91Ó°ÊÓ

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

role="math" localid="1646983518799" $\mathrm{Î\copyright }≈{\left(\frac{Ne}{q}\right)}^q..........\left(1\right)$

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

Mathematically, temperature can be defined as:

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

Where,

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

Total energy of the system is given as:

$U=q\mathrm{Î\mu }$

Where, $q$is number of energy unit.

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

$\mathrm{Î\copyright }≈{\left(\frac{Ne\mathrm{Î\mu }}{U}\right)}^{\frac{U}{\mathrm{Î\mu }}}$

Entropy is given as:

$S=k\ln \mathrm{Î\copyright }$

Here, $k$is Boltzmann constant and $\mathrm{Î\copyright }$is multiplicity.

By substituting the value of $\mathrm{Î\copyright }$in the above equation, we get,

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

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

$S=k\frac{U}{\mathrm{Î\mu }}\ln \left(\frac{Ne\mathrm{Î\mu }}{U}\right)$

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:

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

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

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

Take exponential for both the sides of the above equation,

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

This result is valid only for low temperature as $U=q\mathrm{Î\mu }≪N$.

Hence, the energy as a function of temperature can be solved to obtain$U=N\mathrm{Ï\mu }{e}^{-\mathrm{Ï\mu }/kT}$.