Heavy evaporation fragments in mgdraw.f

Dear professors:
I’m using USDRAW in mgdraw.f to record the productions and reactions for isotopes. I notice that the heavy evaporation fragments are not included in USDRAW according to the manual. So I have written two ways to record in mgdraw.f. Take DEUTERON for example:

  1. IF (KPART(IP) .EQ. -3 ) THEN
    WRITE(96,'(A15, I15, 3ES15.3,4I15)') "DEUproduce", ICODE, XSCO, YSCO, ZSCO, MREG, ICRES, IBRES, ISMRES
    WRITE(96,'(A15, I15)') "DEUparent", JTRACK
    WRITE(96,'(A15,4ES15.3)') "DEUTERON", CXR(IP)*PLR(IP), CYR(IP)*PLR(IP), CZR(IP)*PLR(IP), AGESEC(IP)
    WRITE(96,'(A15)') "DEUproend"
    
  2. IF (KHEAVY(IP) .EQ. 3 ) THEN
    WRITE(96,'(A15, I15, 3ES15.3, 4I15)') "DEUevapo", ICODE, XSCO, YSCO, ZSCO, MREG, ICRES, IBRES, ISMRES
    WRITE(96,'(A15, I15)') "DEUevaparent", JTRACK
    WRITE(96,'(A15, 4ES15.3)') "evaDEUTERON", CXHEAV(IP)*PHEAVY(IP), CYHEAV(IP)*PHEAVY(IP), CZHEAV(IP)*PHEAVY(IP), AGHEAV(IP)
    WRITE(96,'(A15)') "DEUevaend"
    

I wonder if it can involve all the productions and avoid double counting at the same time in this way.
Because I have found the following lines in the result (I have omitted the irrelevant information):

DEUproduce            101      4.938E+01      3.449E+02      2.191E+03     
DEUparent              9
DEUTERON      1.375E-01     -2.696E-01     -4.754E-01      0.000E+00
DEUproend
DEUevapo            101      4.938E+01      3.449E+02      2.191E+03            
DEUevaparent              9
evaDEUTERON      2.748E-01      4.106E-02     -3.734E-02
DEUevaend

These “two” DEUTERON have the same position but different momentum. Can I interpret these lines as that one inelastic interaction occurs with one DEUTERON produced and another DEUTERON is evaporated after the residual nuclei is cooled, or it has simply only one DEUTERON? I have uploaded the files if needed. Thank you for your help!
Best
Hechong Han
muon.txt (151.6 KB)
Te-LS.inp (3.5 KB)
newsource.f (7.3 KB)
mgdraw.f (11.4 KB)
mdstck.f (3.0 KB)

Dear @hanhechong,

It is perfectly reasonable to have multiple particles (e.g. DEUTERONs) produced from the same inelastic interaction (hence why the same position). To be completely certain, you could additionally score the target nucleus just to double check that conservation laws (e.g. that the target is big enough to produce two deuterons) are preserved.

I hope this helps!

Best,
Daniel