Question

Derive an expression for specific heat of solids on the basis of Einstein’s model. How the specific heat does depend on temperature?

Answer 1

Answer:

The specific heat approaches zero exponentially as $T\rightarrow 0.$

Explanation:  

Einstein employed the model of the behaviour of specific heat capacity, $C_{V}$ at low temperatures assuming that all the normal modes oscillate freely at the same frequency, $\omega$.

In Einstein’s theory, the crystal lattice structure of a solid comprising of N atoms is an assemble of 3N distinguishable one-dimensional oscillators.

The mean energy of the solid in terms of 3N distinguishable one-dimensional oscillators is given by

$\bar E=3N\hbar \omega\Big(\frac{1}{2} +\frac{1}{e^{\hbar \omega/kT} -1} \Big)$

where k is the Boltzmann constant and $\hbar$ is the  reduced Planck's constant.

The molar heat capacity at constant volume is given by  

$C_{V} =\Big(\frac{d\bar E}{dT} \Big)_{V}=\Bigg(\frac{d}{dT}\bigg( 3N\hbar \omega\Big(\frac{1}{2} +\frac{1}{e^{\hbar \omega/kT} -1} \Big)\bigg) \Bigg)_{V}$

$=3N\hbar \omega\bigg(0- \frac{1}{({e^{\hbar \omega/kT} -1} )^{2}} \Big(-e^{\hbar \omega/kT}\times \frac{\hbar \omega}{kT^{2} } \Big)\bigg)$

$=3Nk \big(\frac{\hbar \omega}{kT}\big)^{2} \bigg( \frac{e^{\hbar \omega/kT}}{({e^{\hbar \omega/kT} -1} )^{2}} \bigg)$

Replacing $\frac{\hbar \omega}{k}=\theta_{E}$, where  $\theta_{E}$ is called as Einstein temperature and $Nk=R$, the Gas constant

$C_{V}=3R \big(\frac{\theta_{E}}{T}\big)^{2} \bigg( \frac{e^{\theta_{E}/T}}{({e^{\theta_{E}/T} -1} )^{2}} \bigg)$  

Case 1: If the temperature is very high, i.e., $T > > \theta_{E}$,

then $\frac{\theta_{E}}{T} < < 1$ and $e^{\theta_{E}/T} =1$

Therefore, $C_{V}=3R$

Case 2: If the temperature is very low , i.e., $T < < \theta_{E}$,  

then  $\frac{\theta_{E}}{T} > > 1$ and $e^{\theta_{E}/T} -1\approx e^{\theta_{E}/T}$

Therefore, $C_{V}=3R \big(\frac{\theta_{E}}{T}\big)^{2} \big( -e^{\theta_{E}/T}} \big)$

Hence, the specific heat approaches zero exponentially as $T\rightarrow 0.$