Explanation

(a) Since, de-Broglie wavelength is related to momentum by the relation
$\lambda = \frac{h}{P}$ (where $h$=plack’s constant )
For electron $\lambda_{e} = \frac{h}{P_{e}}$
For neutron $\ \lambda_{n} = \frac{h}{p_{n}}$
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{\lambda_{e}}{\ \lambda_{n}} = \frac{P_{n}}{P_{e}}\ \ \ \ldots(i)$
Case I since, $(KE)_{electron} = (KE)_{neutron}$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{p_{e}^{2}}{2m_{e}} = \frac{p_{n}^{2}}{2m_{n}}$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $ $\frac{p_{n}}{p_{e}} = \sqrt{\frac{m_{n}}{m_{e}}}\ \ $…(ii)
From Eqs. (i) and (ii),we get
$\frac{\lambda_{e}}{\ \lambda_{n}} = \sqrt{\frac{m_{n}}{m_{e}}}$
But $m_{n} > m_{e}$
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{m_{n}}{m_{e}} > 1$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{\lambda_{e}}{\ \lambda_{n}} \gg 1$
$\ \ \ \ \ \ \ \lambda_{e} \gg \ \lambda_{n}$
$\mathbf{case\ II}\mathbf{\ }\ If\ momenta\ are\ equal,\ then\ $
${\ \ \ \ \ \ p}_{e} = p_{n}$
From Eq.(i)
$\frac{\lambda_{e}}{\lambda_{n}} = 1$
Case III If speeds are same
$v_{e} = v_{n}$
then $\frac{\lambda_{e}}{\lambda_{n}} = \frac{p_{n}}{p_{e}} = \frac{m_{n}v_{n}}{m_{e}v_{e}} = \frac{m_{n}}{m_{e}}$
Now, $m_{n} \gg m_{e}$
$\mathbf{\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }\frac{m_{n}}{m_{e}}\mathbf{\ \gg}1$
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{\lambda_{e}}{\lambda_{n}}\ \gg 1$
${\ \lambda}_{e} \gg \ \lambda_{n}$