In a planetary atmosphere in hydrostatic equilbium the numbe

In a planetary atmosphere in hydrostatic equilbium, the number density \'n\' of a gas at a height h, n(h) is given by n(h) = n(0) exp [-(h-ho)/H] where H is called the \'scale height (kT/Mg) k is the Boltzmann\'s constant T the tempea(2000 K), M the atomic mass; Mis 4 atomic mass units (2 protons & 2 neutrons) for helium g is the gravity(0.8 x earth\'s gravity, for Venus) n (0) is the number density at a reference height of ho. By what fraction does the density of helium changes between the altitudes of 500 km and 2000 km in the atmosphere of Venus?

Solution

n(h) = n(0) exp [- (h-h0)/H]

H =kT/Mg =1.38*10²³* 2000/4amu*0.8*9.8), 1 amu=931Mev=931*10⁶ *1.6*10^-19 =1489*10^-13 joule

H=(2760*10²³) / (1489*10^-13 *0.8*9.8)=(2760*10²³) /(11673.76*10^-13)=0.2364*10³⁶

for h₁= 500km,

n(500) = n(0) exp [- (500-h0)/H]------------------(1)

n(2000)= n(0) exp [- (2000-h0)/H]............................(2)

dividing 1 by 2

n(0) exp [- (500-h0)/H] / n(0) exp [- (2000-h0)/H]

exp(1500/H) =exp(1500/ 0.2364*10³⁶) =exp(6345.17*10^-36 )

fraction x= exp(6345.17*10^-36 ) answer