Question

Starting with the result of Problem 2.17, find a formula for the temperature of an Einstein solid in the limit $q\ll N$. Solve for the energy as a function of temperature to obtain $U=Nϵe^{-ϵ/kT}$ (where$ϵ$ is the size of an energy unit).

Short answer

The required expression is$U=N\epsilon e^{-\left(\frac{\epsilon }{kT}\right)}$.

Step-by-step solution

Given

Expression for the multiplicity of Einstein solid in low-temperature limit $q\ll N$is given as:

role="math" localid="1646983518799" $\Omega \approx {\left(\frac{Ne}{q}\right)}^q..........\left(1\right)$

Here, $N$is number of oscillator in solid, $q$is number of energy unit.

Explanation

Mathematically, temperature can be defined as:

$\frac{1}{T}=\left(\frac{\partial S}{\partial U}\right)..........\left(2\right)$

Where,

$\partial S$is the change in entropy and $\partial U$is the change in the internal energy of the body.

Total energy of the system is given as:

$U=q\epsilon$

Where, $q$is number of energy unit.

Equation (1) can be written by substituting the values of $q$as:

$\Omega \approx {\left(\frac{Ne\epsilon }{U}\right)}^{\frac{U}{\epsilon }}$

Entropy is given as:

$S=k\ln \Omega$

Here, $k$is Boltzmann constant and $\Omega$is multiplicity.

By substituting the value of $\Omega$in the above equation, we get,

$S=k\ln {\left(\frac{Ne\epsilon }{U}\right)}^{\frac{U}{\epsilon }}$

But $\ln \left(a^b\right)=b\ln \left(a\right)$, therefore, the above equation can be rewritten as:

$S=k\frac{U}{\epsilon }\ln \left(\frac{Ne\epsilon }{U}\right)$

Also, $\ln \left(ab\right)=\ln \left(a\right)+\ln \left(b\right)$and $\ln \left(\frac{a}{b}\right)=\ln \left(a\right)-\ln \left(b\right)$so, the above equation becomes:

$$\begin{gathered} S=\frac{Uk}{\epsilon }\ln \left(N\epsilon \right)+\ln \left(e\right)-\ln \left(U\right) \\ S=\frac{Uk}{\epsilon }\ln \left(N\epsilon \right)+1-\ln \left(U\right) \end{gathered}$$

By substituting this value of $S$in equation (2), we get,

$$\begin{gathered} \frac{1}{T}=\left(\frac{\partial S}{\partial U}\right)=\frac{\partial }{\partial U}\left[\frac{Uk}{\epsilon }\ln \left(N\epsilon \right)+1-\ln \left(U\right)\right] \\ \frac{1}{T}=\frac{k}{\epsilon }\ln \left(N\epsilon \right)+\frac{k}{\epsilon }-\frac{Uk}{U\epsilon }-\frac{k}{\epsilon }\ln \left(U\right) \\ \frac{1}{T}=\frac{k}{\epsilon }\left[\ln \right(N\epsilon )-\ln (U\left)\right] \\ \ln U=\left[\ln \left(N\epsilon \right)-\frac{\epsilon }{kT}\right] \end{gathered}$$

Take exponential for both the sides of the above equation,

$$\begin{gathered} U=e^{\left[\ln \left(N_{\epsilon }\right)-\frac{\epsilon }{kT}\right]} \\ U=e^{\ln \left(N\epsilon \right)}·e^{-\left(\frac{\epsilon }{kT}\right)} \\ U=N\epsilon e^{-\left(\frac{\epsilon }{kT}\right)} \end{gathered}$$

This result is valid only for low temperature as $U=q\epsilon \ll N$.

Final answer

Hence, the energy as a function of temperature can be solved to obtain$U=Nϵe^{-ϵ/kT}$.