# Maxwell-Boltzmann distribution

## Main Question or Discussion Point

Hello, I have a question regarding the Maxwell-Boltzmann distribution.

As you know, the distribution basically looks like
$$n(v) = constant \cdot v^2e^{-\frac{mv^2}{2kT}}$$
, where v is the speed, m is the particle mass, is k the boltzmann constant and T is the absolute temperature.

Now, one can calculate the mean value of the squared velocity $$<v^2>$$ by evaluating the integral
$$\frac{\int_0^\infty v^2 n(v) dv}{\int_0^\infty n(v) dv} = \frac{3kT}{m}$$

From here, we can calculate the average (translational-)kinetic energy of particles,
$$<\frac{mv^2}{2}> = \frac{m}{2} <v^2>=\frac{3kT}{2}$$

Here comes the question, say we have air of temperature T. How can one easily show that the fraction of (for example) oxygen molecules with kinetic energy greater than $$\frac{3kT}{2}$$ is less than 50%?

My idea is basically to create a new distribution, say $$n_1(v^2)$$ by substituting $$v$$ with $$v^2$$ in the original distribution, integrating from 0 to $$\frac{3kT}{m}$$ and then from $$\frac{3kT}{m}$$ to infinity. One can then compare these integrals and conclude whether the fraction of molecules with kinetic energy greater than $$\frac{3kT}{2}$$ is less than 50%. However, this approach seems extremely superflous and I am not even sure the integrals converge (even though I think so). There has to be an easier argument!! Can anyone help please?