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

In summary, a proton ion gun with a 1mm thick Boron-11 sheet was analyzed using data from a table on the IAEA website. After correcting for errors, the calculated stopping power was found to be 58 MeV, which is 50% higher than previous calculations. It is recommended to obtain data from tables on the IAEA website for accurate results.
  • #1
Javier Lopez
75
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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:
2q24z6r.jpg


Then I calculated for 600keV protons at table 5.1eV*cm2/atom.
I calculated for Boron which 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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  • #2
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?
 
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  • #3
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

Also this link could help:
https://www-nds.iaea.org/stopping/

Thanks to Helmut, I will never forget him
 

1. How does the thickness of the Boron-11 sheet affect the stopping power for a proton ion gun?

The thicker the sheet, the greater the stopping power. This is because a thicker sheet provides more material for the protons to interact with, increasing the likelihood of collisions and energy loss.

2. Can the stopping power be calculated for a specific thickness of Boron-11 sheet?

Yes, the stopping power can be calculated using the Bethe-Bloch equation, which takes into account the thickness of the sheet, the energy of the protons, and other factors such as the density of the material.

3. How does the stopping power change as the energy of the protons increases?

As the energy of the protons increases, the stopping power decreases. This is because higher energy protons can penetrate deeper into the material, resulting in fewer interactions and less energy loss.

4. Is the stopping power the same for all types of particles?

No, the stopping power varies depending on the type of particle. For example, the stopping power for protons will be different than that of electrons or alpha particles due to differences in mass and charge.

5. Are there any other factors that can affect the stopping power of a Boron-11 sheet?

Yes, factors such as the temperature and density of the material, as well as the angle of incidence of the protons, can also affect the stopping power. Additionally, the presence of impurities or defects in the material can also impact the stopping power.

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