$(a)$ For what kinetic energy of a neutron will the associated de Broglie wavelength be $1.40 \times 10^{-10} \; m ?$
$(b)$ Also find the de Broglie wavelength of a neutron,in thermal equilibrium with matter,having an average kinetic energy of $\frac{3}{2} k T$ at $300 \; K$.

(N/A) Given: $\lambda = 1.40 \times 10^{-10} \; m$,$m = 1.675 \times 10^{-27} \; kg$,$h = 6.626 \times 10^{-34} \; J \cdot s$.
$(a)$ The kinetic energy $K$ is related to the de Broglie wavelength $\lambda$ by $K = \frac{p^2}{2m} = \frac{h^2}{2m\lambda^2}$.
Substituting the values: $K = \frac{(6.626 \times 10^{-34})^2}{2 \times 1.675 \times 10^{-27} \times (1.40 \times 10^{-10})^2} \approx 6.68 \times 10^{-21} \; J$.
$(b)$ The average kinetic energy is $K = \frac{3}{2} kT$.
Given $T = 300 \; K$ and $k = 1.381 \times 10^{-23} \; J/K$,$K = 1.5 \times 1.381 \times 10^{-23} \times 300 = 6.2145 \times 10^{-21} \; J$.
The de Broglie wavelength is $\lambda = \frac{h}{\sqrt{2mK}}$.
$\lambda = \frac{6.626 \times 10^{-34}}{\sqrt{2 \times 1.675 \times 10^{-27} \times 6.2145 \times 10^{-21}}} \approx 1.45 \times 10^{-10} \; m$ or $0.145 \; nm$.