Knowing the viscosity coefficient of helium under standard conditions, calculate the effective diameter of the helium atom.

Explanation

`eta = (1)/(3) sqrt((8 kT)/(pi m)) (m)/(sqrt(2) pi d^2) = (2)/(3) sqrt((m kT)/(pi^3)) (1)/(d^2)`
or `d = ((2)/(3 eta))^(1//2) ((mkT)/(pi^3))^(1//4) = ((2)/(3 xx 18.9) xx 10^6)^(1//2) ((4 xx 8.31 xx 273 xx 10^-3)/(pi^3 xx 36 xx 10^46))^(1//4)`
=`10^10 ((2)/(3 xx 18.9))^(1//2) ((4 xx 83.1 xx 273)/(pi^3 xx .36))^(1//4) ~~ 0.178 nm`.