From the law of atmosphere
[tex]N_v(y) = n_0*e^{-\frac{mgy}{Kb*T}}[/tex]
Where
[tex]n_0[/tex] = constant and is number density where the height y = 0cm
[tex]n_V[/tex] = Number density at height y=3cm
Kb = Boltzmann constant [tex]= 1.38*10^{-23}J/K[/tex]
[tex]T=20K[/tex]
[tex]m = 10^{-19}kg[/tex]
Re-arranging the equation to have the value of the gravity,
[tex]\frac{N_v(y)}{n_0} = e^{-\frac{mghy}{KbT}}[/tex]
[tex]ln(\frac{N_v(y)}{n_0}) = -\frac{mgy}{KbT}[/tex]
Since it is 30% of value above surface, therefore [tex]N_v = 0.3n_0[/tex]
[tex]ln(\frac{0.3n_0}{n_0}) = -\frac{mgy}{KbT}[/tex]
[tex]g = -\frac{KbT ln(0.3)}{my}[/tex]
[tex]g = -\frac{(1.38*10^{-23}J/K)(20K)(Ln(0.3))}{10^{-19}(3*10^{-2})}[/tex]
[tex]g = \frac{1.38*2*ln(0.3)*10^{-22}}{3*10^{-4}}[/tex]
[tex]g = 1.104*10^{-1}m/s^2[/tex]
[tex]g = 0.1m/s^2[/tex]
Therefore the correct answer is C.
