Ans : (b)
$T = 27^{\circ}C = 300K \: \: \: \: \: P=1 atm = 1.01\times10^5 Pa$
Average energy of a gas at temp. $T$, is:
$E = \large\frac{3}{2} kT$
De Broglie wavelength is :
$ \lambda = \large\frac{h}{ \sqrt2 mE}\: \: \: \: \: where,\: m $= mass of a $He$ atom
$= Atomic\: wt. / N_A$
$= \large\frac{4}{(6.023 \times 10^{23} )}$
$= 6.64 \times 10^{-27}\: kg$
S0, $ \lambda = \large\frac{h}{ \sqrt{3mkT}}$
$\large\frac{= 6.6 \times 10^{-34} }{ \sqrt{[3 \times 6.64 \times 10^{-27} \times 1.38 \times 10^{-23} \times 300]}}$
$= 0.7268 \times 10^{-10}m$
We know that,$ PV=RT$
$PV = kNT$
$\large\frac{V}{N} = \large\frac{kT}{P}\: \: \: where,V$= volume of gas, $N$ = no. of moles of gas
So, Mean separation between two atoms of gas is:
$r = \bigg( \large\frac {V}{N}\bigg)^{\large\frac{1}{3}} = \bigg( \large\frac{kT}{P} \bigg)^{\large\frac{1}{3}}$
$= \bigg[\large\frac{1.38 \times 10^{-23} \times 300} {1.01 \times 10^5 } \bigg]^{\large\frac{1}{3}} = 3.35 \times 10^{-9}m$
Hence, $ r> \lambda$