# Rate of reaction of collision in relation to the geometric cross-section

I am reading "Particles and nuclei : an introduction to the physical concepts / Bogdan Povh" and on page 44 of the book there is a section about the geometric cross section in a idealised scattering experiment. The experiment consists in firing a beam of monoenergetic particle against a target of width $$d$$. The author affirms that the rate of reaction of the collisions of the particles is given by: $$\dot{N} = \Phi_a * N_b *\sigma_b$$ where $$\Phi_a$$ is the flux of incident particles and $$N_b$$ is the number of target particles and $$\sigma_b$$ is the cross sectional area of each target particle. I am not sure how does this follow. I know that $$\Phi_a = n_a \cdot v_a$$, where $$n_a$$ and $$v_a$$ are the density and the velocity of the incident particles respectively. The collision rate is the product of two factors: the rate of particles incident on the target $$R$$, and the average number of collisions $$N_{coll}$$ of a single incident particle as it travels through the target. So $$\dot{N}=RN_{coll}$$.
Let's examine these quantities separately. The rate of particles incident on the target $$R$$ is the product of the flux $$\Phi_a=n_a v_a$$ of particles on the target, which has units of particles/(m$$^2\cdot$$s), and the area $$A$$ of the target. So $$R=\Phi_a A$$, which has the correct units of particles/s.
For a single incident particle, the average number of collisions $$N_{coll}$$ is equivalent to the number of target particles "in the way" of the average incident particle. The number of target particles "in the way" of the average incident particle is equal to the total number of target particles $$N_b$$ multiplied by the probability $$P$$ of a single particle being "in the way". This probability is equivalent to the fraction of the area of the target taken up by a single target particle: after all, if a target particle takes up the entire target, then a collision is inevitable, while a smaller target particle is less likely to be hit. So $$P=\frac{\sigma_b}{A}$$.
$$\dot{N}=RN_{coll}=(\Phi_a A)(N_b P)=\Phi_a A N_b \left(\frac{\sigma}{A}\right)=\Phi_aN_b\sigma_b$$