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In below you showed that the multiplicity of an Einstein
Chapter 3, Problem 25P(choose chapter or problem)
In below Problem 1 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 \(U=q \epsilon\), where \(\epsilon\) 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 \(T \rightarrow \infty\) 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 dimensionlcss variable \(t=k T / \epsilon\), 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. Estimate the value of \(\epsilon\), 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 \((\epsilon / k T)^{2}\) in the final answer. When the smoke clears, you should find \(C=N k\left[1-\frac{1}{12}(\epsilon / k T)^{2}\right]\).
Questions & Answers
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QUESTION:
In below Problem 1 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 \(U=q \epsilon\), where \(\epsilon\) 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 \(T \rightarrow \infty\) 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 dimensionlcss variable \(t=k T / \epsilon\), 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. Estimate the value of \(\epsilon\), 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 \((\epsilon / k T)^{2}\) in the final answer. When the smoke clears, you should find \(C=N k\left[1-\frac{1}{12}(\epsilon / k T)^{2}\right]\).
ANSWER:Step 1 of 9
Given information:
\(\Omega(N, q) \approx\left(\frac{q+N}{q}\right)^{q}\left(\frac{q+N}{N}\right)^{N}\)
(a)
The entropy is given as,
\(S=k \ln \Omega\)
Using equation (1.1)
\(\begin{aligned}
S & =k \ln \left\{\left(\frac{q+N}{q}\right)^{q}\left(\frac{q+N}{N}\right)^{N}\right\} \\
& =k\left[q \ln \left(\frac{q+N}{q}\right)+N \ln \left(\frac{q+N}{N}\right)\right] \\
& =k[q \ln (q+N)-q \ln q+N \ln (q+N)-N \ln N] \\
& =k[(q+N) \ln (q+N)-q \ln q-N \ln N]
\end{aligned}\)
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