If the mass of neutron is $1.7x {10}^{-27} kg$, then the de-Broglie wavelength of neutron of energy 3 eV is: $(\mathrm{h}= 6.6\mathrm{x}{10}^{-34} \mathrm{Js})$

$\begin{array}{l}{m}_n=1.7\times {10}^{-27}kgE=3eV, \lambda =?\\ E=\frac{P^2}{2m}\Rightarrow P=\sqrt{2mE}\\ \lambda =\frac{h}{\sqrt{2mE}}=\frac{6.6\times {10}^{-34}}{\sqrt{2\times 1.7\times {10}^{-27}\times 3\times 1.6\times {10}^{-19}}}\\ =\frac{6.6\times {10}^{-34}}{4.04\times {10}^{-23}}\\ =1.65\times {10}^{-11}m\end{array}$