Given, 
Wavelength of sodium light, $\mathrm{\lambda }$ = 589 nm 

We know that,   $\boxed{\mathrm{\lambda } = \frac{\mathrm{h}}{\sqrt{2\mathrm{mE}}}}$ 

$\Rightarrow$                     $\mathrm{E} = \frac{{\mathrm{h}}^2}{2{\mathrm{\lambda }}^2\mathrm{m}}$ 

(a) For electron, 

Energy, $\mathrm{E} = \frac{(6.63 \times {10}^{-34}{)}^2}{2 \times (589 \times {10}^{-9}{)}^2 \times 9 \times {10}^{-31}}$ 

               $= 7.03 \times {10}^{-25}\mathrm{J}$ 

(b) For neutron,

Energy, $\mathrm{E} = \frac{(6.63 \times {10}^{-34}{)}^2}{2 \times (589 \times {10}^{-9}{)}^2 \times 1.66 \times {10}^{-27}}$ 

               $= 3.81 \times {10}^{-28}\mathrm{J}.$

(a) Given,

$$\begin{gathered} \mathrm{Mass} \mathrm{of} \mathrm{the} \mathrm{bullet}, \mathrm{m} = 0.040 \mathrm{kg} \\ \mathrm{Velocity} \mathrm{with} \mathrm{which} \mathrm{the} \mathrm{bullet} \mathrm{is} \mathrm{travelling}, \mathrm{v} = 1 {\mathrm{kms}}^{-1} = {10}^3 {\mathrm{ms}}^{-1} \\ \mathrm{Using} \mathrm{the} \mathrm{formula} \mathrm{of} \mathrm{momentum}, \\ \boxed{\mathrm{p} = \mathrm{mv}} \\ = 0.040 \times {10}^3 \\ = 40 \mathrm{kg} {\mathrm{ms}}^{-1} \end{gathered}$$
$\therefore$ De-broglie wavelength is, 
                         $$\begin{gathered} \boxed{\mathrm{\lambda } = \frac{\mathrm{h}}{\mathrm{p}}} \\ = \frac{6.62 \times {10}^{-34}}{40} \\ = 1.7 \times {10}^{-35}\mathrm{m} \end{gathered}$$ 


(b) Mass of the ball, $\mathrm{m} = 0.060 \mathrm{kg}$ 
     Velocity with which the ball is moving, $\mathrm{v} = 1.0 {\mathrm{ms}}^{-1}$
     Momentum of the particle,  $\mathrm{p} = \mathrm{mv} = 0.060 \mathrm{kg} {\mathrm{ms}}^{-1}$

Therefore, 

De-broglie wavelength of the particle, $$\begin{gathered} \mathrm{\lambda } = \frac{\mathrm{h}}{\mathrm{p}} \\ = \frac{6.62 \times {10}^{-34}}{0.060} \\ = 1.1 \times {10}^{-32} \end{gathered}$$

(c) Mass of the dust particle, $\mathrm{m} = 1.0 \times {10}^{-9} \mathrm{kg}$
     $$\begin{gathered} \mathrm{Velocity} \mathrm{of} \mathrm{the} \mathrm{particle}, \mathrm{v}= 2.2 \mathrm{m}/\mathrm{s} \\ \mathrm{Momentum} \mathrm{of} \mathrm{the} \mathrm{particle}, \mathrm{p} = \mathrm{mv} = 2.2 \times {10}^{-9} \\ \mathrm{Therefore}, \\ \mathrm{De}-\mathrm{broglie} \mathrm{wavelength}, \mathrm{\lambda } = \frac{\mathrm{h}}{\mathrm{p}} = \frac{6.62 \times {10}^{-34}}{2.2 \times {10}^{-19}} \\ = 3 \times {10}^{-25}\mathrm{m}. \end{gathered}$$