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Q. 3.25

Expert-verifiedFound in: Page 108

Book edition
1st

Author(s)
Daniel V. Schroeder

Pages
356 pages

ISBN
9780201380279

In Problem 2.18 you showed that the multiplicity of an Einstein solid containing N oscillators and q energy units is approximately

$\Omega (N,q)\approx {\left(\frac{q+N}{q}\right)}^{q}{\left(\frac{q+N}{N}\right)}^{N}$

(a) Starting with this formula, find an expression for the entropy of an Einstein solid as a function of N and q. Explain why the factors omitted from the formula have no effect on the entropy, when N and q are large.

(b) Use the result of part (a) to calculate the temperature of an Einstein solid as a function of its energy. (The energy is $U=q\u03f5$, where $\u03f5$ is a constant.) Be sure to simplify your result as much as possible.

(c) Invert the relation you found in part (b) to find the energy as a function of temperature, then differentiate to find a formula for the heat capacity.

(d) Show that, in the limit $T\to \infty $, the heat capacity is $C=Nk$. (Hint: When x is very small, ${e}^{x}\approx 1+x$.) Is this the result you would expect? Explain.

(e) Make a graph (possibly using a computer) of the result of part (c). To avoid awkward numerical factors, plot $C/Nk$ vs. the dimensionless variable $t=kT/\u03f5$, for t in the range from 0 to about 2. Discuss your prediction for the heat capacity at low temperature, comparing to the data for lead, aluminum, and diamond shown in Figure 1.14. Estimate the value of $\u03f5$, in electron-volts, for each of those real solids.

(f) Derive a more accurate approximation for the heat capacity at high temperatures, by keeping terms through ${x}^{3}$ in the expansions of the exponentials and then carefully expanding the denominator and multiplying everything out. Throw away terms that will be smaller than $(\u03f5/kT{)}^{2}$ in the final answer. When the smoke clears, you should find $C=Nk\left[1-\frac{1}{12}(\u03f5/kT{)}^{2}\right]$.

(a) $S=k\left[\right(q+N\left)\mathrm{ln}\right(q+N)-q\mathrm{ln}q-N\mathrm{ln}N]$

(b) $\frac{\u03f5}{kT}=\mathrm{ln}\left(\frac{U+N\epsilon}{U}\right)$

(c) $U=\frac{N\u03f5}{({e}^{\frac{\u03f5}{kT}}-1)}$, $C=\frac{{\u03f5}^{2}N{e}^{\frac{\u03f5}{kT}}}{k{T}^{2}{({e}^{\frac{\u03f5}{kT}}-1)}^{2}}$

(d) It is shown that in the limit $T\to \infty $, the heat capacity is $C=Nk$.

(e) The graph can be made as:

$\begin{array}{rl}& \frac{C}{Nk}=\frac{{\left(\frac{\u03f5}{kT}\right)}^{2}kN{e}^{\frac{\epsilon}{kT}}}{{({e}^{\frac{\epsilon}{kT}}-1)}^{2}}\\ & {\epsilon}_{\text{lead}}=6.875\times {10}^{-3}\mathrm{eV}\\ & {\epsilon}_{\text{aluminum}}=2.587\times {10}^{-2}\mathrm{eV}\\ & {\epsilon}_{\text{diamond}}=0.19233\mathrm{eV}\end{array}$

(f) $C\approx Nk(1-\frac{{\left(\frac{\u03f5}{kT}\right)}^{2}}{12})$

The multiplicity of an Einstein solid is given as:

$\Omega (N,q)\approx {\left(\frac{q+N}{q}\right)}^{q}{\left(\frac{q+N}{N}\right)}^{N}$

The entropy is given as:

$S=k\mathrm{ln\Omega}$

By substituting the given value in the above equation, we get,

$S=k\mathrm{ln}\left\{{\left(\frac{q+N}{q}\right)}^{q}{\left(\frac{q+N}{N}\right)}^{N}\right\}\phantom{\rule{0ex}{0ex}}S=k[q\mathrm{ln}\left(\frac{q+N}{q}\right)+N\mathrm{ln}\left(\frac{q+N}{N}\right)]\phantom{\rule{0ex}{0ex}}S=k\left[q\mathrm{ln}\right(q+N)-q\mathrm{ln}q+N\mathrm{ln}(q+N)-N\mathrm{ln}N]\phantom{\rule{0ex}{0ex}}S=k\left[\right(q+N\left)\mathrm{ln}\right(q+N)-q\mathrm{ln}q-N\mathrm{ln}N]$

The multiplicity for Einstein solid can be given as:

$\mathrm{\Omega}=\frac{\lfloor q+N-1}{\lfloor q\lfloor N-1}\phantom{\rule{0ex}{0ex}}\mathrm{\Omega}=\frac{\lfloor q+N-1}{\lfloor q\lfloor N-1}\times \frac{N(q+N)}{N(q+N)}\phantom{\rule{0ex}{0ex}}\mathrm{\Omega}=\frac{\lfloor q+N}{\lfloor q\lfloor N}\times \frac{N}{(q+N)}$

But,

Stirling approximation of large n is given as:

localid="1647403887405" $\lfloor n=\sqrt{2\pi n}{n}^{n}{e}^{-n}$

Hence,

$\mathrm{\Omega}=\frac{N}{(q+N)}\times \frac{\sqrt{2\pi (q+N)}(q+N{)}^{q+N}{e}^{-(q+N)}}{\sqrt{2\pi \left(q\right)}\left(q{)}^{q}{e}^{-\left(q\right)}\sqrt{2\pi \left(N\right)}\right(N{)}^{N}{e}^{-\left(N\right)}}\phantom{\rule{0ex}{0ex}}\mathrm{\Omega}=\frac{N}{(q+N)}\times \sqrt{\frac{q+N}{qN}}\times \frac{(q+N{)}^{q+N}{e}^{-(q+N)}}{{q}^{q}{N}^{N}{e}^{-(q+N)}}\phantom{\rule{0ex}{0ex}}\mathrm{\Omega}=\sqrt{\frac{N}{2\pi q(q+N)}{\left(\frac{q+N}{q}\right)}^{q}{\left(\frac{q+N}{N}\right)}^{N}}$

For large $q$ and $N$, when compared to the power terms, the square root terms are small. As they have a negligible effect, they can be neglected. Hence, the approximate value can be written as:

$\mathrm{\Omega}=\sqrt{\frac{N}{2\pi q(q+N)}}{\left(\frac{q+N}{q}\right)}^{q}{\left(\frac{q+N}{N}\right)}^{N}\phantom{\rule{0ex}{0ex}}\mathrm{\Omega}\approx {\left(\frac{q+N}{q}\right)}^{q}{\left(\frac{q+N}{N}\right)}^{N}$

Hence, the entropy of an Einstein solid containing $N$ oscillators and $q$ energy units is given as:

$S=k\left[\right(q+N\left)\mathrm{ln}\right(q+N)-q\mathrm{ln}q-N\mathrm{ln}N]$

For large values of $q$ and $N$, the square root terms are small when compared to the power terms. Hence they are neglected as they have a very negligible effect on the entropy. Therefore the approximate value of multiplicity is given as:

$\mathrm{\Omega}\approx {\left(\frac{q+N}{q}\right)}^{q}{\left(\frac{q+N}{N}\right)}^{N}$

The total energy of the system is given as:

$U=q\epsilon $

The entropy of the system is given as:

$S=k\left[\right(q+N\left)\mathrm{ln}\right(q+N)-q\mathrm{ln}q-N\mathrm{ln}N]$

The temperature can be given as:

$\frac{1}{T}=\frac{\partial S}{\partial U}\phantom{\rule{0ex}{0ex}}\frac{1}{T}=\frac{\partial \left[k\right[(q+N)\mathrm{ln}(q+N)-q\mathrm{ln}q-N\mathrm{ln}N\left]\right]}{\epsilon \partial q}\phantom{\rule{0ex}{0ex}}\frac{1}{T}=\frac{k}{\u03f5}\left[\mathrm{ln}\right(q+N)+1-\mathrm{ln}q-1-0]\phantom{\rule{0ex}{0ex}}\frac{1}{T}=\frac{k}{\u03f5}\mathrm{ln}\left[\frac{q+N}{q}\times \frac{\u03f5}{\u03f5}\right]\phantom{\rule{0ex}{0ex}}\frac{\u03f5}{kT}=\mathrm{ln}\left(\frac{U+N\epsilon}{U}\right)$

The temperature as a function of its energy can be given as:

$\frac{\u03f5}{kT}=\mathrm{ln}\left(\frac{U+N\epsilon}{U}\right)$

The temperature as a function of energy is given as:

$\frac{\u03f5}{kT}=\mathrm{ln}\left(\frac{U+N\epsilon}{U}\right)$

As given:

$\begin{array}{r}\frac{\u03f5}{kT}=\mathrm{ln}\left(\frac{U+N\epsilon}{U}\right)\end{array}$

On simplifying, we get,

${e}^{\frac{\u03f5}{kT}}=\frac{U+N\u03f5}{U}\phantom{\rule{0ex}{0ex}}U{e}^{\frac{\u03f5}{kT}}-U=N\epsilon \phantom{\rule{0ex}{0ex}}U=\frac{N\u03f5}{\left({e}^{\frac{\u03f5}{kT}}-1\right)}$

Now, differentiating the energy with respect to temperature to get the heat capacity as:

$C=\frac{\partial U}{\partial T}\phantom{\rule{0ex}{0ex}}C=\frac{\partial}{\partial T}\left[\frac{N\epsilon}{\left({e}^{\frac{\u03f5}{kT}}-1\right)}\right]\phantom{\rule{0ex}{0ex}}C=N\epsilon \left(-{\left({e}^{\frac{\u03f5}{kT}}-1\right)}^{-2}\right){e}^{\frac{\u03f5}{kT}}\left(\frac{\u03f5}{k{T}^{2}}\right)\phantom{\rule{0ex}{0ex}}C=\frac{{\u03f5}^{2}N{e}^{\frac{\u03f5}{kT}}}{k{T}^{2}{\left({e}^{\frac{\u03f5}{kT}}-1\right)}^{2}}$

The energy as a function of temperature can be given as:

$U=\frac{N\u03f5}{\left({e}^{\frac{\u03f5}{kT}}-1\right)}$

Expression of Heat capacity is:

$C=\frac{{\u03f5}^{2}N{e}^{\frac{\u03f5}{kT}}}{k{T}^{2}{\left({e}^{\frac{\u03f5}{kT}}-1\right)}^{2}}$

Heat capacity in terms of temperature is given as:

$C=\frac{{\u03f5}^{2}N{e}^{\frac{\u03f5}{kT}}}{k{T}^{2}{\left({e}^{\frac{\u03f5}{kT}}-1\right)}^{2}}$

For very small values of $x,{e}^{x}\approx 1+x$.

By expanding the exponential for higher temperatures, we get

${e}^{\frac{\u03f5}{kT}}\approx 1+\frac{\epsilon}{kT}$

The heat capacity for very large values of $\mathrm{T}$ can be given as:

$C=\frac{{\u03f5}^{2}N{e}^{\frac{\u03f5}{kT}}}{k{T}^{2}{({e}^{\frac{\u03f5}{kT}}-1)}^{2}}\phantom{\rule{0ex}{0ex}}C\approx \frac{{\u03f5}^{2}N(1+\frac{\u03f5}{kT})}{k{T}^{2}{\left(\frac{\u03f5}{kT}\right)}^{2}}\phantom{\rule{0ex}{0ex}}C\approx Nk(1+\frac{\u03f5}{kT})\phantom{\rule{0ex}{0ex}}C\approx Nk$

The derived expression is analogous to the expression of heat capacity at high temperatures.

So, $C\approx Nk$ is approximately the expected result.

The heat capacity is given as:

$C=\frac{{\u03f5}^{2}N{e}^{\frac{\u03f5}{kT}}}{k{T}^{2}{\left({e}^{\frac{\u03f5}{kT}}-1\right)}^{2}}$

The heat capacity for low temperature can be given as:

$C=\frac{{\u03f5}^{2}N{e}^{\frac{\u03f5}{kT}}}{k{T}^{2}{\left({e}^{\frac{\epsilon}{kT}}-1\right)}^{2}}\phantom{\rule{0ex}{0ex}}C=\frac{{\u03f5}^{2}kN{e}^{\frac{\epsilon}{kT}}}{{k}^{2}{T}^{2}{\left({e}^{\frac{\epsilon}{kT}}-1\right)}^{2}}\phantom{\rule{0ex}{0ex}}\frac{C}{Nk}=\frac{{\left(\frac{\epsilon}{kT}\right)}^{2}{e}^{\frac{\epsilon}{kT}}}{{\left({e}^{\frac{\epsilon}{kT}}-1\right)}^{2}}$

Based on the derived equation, the graph of $\frac{C}{Nk}$ versus $t=\frac{kT}{\u03f5}$ can be made as follows:

For lead, heat capacity is maximum at $T=80\mathrm{K}$

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

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

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

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

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

The heat capacity for low temperatures can be given as:

$\frac{C}{Nk}=\frac{{\left(\frac{\epsilon}{kT}\right)}^{2}{e}^{\frac{\epsilon}{kT}}}{{\left({e}^{\frac{\epsilon}{kT}}-1\right)}^{2}}$

The graph can be made as:

The values of $\epsilon $ are as follows:

$\begin{array}{l}{\epsilon}_{\text{lead}}=6.875\times {10}^{-3}\mathrm{eV}\\ {\epsilon}_{\text{aluminum}}=2.587\times {10}^{-2}\mathrm{eV}\\ {\epsilon}_{\text{diamond}}=0.19233\mathrm{eV}\end{array}$

The graph of heat capacity versus temperature for lead, aluminum, and diamond is approximately anomalous as in figure 1.14.

The heat capacity is given as:

Let,

$x=\frac{\epsilon}{kT}$

The heat capacity can be modified as:

$\begin{array}{l}C=\frac{{\epsilon}^{2}N{e}^{\frac{\epsilon}{kT}}}{k{T}^{2}{({e}^{\frac{\u03f5}{kT}}-1)}^{2}}\\ \frac{C}{Nk}=\frac{{\left(\frac{\u03f5}{kT}\right)}^{2}{e}^{\frac{\u03f5}{kT}}}{{({e}^{\frac{\u03f5}{kT}}-1)}^{2}}\\ \frac{C}{Nk}=\frac{(x{)}^{2}{e}^{x}}{{({e}^{x}-1)}^{2}}\end{array}$

Now,

The Taylor expansion is given as:

${e}^{x}=1+x+\frac{{x}^{2}}{2!}+\frac{{x}^{3}}{3!}+....$

Hence, the above equation becomes:

$\begin{array}{l}\frac{C}{Nk}=\frac{(x{)}^{2}(1+x+\frac{{x}^{2}}{2!}+\frac{{x}^{3}}{3!})}{{(x+\frac{{x}^{2}}{2!}+\frac{{x}^{3}}{3!})}^{2}}\\ \frac{C}{Nk}=\frac{(1+x+\frac{{x}^{2}}{2!}+\frac{{x}^{3}}{3!})}{{(1+\frac{x}{2!}+\frac{{x}^{2}}{3!})}^{2}}\\ \frac{C}{Nk}=(1+x+\frac{{x}^{2}}{2!}+\frac{{x}^{3}}{3!}){(1+\frac{x}{2!}+\frac{{x}^{2}}{3!})}^{-2}\end{array}$

Using the Taylor series, the above equation becomes:

$\frac{C}{Nk}=\left(1+x+\frac{{x}^{2}}{2}+\frac{{x}^{3}}{6}\right)\left(\frac{5{x}^{2}}{12}-x+1\right)$

Solving further and neglecting the higher powers than 2, we get,

$C\approx Nk\left(1-\frac{{\left(\frac{\u03f5}{kT}\right)}^{2}}{12}\right)$

Hence, the required expression is:

$C\approx Nk\left(1-\frac{{\left(\frac{\u03f5}{kT}\right)}^{2}}{12}\right)$

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