a target with

• thickness d
• scattering center (particle) density be $n_b$
• cross section area $\sigma_b$

the cross section area is the cross section that the incoming particle will react with the target particle. here, we assume the cross section areas does not overlap. or to say, the incoming particle can only be reflectecd by 1 scattering center at 1 time.

the target was bombard by a monoenergy beam with

• cross area A
• beam particle density $n_a$
• speed $v_a$

the flux of the beam (particle per unit time per unit area) is:

$\Phi_a =n_a v_a$

The total number of target within the beam area is the density of target times the thickness times the cross area of the beam:

$N_b = n_b d A$

The total reaction rate will be proprtion to the flux and number of particle with in the beam area and area of each scatter center:

$\dot{N} = \Phi_a N_b \sigma_b$

since the flux can also be realted to the beam particle rate ( particle per unit time ):

$\dot{N}_a = \Phi_a A$

Thus, the total reaction rate can be rewrtiten as:

$\dot{N} = \dot{N}_a N_b \frac{\sigma_b}{A}$

The cross section will be reaction rate per flux per number of scattering center:

$\sigma_b = \dot{N} / ( \phi_a N_b)$

in most cases, the detector with surface area S at distance r, the solid angle it covers :

$\Delta \Omega = S/r^2$

thus the reaction rate it detected is

$\dot{N}=\Phi_aN_b\frac{d\sigma_b}{d\Omega}\Delta\Omega$

If the detector can resolve energy of the detected particle, then we have

$\dot{N}=\Phi_aN_b\frac{d^2\sigma_b}{d\Omega d E}\Delta\Omega\Delta E$