Number of secondaries generated in a volume MCNP?

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SUMMARY

The discussion focuses on calculating the number of secondary protons generated in a lead target when bombarded by a neutron beam using MCNP (Monte Carlo N-Particle Transport Code). The solution involves utilizing the F4 tally with a tally multiplier to obtain the flux averaged over the volume, which is then multiplied by the atom density of lead and the microscopic cross section for the (n,p) reaction. The specific MCNP card setup includes F4:N 1 and FM4 C M (103) (203), where C represents the atom density and M is the material number for lead. For further reference, users can consult Appendix G of the MCNP5 manual Vol I and Page 4-39 of the MCNP5 manual Vol II for similar examples.

PREREQUISITES
  • Understanding of MCNP5 software and its functionalities
  • Familiarity with tally types, specifically F4 tally
  • Knowledge of atomic density and microscopic cross sections
  • Basic principles of neutron interactions with matter
NEXT STEPS
  • Study the MCNP5 manual, particularly Appendix G for reaction types and numbers
  • Learn about the implementation of tally multipliers in MCNP
  • Explore neutron interaction cross sections for various materials
  • Review examples of similar calculations in MCNP5 documentation
USEFUL FOR

Researchers, nuclear engineers, and physicists working with MCNP for neutron transport simulations, particularly those interested in secondary particle generation in nuclear reactions.

Neo Tran
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Hello everyone,
I am having a problem with MCNP. My question is how to get number of secondaries in a certain volume. For example I have a neutron beam bombarded Pb target, and I want to count all of proton formed in the target. I considered tally F4, but the unit is 1/cm**2. Who can explain the tally to me? More over, help me solve this problem, please.
Thank you very much,Neo.
 
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For this, you will need to use the tally multiplier functionality with the F4 tally. F4 gives a flux averaged over the volume of the cell in units of particles/cm**2 per source particle. Using a tally multiplier card, you can get MCNP to multiply this flux with the atom density of the target nuclide and the microscopic cross section of the desired reaction (in this case, (n,p) reaction). If you multiply this value with the source term, you'll get the total number of protons formed in the target.
This is what your MCNP cards with look like (assuming a target cell 1):

F4:N 1
FM4 C M (103) (203)

Where C is the atom density of lead in your target, M is the corresponding material number in your input file for lead (assuming it is pure elemental lead)
103 is the reaction number for (n,p) reaction and 203 is a reaction number for total number of protons produced by all reactions. You can find a complete list of all reaction types and numbers in Appendix G of the MCNP5 manual Vol I
You can also find an example of the similar problem on Page 4-39 of the MCNP5 manual Vol II
 
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quarkle said:
For this, you will need to use the tally multiplier functionality with the F4 tally. F4 gives a flux averaged over the volume of the cell in units of particles/cm**2 per source particle. Using a tally multiplier card, you can get MCNP to multiply this flux with the atom density of the target nuclide and the microscopic cross section of the desired reaction (in this case, (n,p) reaction). If you multiply this value with the source term, you'll get the total number of protons formed in the target.
This is what your MCNP cards with look like (assuming a target cell 1):

F4:N 1
FM4 C M (103) (203)

Where C is the atom density of lead in your target, M is the corresponding material number in your input file for lead (assuming it is pure elemental lead)
103 is the reaction number for (n,p) reaction and 203 is a reaction number for total number of protons produced by all reactions. You can find a complete list of all reaction types and numbers in Appendix G of the MCNP5 manual Vol I
You can also find an example of the similar problem on Page 4-39 of the MCNP5 manual Vol II
You was right. Thank you very much Quarkle. It is very useful for me in this time.
 

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