De-Broglie wavelength $$ \lambda=\frac{\eta}{m v}=\frac{h}{\sqrt{2 m E}} $$ Where E is kinetic energy $$ \begin{aligned} & E=\frac{3 \mathrm{kT}}{2} \text { for gas } \\ & \lambda=\frac{\mathrm{h}}{\sqrt{3 \mathrm{mkT}}}=\frac{6.6 \times 10^{-34}}{\sqrt{3 \times 9 \times 10^{-31} \times 1.38 \times 10^{-23} \times 300}} \\ & \lambda=6.26 \times 10^{-9} \mathrm{~m}=6.26 \mathrm{~nm} \end{aligned} $$ Option (1)