A tall vertical vessel contains a gas composed of two kinds of molecules of masses `m_1` and `m_2` . With `m_2 gt m_1`. The concentrations of these molecules at the bottom of the vessel are equal to `n_1` and `n_2` respectively, with `n_2 gt n_1`. Assuming the temperature `T` and the free-fall acceleration `g` to be independent of the height, find the height at which the concentrations of these kinds of molecules are equal.

Explanation

`n_1(h) = n_1 e^(-m_1 gh//kt), n_2 (h) = n_2 e^(gh(m_1 - m_2)//kT)`
They are equal at a height `h` where `(n_1)/(n_2) = e^(gh(m_1 - m_2)//kT)`
or `h = (kT)/(g) (1 n n_1 - 1n n_2)/(m_1 - m_2)`.