At standard temperature and pressure, the mean free path of He gas is 300 mm.
(a) Determine the effective diameter of He atom
(b) the number of atoms per cubic metre.

Answer

`1.67 xx 10^(10)m, 2.69 xx 10^(25)m^(-3)`

Explanation

Here `T = 0^(@)C = 273 K`,
`P = 1 atm = 1.013 xx 10^(5) Nm^(-2)`
`lambda = 300 nm = 3xx10^(-7) m, d = ? , n=?`
(a) As `lambda = (kT)/(sqrt(2) pi d^(2)P)`
`:. D = [(kT)/(sqrt(2) pi lambda P)]^(1//2`
`=[(1.38 xx 10^(-23 xx 273))/(1.414 xx 3.14 xx 3 xx 10^(-7) xx 1.013 xx 10^(5))]^(1//2)`
`=1.67 xx 10^(10)m`
(b) From `lambda = (1)/(sqrt(2)pi d^(2)n)`
`n = (1)/(sqrt(2)pi d^(2)n)`
`(1)/(1.414 xx 3.14(1.67 xx 10^(-10))^(2) xx 3 xx 10^(-7))`
`n=2.69 xx 10^(25)m^(-3)`.