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Q. 3.8

Expert-verifiedFound in: Page 93

Book edition
1st

Author(s)
Daniel V. Schroeder

Pages
356 pages

ISBN
9780201380279

Starting with the result of Problem 3.5, calculate the heat capacity of an Einstein solid in the low-temperature limit. Sketch the predicted heat capacity as a function of temperature.

The required expression is ${C}_{v}=\frac{N{\epsilon}^{2}}{k{T}^{2}}{e}^{-\left(\frac{\epsilon}{kT}\right)}$.

The graph of the predicted heat capacity as a function of temperature can be sketched as follows:

The equation for Einstein solid at low temperature is calculated as:

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

Here, $N$ is number of oscillator, $\epsilon $ is the amount of energy quanta, $k$ is Boltzmann constant and $T$ is temperature.

Heat capacity at constant volume is given as:

${C}_{v}={\left(\frac{\partial U}{\partial T}\right)}_{N,V}$

Where, $U$ is internal energy.

By substututing the value of $U$ in the above equation, we get,

${C}_{v}=\frac{\partial}{\partial T}\left(N\epsilon {e}^{-\left(\frac{\epsilon}{kT}\right)}\right)\phantom{\rule{0ex}{0ex}}{C}_{v}=\frac{N{\epsilon}^{2}}{k{T}^{2}}{e}^{-\left(\frac{\epsilon}{kT}\right)}$

Consider the equation which gives the relation of the heat capacity as a function of temperature:

${C}_{v}=\frac{N{\epsilon}^{2}}{k{T}^{2}}{e}^{-\left(\frac{\epsilon}{kT}\right)}$

Now, by considering the rest other factors as a constant, heat capacity as a function of temperature can be given as:

${C}_{v}\propto \frac{1}{{T}^{2}}{e}^{-\left(\frac{1}{T}\right)}$

Based on the above relation, the graph can be plotted as below:

The required expression is derived as ${C}_{v}=\frac{N{\epsilon}^{2}}{k{T}^{2}}{e}^{-\left(\frac{\epsilon}{kT}\right)}$ and the graph showing the heat capacity as a function of temperature can be made as follows:

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