathrm{Î\copyright } \\ \frac{S}{k}=\ln \left(\frac{q+N-1}{⌊q⌊N-1}\right) \end{gathered}$$

The temperature is given as:

$$\begin{gathered} \frac{1}{T}=\frac{∂S}{∂U} \\ T=\frac{\mathrm{Δ}U}{\mathrm{Δ}S} \\ T=\frac{\mathrm{Δ}\left(\mathrm{Î\mu }q\right)}{\mathrm{Δ}\left(k\ln \mathrm{Î\copyright }\right)} \\ \frac{kT}{\mathrm{Ï\mu }}=\frac{\mathrm{Δ}\left(q\right)}{\mathrm{Δ}\left(\ln \mathrm{Î\copyright }\right)} \end{gathered}$$

Heat capacity per oscillation is given as:

$$\begin{gathered} \frac{C}{N}=\frac{\mathrm{Δ}U}{N\mathrm{Δ}T} \\ \frac{{\displaystyle C}}{{\displaystyle N}}=\frac{\mathrm{Δ}\left(\mathrm{Î\mu }q\right)}{N\mathrm{Δ}\left(\frac{\mathrm{Δ}\left(\mathrm{Î\mu }q\right)}{\mathrm{Δ}\left(k\ln \mathrm{Î\copyright }\right)}\right)} \\ \frac{C}{Nk}=\frac{\mathrm{Δ}\left(q\right)}{N\mathrm{Δ}\left(\frac{\mathrm{Δ}\left(q\right)}{\mathrm{Δ}\left(\ln \mathrm{Î\copyright }\right)}\right)} \end{gathered}$$

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:

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.

For $q=1,\mathrm{Î\copyright }=1$

Solving for $\mathrm{Î\mu }$,we get,

$$\begin{gathered} U=\mathrm{Ï\mu }q \\ \mathrm{Î\mu }=\frac{U}{q} \\ \mathrm{Ï\mu }=\frac{Tk\left(\ln \mathrm{Î\copyright }\right)}{q} \\ \mathrm{Î\mu }=kT \end{gathered}$$

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

$$\begin{gathered} {\mathrm{Î\mu }}_{\text{lead}}=kT \\ {\mathrm{Î\mu }}_{\text{lead}}=\left(1.38×{10}^{-23}\right)×80 \\ {\mathrm{Î\mu }}_{\text{lead}}=1.1×{10}^{-21}\mathrm{J} \\ {\mathrm{Î\mu }}_{\text{lead}}=6.875×{10}^{-3}\mathrm{eV} \end{gathered}$$

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

$$\begin{gathered} {\mathrm{Î\mu }}_{alu\min um}=kT \\ {\mathrm{Î\mu }}_{alu\min um}=\left(1.38×{10}^{-23}\right)×300 \\ {\mathrm{Î\mu }}_{alu\min um}=4.14×{10}^{-21}\mathrm{J} \\ {\mathrm{Î\mu }}_{alu\min um}=2.587×{10}^{-2}\mathrm{eV} \end{gathered}$$

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

$$\begin{gathered} {\mathrm{Î\mu }}_{\text{diamond}}=kT \\ {\mathrm{Î\mu }}_{\text{diamond}}=\left(1.38×{10}^{-23}\right)×2230 \\ {\mathrm{Î\mu }}_{\text{diamond}}=3.0774×{10}^{-20}\mathrm{J} \\ {\mathrm{Î\mu }}_{\text{diamond}}=0.19233\mathrm{eV} \end{gathered}$$

The table of the values can be prepared as:

The graphs can be prepared as:

For 50 oscillations:

For 5000 oscillations:

The value of $\mathrm{Î\mu }$are:

$$\begin{gathered} {\mathrm{Ï\mu }}_{\text{lead}}=6.875×{10}^{−3}\mathrm{eV} \\ {\mathrm{Ï\mu }}_{\text{aluminium}}=2.587×{10}^{−2}\mathrm{eV} \\ {\mathrm{Ï\mu }}_{\text{diamond}}=0.19233\mathrm{eV} \end{gathered}$$