Explanation

(c) de- Broglie wavelength of electron is given by
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \lambda = \frac{h}{mv} = \frac{h}{\sqrt{2mE}}\ $
Substituting the value of $E,$ we get
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \lambda = \frac{h}{\sqrt{2meV}}$
Here m=$9.1 \times 10^{- 31}\ kg;e = 1.6 \times 10^{- 19}$C
and $h = 6.6 \times 10^{- 34}Js$
we get $\lambda = \frac{12.27}{\sqrt{V}} \times 10^{- 10} = \frac{12.27}{\sqrt{V}}\mathring{\mathrm{A}}$
The de-Broglie wavelength of electrons, when accelerated through a potential difference of 100 V will be
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \lambda = \frac{12.27}{\sqrt{100}} = 1.227\mathring{\mathrm{A}}\ $
Moreover, $\lambda = \frac{h}{p}$
$\Rightarrow P = \frac{6.6 \times 10^{- 34}}{1.227 \times 10^{- 10}} = 5.5 \times 10^{- 24}\ kg$-${ms}^{- 1}$