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 $.

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.