Distribution of speeds in a molecular beam.

[Narcol2000]

I'm struggling to understand the basis for the following formula.

The goal is to find the distribution of molecular speeds emerging from a smal hole in an oven where molecules are allowed to come to thermal equalibrium with the oven walls before exiting through the small whole.

The book states:

'Suppose we consider particles with speed in the range u to u+du which cross an area A at an angle $$\theta$$ to the normal to the area. In a time t they travel a distance ut and sweep out a volume $$Autcos(\theta)$$. The number of particles in this volume with speeds in the range u to u+du and whose direction of motion lies in the range $$\theta$$ to $$\theta + d\theta$$ and $$\phi$$ to $$\phi + d\phi$$ is:

$$Autcos(\theta)\frac{n(u)du}{V}\frac{d\theta sin(\theta)d\phi}{4\pi}$$

where n(u) is the Maxwell distribution and V i assume is the total volume of the oven (i think... its not actually stated what the volume V is.)

I don't see where this formula comes from since it is just stated with no derivation, and would like to have some idea of where it comes from so any help would be appreciated.

 

[azntoon]

The formula is derived from the Maxwell-Boltzmann distribution, which describes the probability of finding particles with a given energy at thermal equilibrium. The formula you've provided takes into account the total number of particles in the volume V of the oven, and the number of particles that cross an area A in a given time t. The d\theta and d\phi terms are used to account for the direction of the particles, while the cos(\theta) term accounts for the angle at which they cross the area A. Finally, the n(u)du term takes into account the number of particles in the range u to u+du. So the entire formula gives the probability of finding a particle with speed in the range u to u+du, crossing an area A at an angle \theta to the normal to the area, and whose direction of motion lies in the range \theta to \theta + d\theta and \phi to \phi + d\phi.