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.

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Brainiac online

Mar 07, 2011@ 00:54:50The best comments in this thread do not differ much from the worst. Is this a good article does not deserve anything better?