Given $\lambda=589 \mathrm{~nm}=589 \times 10^{-9} \mathrm{~m}$
De Broglie wavelength $\lambda=\frac{\mathrm{h}}{\mathrm{p}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mE}}}$
$$
\begin{aligned}
&\left(\because \frac{1}{2} \mathrm{mv}^2=\mathrm{E} \Rightarrow \mathrm{m}^2 \mathrm{v}^2=2 \mathrm{mE}\right. \\
&\Rightarrow \mathrm{p}=\sqrt{2 \mathrm{mE}}) \Rightarrow \mathrm{E}=\frac{\mathrm{h}^2}{2 \mathrm{~m} \lambda^2} \\
&\text { (a) } \begin{array}{l}
\text { For } \mathrm{e}^{-}: \lambda=589 \mathrm{~nm}=589 \times 10^{-9} \mathrm{~m} \\
\mathrm{~m}=9.1 \times 10^{-27} \mathrm{~kg}
\end{array}
\end{aligned}
$$
$\mathrm{E}=\frac{\mathrm{h}^2}{2 \mathrm{~m} \lambda^2}$
$=\frac{\left(6.6 \times 10^{-34}\right)^2}{2 \times 9.1 \times 10^{-31} \times\left(589 \times 10^{-9}\right)^2}$
$=\frac{43.56 \times 10^{-19}}{6313962}=6.94 \times 10^{-25} \mathrm{~J}$.
(b) For neutron, $\mathrm{n}^0: \lambda=589 \mathrm{~nm}=589 \times 10^{-9} \mathrm{~m}$,
$\mathrm{m}=1.66 \times 10^{-27} \mathrm{~kg}$
$\Rightarrow E=\frac{h^2}{2 m \lambda^2}=\frac{\left(6.6 \times 10^{-34}\right)^2}{2 \times 1.66 \times 10^{-27} \times\left(589 \times 10^{-9}\right)^2}$
$=\frac{43.56 \times 10^{-68}}{1151777.7 \times 10^{-45}}=3.81 \times 10^{-28} \mathrm{~J}$.