a single diagram can illustrate everything — the relation between the yield ( the number of particle detected per second in the detector ) and the differential cross section.
the upper drawing is a big view, and the lower one is a zoom-in. the red-circles on the upper drawing are same as the lower one.
From the upper one, the number of particles that scatted by the target and go to direction (θ,φ) is :
the left hand side can be interpolated at follow:
is the number of particle per second per unit area, or the flux.
is the total area of the cross section that deflect or scatter particle to direction (θ,φ).
The left hand side can also be viewed as :
where the fraction after multiplication is the chance of getting scattered to the direction (θ,φ). ( the requirement of the flux is in HERE. )
and the right hand side is the number of particle detected at direction (θ,φ).
since both and depend on (θ,φ). thus we can differential it and get the angle dependent of these 2.
if we set the detector moving as radius R. thus the detector area is:
Therefore, the number of particles will be detected per second on the detector with some area (= Yield ) is:
finish. Oh, the unit of the differential cross section is barn = , recall that a nucleus radius is about .
the relation between the differential cross section to the nuclear potential was discussed on HERE.