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ANSWER :
In collision of electrons only a fraction close to the Fermi surface will undergo a change in energy. That fraction is \(\sim n\Big({\frac{kT}{\epsilon_F}}\Big)\) of the electrons. Thus
\[ \text{Mean free path } \lambda = \frac{\epsilon_F}{n(k)\sigma^2} >> \frac{1}{n\sigma^2}.\]
Further \(\lambda \propto \frac{1}{T}\), as \(T\) increases \(\lambda\) will decrease.
Consider an isolated system of $N$ non-interacting particles occupying two states of energies $-\epsilon$ and $+\epsilon$. The energy of the system is $E$. Let $x=\displaystyle{\frac{E}{N\epsilon}}.$
- Show that the entropy of the system is given by\footnote{HINT : Let $n_1$ and $n_2$ denote the number of particles in the two states of energy $-\epsilon$ and $+\epsilon$ respectively. We have $\widetilde{\Omega}=N!/(n_1!n_2!)$; $S=k_B\ln\widetilde{\Omega}$; Calculate $n_1$ and $n_2$ by solving : $n_1+n_2=N$ and $n_2\epsilon-n_1\epsilon=E$.} $$ S(E)=Nk_B\left[\left(\frac{1+x}{2}\right)\ln\left(\frac{2}{1+x} \right)+\left(\frac{1-x}{2}\right)\ln\left(\frac{2}{1-x}\right)\right] $$
- Show that ${\displaystyle \beta=\frac{1}{k_BT}=\frac{1}{2\epsilon}\ln\left(\frac{1-x}{1+x}\right)}$
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