Answer: [tex]0.10233nm[/tex]
Explanation:
The mean free path [tex]\lambda[/tex] of an atom is given by the following formula:
[tex]\lambda=\frac{RT}{\sqrt{2} \pi d^{2}N_{A}P}[/tex] (1)
Where:
[tex]\lambda=0.2\mu m=0.2(10)^{-6}m[/tex]
[tex]R=8.3145J/mol.K[/tex] is the Universal gas constant
[tex]T=0\°C=273.115K[/tex] is the absolute standard temperature
[tex]d[/tex] is the diameter of the helium atom
[tex]N_{A}=6.0221(10)^{23}/mol[/tex] is the Avogadro's number
[tex]P=1atm=101.3kPa=101.3(10)^{3}Pa=101.3(10)^{3}J/m^{3}[/tex] absolute standard pressure
Knowing this, let's find [tex]d[/tex] from (1), in order to find the radius [tex]r[/tex] of the helium atom:
[tex]d=\sqrt{\frac{RT}{\sqrt{2}\pi\lambda N_{A}P}}[/tex] (2)
[tex]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})}}[/tex] (3)
[tex]d=2.0467(10)^{-10}m[/tex] (4)
If the radius is half the diameter:
[tex]r=\frac{d}{2}[/tex] (5)
Then:
[tex]r=\frac{2.0467(10)^{-10}m}{2}[/tex] (6)
[tex]r=1.0233(10)^{-10}m[/tex] (7)
However, we were asked to find this radius in nanometers. Knowing [tex]1nm=(10)^{-9}m[/tex]:
[tex]r=1.0233(10)^{-10}m.\frac{1nm}{(10)^{-9}m}=0.10233nm[/tex] (8)
Finally:
[tex]r=0.10233nm[/tex] This is the radius of the helium atom in nanometers.
