# Accurate value of energy deposition in target

Dear FLUKA experts,

I am a beginner of FLUKA and I tried to simulate energy deposition of protons. My purpose was to get the accurate value of energy deposition in the target, e.g., the energy deposition of a 100 MeV proton in water is 90 MeV (just an example, may not be accurate).

So I set a 20cm* 20cm* 20cm target and let the proton beam penetrate the target. The data in my .out file shown below.

``````1Region # name     volume         ALL-PART Star Density   BEAMPART Star Density   ENERGY        Density   EM-ENRGY      Density
in cubic cm       Stars/cm**3             Stars/cm**3               GeV/cm**3               GeV/cm**3
/one beam particle      /one beam particle      /one beam particle      /one beam particle

1 BLKBODY  1.000000000D+00         0.000000000D+00         0.000000000D+00         6.459018975D-04         2.414527592D-04
2 VOID     1.000000000D+00         0.000000000D+00         0.000000000D+00         0.000000000D+00         0.000000000D+00
3 TARGET   1.000000000D+00         9.910000000D-02         9.450000000D-02         9.764148484D-02         1.508375133D-03

Total (integrated over volume):         9.910000000D-02         9.450000000D-02         9.828738674D-02         1.749827892D-03
``````

After that, I learned this lecture and found

1. Does this mean the value of energy deposition in my .out file is “9.764148484D-02”?
2. I just set 1 bin in USRBIN card, would I get the accurate data of energy deposition?

Here is my .out file and .inp file.
10MeVproton.inp (1.1 KB)
10MeVproton001.out (90.5 KB)

Thank you so much for your attention!
Zong

Yes, i.e. 97.6 MeV per 100 MeV primary proton, as an average over that single cycle. With no statistical error associated, though, since it comes from a single cycle. The statistical error can be gotten from USRBIN.

Yes, but in this case divided by the 8,000 cm^3 volume, since you asked for a Cartesian mesh, giving energy density (GeV/cm^3). If you ask instead for a USRBIN region mesh (WHAT(1)=12 instead of 10), you will directly get the total energy deposition in the TARGET region (as an average over multiple cycles) and the associated statistical error.

Dear Francesco,