Need help with fast neutron absorbed dose rate

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The discussion revolves around calculating the fast neutron absorbed dose rate in an ionization chamber filled with ethylene gas and surrounded by polyethylene walls. Key factors include the chamber's active volume, the atom density of hydrogen in the walls, and the elastic scattering cross-section for hydrogen. The absorbed dose rate depends on the reaction rate, which is influenced by atomic density, neutron density, and the microscopic cross-section. A Monte Carlo method is suggested for accurate calculations due to the multiple interactions of fast neutrons before scattering. The user seeks confirmation on their solution approach and appreciates the community's assistance.
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Hello everyone. I need a hint of how to do the following question. I spent long time trying to solve and I could not do it.

An ionization chamber that will be used to measure fast neutron absorbed dose rate consists of
ethylene (C2H4) gas at a density of (1.2×10^-4 g/cm^3) enclosed by polyethylene (C2H4)n walls in a parallel-plate geometry, where the active volume of the chamber is 5 cm^3. The atom density for hydrogen in polyethylene is 8.6×1022 atoms per gram.The chamber will be exposed to a uniform beam of fast neutrons with a fluence rate per unit energy of (10^5 n /cm^2 s MeV) in the energy range of 0.1 to 10 MeV. The dominant reaction in the chamber walls is elastic scatter with hydrogen, and the microscopic cross-section for hydrogen scatter with neutrons in the energy range 0.1 to 10 MeV is approximated by:

σ(E)= (4.83/E^0.5)-0.578​

where neutron energy E has units of MeV and σ(E) has units of barns. For a given neutron
energy E, the average energy for a scattered proton is E / 2. The chamber walls are thick enough
to achieve charged particle equilibrium for the scattered protons, and offer negligible attenuation
to the incident fast neutrons. The stopping powers for scattered protons in the chamber wall and
in the chamber gas are equal.

What is the expected absorbed dose rate (in units of MeV/g*s ) to the chamber wall
due to the fast neutron irradiation as described above?
 
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Try to write a description for 'absorbed dose' and then write an equation for 'absorbed dose'. This is why we request students to write problem statements and relevant equations.

In computing an absorbed dose, one must consider the 'reaction rate' and the energy absorbed per reaction. The reaction rate is a function of atomic density, neutron density and microscopic cross-section.

Dose rate implies dose per unit time, and since there is an energy spectrum, one must integrate over the energy range, or use the average energy per interaction.


Typically one would use a Monte Carlo method since a fast neutron may have one, two or more interactions before it scatters out of the fast energy range.
 
Thank you so much for your quick response. I will try to do it and i will response later. Thank you again
 
Mr.Astronuc.

Hello again Mr.Astronuc.
Can you check if this is the correct way to day as I attached my solution.
Thank you again
 

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