Thickness of a target bombarded with deuterium

Question

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)

(http://www.personal.soton.ac.uk/ab1u06/teaching/phys3002/course/02_rutherford.pdf)

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.

Answer 1

I found out that I had to integrate the function $dN=N_i \frac{\rho L}{m} \frac{d\sigma}{d\Omega}d\Omega $.

By doing this I got:

$$L=\frac{N}{N_i} \frac{m}{4\pi D \rho}$$

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.