Question

The mean free path of a helium atom in helium gas at standard temperature and pressure is 0.2 um.What is the radius of the helium atom in nanometers?

Answer 1

Answer: $0.10233nm$

Explanation:

The mean free path $\lambda$   of an atom is given by the following formula:

$\lambda=\frac{RT}{\sqrt{2} \pi d^{2}N_{A}P}$    (1)

Where:

$\lambda=0.2\mu m=0.2(10)^{-6}m$

$R=8.3145J/mol.K$ is the Universal gas constant

$T=0\°C=273.115K$ is the absolute standard temperature

$d$ is the diameter of the helium atom

$N_{A}=6.0221(10)^{23}/mol$ is the Avogadro's number

$P=1atm=101.3kPa=101.3(10)^{3}Pa=101.3(10)^{3}J/m^{3}$ absolute standard pressure

Knowing this, let's find $d$ from (1), in order to find the radius $r$ of the helium atom:

$d=\sqrt{\frac{RT}{\sqrt{2}\pi\lambda N_{A}P}}$    (2)

$d=\sqrt{\frac{(8.3145J/mol.K)(273.115K)}{\sqrt{2}\pi(0.2(10)^{-6}m)(6.0221(10)^{23}/mol)(101.3(10)^{3}J/m^{3})}}$    (3)

$d=2.0467(10)^{-10}m$    (4)

If the radius is half the diameter:

$r=\frac{d}{2}$  (5)

Then:

$r=\frac{2.0467(10)^{-10}m}{2}$  (6)

$r=1.0233(10)^{-10}m$  (7)

However, we were asked to find this radius in nanometers. Knowing $1nm=(10)^{-9}m$:

$r=1.0233(10)^{-10}m.\frac{1nm}{(10)^{-9}m}=0.10233nm$  (8)

Finally:

$r=0.10233nm$ This is the radius of the helium atom in nanometers.