- #1
atomicpedals
- 209
- 7
Hello,
I'm attempting to work out the number of ion pairs produced by a 5MeV alpha particle in UF6 (in each of the solid, liquid, and gas states). Using the technique from my old radiation physics book (Turner 3rd ed) I should simply apply the Bethe formula
[tex]-\frac{dE}{dx}=\frac{4\pi k^{2}z^{2}e^{4}n}{mc^{2}\beta^{2}}[ln\frac{2mc^{2}\beta^{2}}{I(1-\beta^{2}}-\beta^{2}][/tex]
Then divide by the ion production energy to arrive at ion-pairs per unit distance. However, don't I need to account for the density of the target material somewhere in the Bethe formula (and thus the differences between -dE/dx in solid, liquid, or gaseous UF6)?
Thanks for any pointers on where I'm going astray!
I'm attempting to work out the number of ion pairs produced by a 5MeV alpha particle in UF6 (in each of the solid, liquid, and gas states). Using the technique from my old radiation physics book (Turner 3rd ed) I should simply apply the Bethe formula
[tex]-\frac{dE}{dx}=\frac{4\pi k^{2}z^{2}e^{4}n}{mc^{2}\beta^{2}}[ln\frac{2mc^{2}\beta^{2}}{I(1-\beta^{2}}-\beta^{2}][/tex]
Then divide by the ion production energy to arrive at ion-pairs per unit distance. However, don't I need to account for the density of the target material somewhere in the Bethe formula (and thus the differences between -dE/dx in solid, liquid, or gaseous UF6)?
Thanks for any pointers on where I'm going astray!