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

$1.6×10^{-10}m$ $1.6×10^{-11}m$ $1.4×10^{-10}m$ $1.4×10^{-11}m$

$1.6×10^{-11}m$

Here, $E= 3eV=3×1.6×10^{-19}J$ de Broglie wavelength, $λ=\frac{h}{\sqrt{2mE}}$ $=\frac{6.6×10^{-34}}{\sqrt{2×1.7×10^{-27}×3×1.6×10^{-19}}}=1.6×10^{-11}m$