Physics Group - II (3110018)

Here, $\mathrm{h}$ = Plank constant $= 6.62 \times {10}^{-34}$

Here, $\mathrm{m}$ = Mass of the electron $= 9.1 \times {10}^{-31}$

Here, $\mathrm{a}$ = Side of the lattice $= 3 \times {10}^{-10}$

For first Brillouin zone $\mathrm{K} =\pm \frac{\mathrm{\pi } }{ \mathrm{a} }$

We know, $\mathrm{p} = \hslash \mathrm{k}$

$\Rightarrow \mathrm{p} = \pm \frac{{\mathrm{h}}^{ }}{2 \mathrm{\pi } }·\frac{\mathrm{\pi }}{\mathrm{a}}$

$\Rightarrow \mathrm{p} = \frac{6.62 \times {10}^{-34}}{ 2 \times 3 \times {10}^{-10} }$

$\Rightarrow \mathrm{p} =1.1 \times {10}^{-24} \mathrm{kg}·\mathrm{m} / \sec$

Now, energy of free electron:

$\mathrm{E} = \frac{{\mathrm{P}}^2}{ 2 \mathrm{m} }$

$\Rightarrow \mathrm{E} = \frac{{\left( 1.1 \times {10}^{-24}\right)}^2}{ 2 \times 9.1 \times {10}^{-31} }$

$\Rightarrow \mathrm{E} = 6.6 \times {10}^{-19} \mathrm{Joule}$

$\Rightarrow \mathrm{E} = \frac{6.6 \times {10}^{-19}}{1.6 \times {10}^{-19}}$

$\Rightarrow \mathrm{E} = 4.2 \mathrm{eV}$