Ans : (b)
$T=300\: K\: \: \: \: \: \: k=1.38 \times 10^{-23}kg m^2 / s^2K$
Average $KE,\: K’ = \large\frac{3}{2} kT$
$= \large\frac{3}{2} \times 1.38 \times 106{-23} \times 300 = 6.21 \times 10^{-21} J$
de Broglie wavelength, $ \lambda’ = \large\frac{h}{ \sqrt {(2K’mn)}}$
$= \large\frac{6.63 \times 10^{-34}}{\sqrt{(2 \times 6.21 \times 10^{-21} \times 1.66 \times 10^{-27 }) }}$
$= 1.46 \times 10^{-10}m = 0.146\: nm$