As given that $\mathrm{d}=0.1 \mathrm{~nm}$, (spacing distance).
Now, by Bragg's law of diffraction
$$
\theta=30^{\circ} \Rightarrow \mathrm{n}=1
$$
$2 \mathrm{~d} \sin \theta=\mathrm{n} \lambda$ condition for $\mathrm{n}$ th maxima is
$2 \mathrm{~d} \sin \theta=\mathrm{n} \lambda \Rightarrow 2 \times 0.1 \times \sin 30=1 \lambda$
$$
\Rightarrow \lambda=0.1 \mathrm{~nm} \Rightarrow 10^{-10} \mathrm{~m}
$$
Now, $\lambda=\frac{\mathrm{h}}{\mathrm{mv}}=\frac{\mathrm{h}}{\mathrm{p}} \quad\left(\because \mathrm{p}=\frac{\mathrm{hc}}{\lambda}\right.$ or $\left.\lambda=\frac{\mathrm{hc}}{\mathrm{p}}\right)$
$$
\begin{aligned}
&\Rightarrow \quad \mathrm{p}=\frac{\mathrm{h}}{\lambda}=\frac{6.62 \times 10^{-34}}{10^{-10}} \quad(\because \mathrm{p}=\mathrm{mv}) \\
&\Rightarrow \quad=6.62 \times 10^{-24} \mathrm{~kg}-\mathrm{m} / \mathrm{s}
\end{aligned}
$$

$$
\begin{aligned}
\text { Now, } \mathrm{KE} &=\frac{1}{2} \mathrm{mv}^2=\frac{1}{2} \frac{\mathrm{m}^2 \mathrm{v}^2}{\mathrm{~m}}=\frac{1}{2} \frac{\mathrm{p}^2}{\mathrm{~m}} \\
&=\frac{1}{2} \times \frac{\left(6.62 \times 10^{-24}\right)^2}{1.67 \times 10^{-27}} \mathrm{~J} \\
&=\frac{1}{2} \times \frac{\left(6.62 \times 10^{-24}\right)^2}{1.67 \times 10^{-27} \times 1.6 \times 10^{-19}} \mathrm{eV} \\
\Rightarrow \lambda &=0.082 \mathrm{eV}
\end{aligned}
$$