Answer
$\frac{5}{3}$
Work Step by Step
$T=20^{\circ}C = 293.15K$
$p=1 torr = 133.322 Pa$
$λ_{Ar} = 9.9\times10^{-6}cm=9.9\times10^{-8}m$
$λ_{N2} = 27.5\times10^{-6}cm=2.75\times10^{-7}m$
Boltzmann constant $k=1.38\times10^{−23}$
Using the mean free path equation $ λ=\frac{kT}{π\times\sqrt 2\times d^{2}\times p}$, we can calculate the diameter of these gases $d=\sqrt {\frac{kT}{λ\times π\times \sqrt 2 \times p}}$
Diameter of an Ar atom
$d_{Ar}=\sqrt {\frac{1.38\times10^{−23}\times293.15}{9.9\times10^{-8}\timesπ\times\sqrt 2 \times133.322}}$
$d_{Ar}=8.3\times10^{-9}$
Diameter of an $N_2$ molecule
$d_{{N_{2}}}=\sqrt {\frac{1.38\times10^{−23}\times293.15}{2.75\times10^{-7}\timesπ\times\sqrt 2 \times133.322}}$
$d_{{N_{2}}}=4.98\times10^{-9}$
Ratio of the diameter of an Ar atom to that of an $N_2$ molecule = $\frac{d_{Ar}}{d_{N_2}}$ =$ \frac{8.3\times10^{-9}}{4.98\times10^{-9}}$ = $\frac{5}{3}$
