Answer
$1.455\times 10^{-10}\;m$
Work Step by Step
The de Broglie wavelength $(\lambda)$ is express by the formula,
$\lambda=\frac{h}{p}$,
where $h$ is the Planck's constant and $p$ is the momentum of the moving particle.
$p$ can be expressed in terms of kinetic energy $E_K$
$p=\sqrt {2mE_K}$, where $m$ is the mass of neutron.
Therefore,
$\lambda=\frac{h}{\sqrt {2mE_K}}$
The average kinetic energy of the neutron is
$E_K=6.21\times 10^{-21}\;J$
Mass of the neutron is
$m=1.67\times 10^{-27}\;kg$
Therefore, the de Broglie wavelength the neutron is given by
$\lambda=\frac{6.63\times 10^{-34}}{\sqrt {2\times1.67\times 10^{-27}\times6.21\times 10^{-21}}}\;m$
$\lambda=1.455\times 10^{-10}\;m$
