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Using the Trim software to generate a cross section which is used to calcualate DPA.
Typology: Assignments
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1. A W (tungsten) target is bombarded with 2 MeV Neutrons. Use the simple Kinchin-Pease Model to calculate the number of displacement per cm^3 and per atom in 10 days. Use Ed= 35 eV. The flux for 2 MeV Neutron is 3x10^13 n/cm^2 s. The scattering cross section of W is 4. barn, and the atomic density of W is 19.25 g/cm3. (30 point).
Solution:
2. Repeat problem 1 for protons of the same flux and energy ( 3 0 points).
We can solve the problem by estimating the displacement cross section (displacements per ion- Å) in the 1- 2 m range as ~ 2x10-^4 Number/Ion-Ang. 1 dpa= 2x10-^4 Number/Ion-ang x fluence(ion/cm^2 ) x 10x10^8 (ang./cm)/Atom Density 8.59x10^22 at/cm^3 Therefore, fluence = 4.29x10^18 (ion/cm^2 )
Below is a plot of the distribution incident ion ranges throughout the target. It is obvious that the peak located at approximately 6.5 m is the peak of the ion range. This peak corresponds to a value of approximately 22,000 (atoms/cm^3 )/(atoms/cm^2 ). These units are basically the ion number density divided by the fluence. Multiplying this value by the fluence computed above, we obtain an H number density of 9.44x10^22 ions/cm^3. Therefore, the hydrogen percentage is: 9.44x10^22 /(9.44x10^22 +8.59x10^22 ) = 52.3%.
