Answer
$7.30\times 10^{-11}\;m$
Work Step by Step
The average kinetic energy of an atom is equal to $E_K=\frac{3}{2}kT$, where $k$ is the Boltzmann constant.
Therefore, the average kinetic energy of a helium at room temperature $(T=300\;K)$ is given by
$E_K=\frac{3}{2}\times 1.38\times 10^{-23}\times300\;J=6.21\times 10^{-21}\;J$
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 a helium atom is
$E_K=6.21\times 10^{-21}\;J$
Mass of the helium atom is
$m=4u=4\times1.66\times 10^{-27}\;kg=6.64\times 10^{-27}\;kg$
Therefore, the de Broglie wavelength the helium gas is given by
$\lambda=\frac{6.63\times 10^{-34}}{\sqrt {2\times6.64\times 10^{-27}\times6.21\times 10^{-21}}}\;m$
$\lambda=7.30\times 10^{-11}\;m$
