Eγ vs probability with 5 cm of D20

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[math]Probability = \sigma \times \rho \times thickness[/math]

[math]\rho of D _2 0 = 1 \frac{g}{cm^{3}} \times \frac {20}{18} \times \frac{6.022 \cdot 10^{23}}{20g}\times 2 = 6.6242 \cdot 10^{22}[/math]


[math]6 MeV = 2200 \cdot 10^{-30} \times 6.6242 \cdot 10^{22} \times 5 cm = 7.36 \cdot 10^{-4}[/math]


[math]8 MeV = 1776 \cdot 10^{-30} \times 6.6242 \cdot 10^{22} \times 5 cm =5.94 \cdot 10^{-4}[/math]


[math]10 MeV = 1409 \cdot 10^{-30} \times 6.6242 \cdot 10^{22} \times 5 cm =4.71 \cdot 10^{-4}[/math]


[math]12 MeV = 1161 \cdot 10^{-30} \times 6.6242 \cdot 10^{22} \times 5 cm =3.88 \cdot 10^{-4}[/math]


[math]13 MeV = 1058 \cdot 10^{-30} \times 6.6242 \cdot 10^{22} \times 5 cm =3.55 \cdot 10^{-4}[/math]


[math]14 MeV = 963 \cdot 10^{-30} \times 6.6242 \cdot 10^{22} \times 5 cm =3.22 \cdot 10^{-4}[/math]

Probability d20 5 cm.jpg

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