$\begin {array} {1 1} (a)\;\lambda > r & \quad (b)\;\lambda < r \\ (c)\;\lambda = r & \quad (d)\;none \: of \: these \end {array}$

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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$

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