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 :

\frac {N_0}{S} \times ( \Delta \sigma \times N) = n

the left hand side can be interpolated at follow:

\frac{N_0}{S} is the number of particle per second per unit area, or the flux.

\Delta \sigma \times N 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 :

N_0 \times \frac { \Delta \sigma \times N } {S}

where the fraction after multiplication is the chance of getting scattered to the direction (θ,φ). ( the requirement of the flux N_0 is in HERE. )

and the right hand side is the number of particle detected at direction (θ,φ).

since both n and \Delta \sigma depend on (θ,φ). thus we can differential it and get the angle dependent of these 2.

\frac { N_0 N}{S} \frac {d \sigma}{d \Omega} = \frac {d n}{d \Omega}

if we set the detector moving as radius R. thus the detector area is:

D_A = R^2 d \Omega

Therefore, the number of particles will be detected per second on the detector with some area (= Yield ) is:

Y = \frac {d n}{d \Omega} R^2 d \Omega = \frac { N_o N}{S} R^2 \frac{ d \sigma}{d \Omega} d \Omega

finish. Oh, the unit of the differential cross section is barn = 10^{-28} m  , recall that a nucleus radius is about 10^{-14} m

the relation between the differential cross section to the nuclear potential was discussed on HERE.

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