(a) According to kinetic interpretation of temperature
$\ \ \ \ \ \ \ \ \ \ \ \ \ Ek\ = \left( = \frac{1}{2}{mv}^{2} \right) = \frac{3}{2}\ kT$
Given :$\ \ \ \ E_{i}$= $10.2\ eV$ = 10.2 $\times$ 1.6 $\times 10^{- 19}\ J$
So, $\ \ \ \ \frac{3}{2}\ kT = 10.2\ \times 1.6\ \times \ 10^{- 19}\ J$
Or $\ \ \ \ \ \ \ \ \ \ T =$ $\frac{2}{3} \times \frac{10.2\ \times \ 1.6\ \times 10^{- 19}}{k}$
= $\frac{2}{3} \times \frac{10.2\ \times \ 1.6\ \times 10^{- 19}}{1.38\ \times 10^{- 23}}$ $= \ 7.9\ \times 10^{4}\ K$