Explanation

(a) De-Broglie wavelength of a particle is given by
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \lambda = \frac{h}{mv}\ \ $...(i)
Where h is Planck’s constant.
If kinetic energy of particle of mass m is $v$, then
$K = \frac{1}{2}mv^{2}$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ v = \ \sqrt{\frac{2K}{m}}$…(ii)
Combining Eqs. (i) and (ii), we get
$\ \lambda = \frac{h}{m\sqrt{\frac{2K}{m}}} = \frac{h}{\sqrt{2mK}}\ \ \ $…(iii)
Given $m = 9.1 \times 10^{- 31}$ kg
$\ K = 10keV = 10 \times 10^{3} \times 1.6 \times 10^{- 19}$ J
$h = 6.6 \times 10^{- 34}$ J-s
Substituting the above values in Eq. (iii), we get
$\ \lambda = \frac{6.6 \times 10^{- 34}}{\sqrt{2 \times 9.1 \times 10^{- 31} \times 10 \times 10^{3} \times 1.6 \times 10^{- 19}}}$
= $1.22 \times 10^{- 11} \approx 0.12\mathring{\mathrm{A}}\ $