Ans : (c)
$T=300\: K$
molecular mass of $N_2=2 \times 14.0076\: u = 28.0152 \times 1.66 \times 10^{-27} kg$
$= 28.0152 \times 1.66 \times 10^{-27} kg$
So, $\large\frac{1}{2} mv^2_{rms} = \large\frac{3}{2} kT$
$ \Rightarrow V_{rms} = \sqrt{ \bigg( \large\frac{3kt}{m} \bigg)}$
Hence, de Broglie wavelength, $ \lambda = h/mv_{rms}$
$= \large\frac{h}{\sqrt{3mkT}}$
$= \large\frac{6.63 \times 10^{-34}}{\sqrt{[3 \times 28.0152 \times 1.66 \times 10^{-27} \times 1.38 \times 10^{-23} \times 300]}}$
$= .028 \times 10^{-9}\: m$