Explanation

(b) Kinetic energy is given by
$K = \frac{1}{2}mv^{2} = \frac{p^{2}}{2m}$ ($\because p = mv)$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ p = \sqrt{2mK}$…(i)
Again de-Broglie wavelength is given by
$\lambda = \frac{h}{p}$…(ii)
Substituting for $p$ from Eq. (i) into Eq. (ii), we have
$\lambda = \frac{h}{\sqrt{2mK}}$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ K = \frac{h^{2}}{2m\lambda^{2}}$
Given, $h = 6.6 \times 10^{- 34}$J-s,$\ m = 9.1 \times 10^{- 3}$ kg
$\lambda = 0.3\ $nm=$0.3 \times 10^{- 9}$ m
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ K = \frac{\left( 6.6 \times 10^{- 34} \right)^{2}}{2 \times 9.1 \times 10^{- 31} \times \left( 0.3 \times 10^{- 9} \right)^{2}}$
=$\frac{26.6 \times 10^{- 19}}{1.6 \times 10^{- 19}}e$V=16.6eV