91Ó°ÊÓ

a) The fraction of conduction electrons having energies greater than the Fermi energy, $frac=0.013$

b) Fermi energy of lithium, $E_F=4.70 eV$

The electrons that have jumped to the conduction band from the valence band and are now free to move within the walls of the substance are known as conduction electrons.

Formula:

The fraction of electrons with energies greater than the Fermi energy is

$frac=\frac{3kT}{2E_F}$ ……( i )

$\mathrm{where}\mathrm{k}=8.62×{10}^{-5} \mathrm{eV}/\mathrm{K}$

Using the given data in equation (i), we get the temperature at which the 1.30% energy is greater than the Fermi energy can be calculated as follows:

$$\begin{gathered} T=\frac{2\left(frac\right)E_F}{3k} \\ =\frac{2\left(0.013\right)\left(4.70eV\right)}{8.62×{10}^{-5}eV/K} \\ =472.5K \end{gathered}$$

Hence, the value of the temperature is 472.5 K.