The mean free path of a molecule of diameter $$5\times 10^{-10}m$$ at the temperature 41 °C and pressure $$1.38\times 10^{5} Pa$$, is given as ___ m (Given $$k_{B}=1.38\times 10^{-23}J/K$$.)

Find the mean free path of a molecule with diameter $$d = 5 \times 10^{-10}$$ m at temperature $$T = 41°C = 314$$ K and pressure $$P = 1.38 \times 10^5$$ Pa.

We apply the mean free path formula $$\lambda = \frac{k_B T}{\sqrt{2}\pi d^2 P}$$.

Substituting the values gives $$\lambda = \frac{1.38 \times 10^{-23} \times 314}{\sqrt{2} \times \pi \times (5 \times 10^{-10})^2 \times 1.38 \times 10^5}$$.

Calculating the numerator yields $$1.38 \times 10^{-23} \times 314 = 433.32 \times 10^{-23} = 4.3332 \times 10^{-21}$$.

For the denominator, we first compute $$d^2 = 25 \times 10^{-20}$$, so

$$\sqrt{2} \times \pi \times 25 \times 10^{-20} \times 1.38 \times 10^5 = \sqrt{2} \times \pi \times 25 \times 1.38 \times 10^{-15}$$

$$= \sqrt{2} \times 3.14159 \times 34.5 \times 10^{-15} = 1.4142 \times 108.34 \times 10^{-15} = 153.18 \times 10^{-15} = 1.5318 \times 10^{-13}$$.

Finally, $$\lambda = \frac{4.3332 \times 10^{-21}}{1.5318 \times 10^{-13}} = 2.828 \times 10^{-8} = 2\sqrt{2} \times 10^{-8}$$ m.

Hence, the mean free path is $$2\sqrt{2} \times 10^{-8}$$ m, corresponding to option 3.