# Thickness of a target bombarded with deuterium

I need to calculate de thickness of a sheet of Zinc that is being bombarded by deuterium nuclei. I'm given that the fraction of nuclei dispersed below $$\theta=90º$$ with $$T=8MeV$$ is $$0.9999$$ and the density of the sheet $$\rho=7.14 g/cm^3$$.

From this pdf (page 7)

I found a relation with a few approximations that could give me the thickness of the sheet. I found that:

$$L=\Bigg(\frac{16\pi \epsilon_0 T }{Z_1 Z_2}\Bigg)^2\frac{M_{Zn}\Delta n \sin^4(\theta/2) }{\rho }$$

where I called $$M_{Zn}$$ the atomic mass of Zinc and $$\Delta n$$ the fraction of atoms meassured. The problem is that with this equation I get an $$L$$ of about $$10^{-20}m$$ which is odd to say the least.

The most probable thing is that I'm not undersanding the relations shown in the article right.

I found out that I had to integrate the function $$dN=N_i \frac{\rho L}{m} \frac{d\sigma}{d\Omega}d\Omega$$.
$$L=\frac{N}{N_i} \frac{m}{4\pi D \rho}$$
where $$D=(\frac{Z_1 Z_2 e^2}{4\pi \epsilon_0 4 T})^2$$
And thats it, I got a thickness of $$L=66.9cm$$, which is a bit too long to hold with this approximation.