Production cross section of 67Ga and 68Ga from interaction of Li7 beam on Cu target

Dear @ceruttif,

Continuing the discussion from Neutron yield for Li 7 beam on thick Cu target:

I tried to implement your suggestions. I am not very much sure if I modified it properly.
For example,

  1. in AUXSCORE card, I have included A and Z of Ga 67. If we want the information for Ga 68 as well, do we have to add another set of USEYIELD (with different unit) and AUXSCORE card. Or in a single run, only one heavy ion can be added?

  2. The following point is not very clear to me, where should we add AIR and COPPER.

  1. In the LAM BIAS card, there is no separate option for Li7. Do we have to choose HEAVYION option there?

  2. Also is the particle charge equivalent to atomic number in this case? Since in AUXSCORE, I have already added Z = 31, is it still required to use partice charge in USRYIELD card?

  3. Should we keep PART and SET option in AUXSCORE blank ?

  4. Fig 4 in the associated paper represents total cross section variation with respect to energy. So in this case we have to do multiple simulations with different incident Li7 beam energy, right ?

I am sorry for my numerous questions, it will be helpful if you can kindly guide.

Li_42MeV.inp (2.0 KB)


  1. Add another pair of cards (it can also be on the same unit, with a different detector name).
  2. The card to be added is exactly the one I wrote down: a MATERIAL card redefining the pre-defined COPPER material in order to select AIR as alternate material for ionisation processes. COPPER is the SDUM (unlabelled field in Flair) and AIR is WHAT(5) (dE/dx in Flair), see the manual.
  3. Yes.
  4. Particle charge coincides indeed with the ion atomic number. The point is that USRYIELD requires to specify the scoring range of two selected physical quantities. I suggested to use particle charge since it’s easy to understand and convenient from the normalization point of view. For your purposes, you are not interested in the differential information, rather to the integral over the two quantities’ ranges, but you need to make sure that, whichever physical quantity you select, the input ranges do not cut out part of the desired isotope yield. The choice of particle charge easily fulfils this requirement and avoids an additional normalization factor, since the suggested particle charge range limits define an unit interval.
  5. Yes.
  6. Yes.
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Thank you @ceruttif for the guidance. During interpretation of results,

  1. In the BEAMPART star density, it is showing very less value, not close to 100%, does this mean that there are many secondary interactions? I have used x lambda inelastic = 0.05.

  1. In the sum.lis file, here we are looking for the parameter Cumulative yields as a function of x1 (integrated over x1, differ in X2) i.e. the last line in the sum.lis here. Is it correct? (then we have to divide this number by no of stars). How does this number (printed in the top of sum.lis file) 1821.60620 mb is appearing?

Here is the card I used:


  1. You are looking at a different place. Please search for Number of stars generated per beam particle and check the HEAVYION prompt percentage, which should be very close to 100%. Then take the corresponding prompt value, which is likely close to what you saw in the table above.

  2. Well, I’d rather look at the total response highlighted in blue, but it’s the same thing as the cumulative yield.
    The cross section value you see at the top is the microscopic reaction cross section of beam particles in COPPER, which you requested in USRYIELD and is already embedded in the total response. The resulting production cross section of a specific isotope, obtained after dividing the total response by the 7Li reaction probability (i.e., the number of stars generated by HEAVYION discussed in 1.), shall represent a fraction of it.

Thank you @ceruttif ,

  1. I could find the no of stars generated per beam. I think instead of printing **, it should print 100%, although from the top row, it is clear that the value is 100%.

However, to get total cross section for the production of Ga-67, if we divide the cumu yield 2.65899E-2 by 1.3669E-4 (since the probability is 1 as no of stars is 100% ). we get a value 194.5. But, the paper gives a value of around 20 mb.


  1. the microscopic reaction cross section of beam particles in COPPER: This is the total response right ? (the value 1821 mb). This includes all types of (x,n) reaction channel right ? (for e.g. Li7, n; Li7, 3n; Li7, 5n etc.)

  2. The figure shows separate lines for 63Cu(7Li,3n) and 65Cu(7Li,5n), both of which produce Ga 67. But what we are getting in FLUKA output, is the total value. Is it so ?

  3. Also can you please explain what is meant by ‘star’?


  1. Please note that 67Ga and 67Ge are not the same nucleus.
    The FLUKA result for the latter is as follows:


exceeding at most by a factor 2 the data you report.

  1. Not sure what you mean with total response (obviously it’s not the total response of your USRYIELD, which was limited to 67Ga). By definition of reaction cross section, it includes all reaction channels.

  2. No. Since your target is made of natural copper (as in reality), the resulting cross section is the weighted average of the two.

  3. What is written in the manual (excluding low energy neutron reactions).

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Thank you @ceruttif

I am sorry for posting the wrong image. For Ga 67, the paper gave a value of nearly 30, But I got 194. I think I need to investigate more.

However, I am able to get similar data for Ge 67 as yours.

Thank you for providing the guidance. I could learn this new topic.



Actually, just above 40 MeV, which is the energy you indicated above, I see an experimental value around 200 mb, in very good agreement with FLUKA:


Thank you @ceruttif for the help. Yes, I was looking at the EMP_EGSM plot. I am sorry for my mistake. I could generate the experimental data pattern using FLUKA. Thank you once again for your patience and explanation.


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