1. Nov 21, 2016

### Nguyen Ngoc Anh

Dear all,

As we know, the reaction rate can be calculated as following:

R = N * σ * Φ (1)

Where R is reaction rate (events/s/cm3)
σ is cross section (cm2)
Φ is flux of incident particle beam (particles/s/cm2
N is density of target (atoms/cm3)

Logically, there is a limitation of R, because R cannot larger than the number of incident particles. But as equation (1), if Φ is constant, that means the number of incident particles is unchanged, and N increases, R will increase without any limitation. So if N is big enough, R will pass the limit that I talked above.

I hope that you can understand my idea well, because, my English is so bad.

Thank you very much,

2. Nov 21, 2016

### Staff: Mentor

If you increase N too much, the incident particle beam won't penetrate the whole target, so you get regions with lower flux.

3. Nov 21, 2016

### Orodruin

Staff Emeritus
To expand on this, what you have given in post #1 is true only as long as the target can be considered thin. When this is no longer the case, you will get an exponential attenuation of the flux as it penetrates into the target material.

4. Nov 21, 2016

### Nguyen Ngoc Anh

I thought about it. But we can think as following:

You have a target of mass m, therefore N atoms. You have I particles come to the target. In this case, R ~ I and N, so if you increase N, R increase, even if particle can not penetrate the whole target.

5. Nov 21, 2016

### Nguyen Ngoc Anh

That's clever, could you please send me a reference with detailed formula?

I'm so sorry for my all stupid questions.

Thank you,

6. Nov 21, 2016

### Orodruin

Staff Emeritus
You have to look at a differential equation for the flux. Looking at a target with number density $n$ and cross section per target $\sigma$, the probability of a given particle interacting in a thin layer of thickness $dx$ is given by $p = \sigma n \, dx$ and the change in the flux over the this distance is therefore
$$d\Phi = - \Phi \sigma n \, dx \quad \Longrightarrow \quad \frac{d\Phi}{dx} = - n \sigma \Phi.$$
Solving this differential equation for constant $n$ and $\sigma$ gives
$$\Phi = \Phi_0 e^{-n\sigma x},$$
where $\Phi_0$ is the flux at $x = 0$.