# Stopping power for a proton ion gun through a Boron-11 sheet 1mm thick

• I
Hello, I need the stopping power for a proton ion gun through Boron-11 sheet 1mm thick.

I have the following table obtained from here:
https://www-nds.iaea.org/stopping/stopping_hydr.html
Where the unit is in 1-15*eV*cm2/atom:

Then I calculated for 600keV protons at table 5.1eV*cm2/atom.
I calculated for Boron wich density is 2080kg/m3 and atomic number 11 a density of 1.139atoms/m3
So the stopping power should be:
5.1*1e-6 (MeV/eV) * 1e-4 (cm2/m2) /1.1389 *.001=58e6 MeV

That is too much (at least 1e6 times), what I did wrong?

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mjc123
Homework Helper
It is not 5.1 eV cm2/atom, it is 5.1x10-15 eV cm2/atom.
It is not 1.139 atoms/m3 (that's a bloody big atom!) What is it?
It is not 1e-4 cm2/m2
Where does 0.001 come from?

berkeman
I am very sorry, I am surprised about lot errors I made yesterday late ( I have to sleep sometimes),
I introducted in the formula 5.1e-15.
The target density was 1.139e29 atoms/m3 (Boron 11 density is 2080kg/m3 so 2080/11amus/1.660538921E-27 kg/amu = 1.139e29)
The 1e-4 is Ok but I had to write 1e-4 (m^2/cm^2)
0.001 was the target thick that I forgot to advise (1mm)
At the end I obtain:
5.1e-15 (eV*cm2/atom)*1e-6 (MeV/eV) * 1e-4 (m2/cm2) *1.1389 e29 (atom/m3) * 1e-3 mm=58 MeV, that is 50% over other calculus I made but more reasonable

If somebody want to calculate the stopping power I should recommend (moreover not follow my calculus) to obtain data from tables at this site:
https://www-nds.iaea.org/stopping/stopping_hydr.html