Using Carbon or Aluminum to block photons

We're looking to see which is better for letting photons through, Carbon or Aluminum.

20 MeV for Carbon

range is [math]10.49 \frac{g}{cm^{3}}[/math]

density of Carbon = [math]~2.3 \frac{g}{cm^{3}}[/math]

thickness = [math]\frac{range}{density} = \frac{10.49}{2.3} = 4.56 cm[/math]

Therefore, the thickness of our Carbon is 4.56 cm

10 MeV hitting 4.56 cm of Carbon

n = [math]2.3 \frac{g}{cm^{3}} \times \frac{6.022 \cdot 10^{23} atoms}{12 g} = 1.2 \cdot 10^{23} \frac{atoms}{cm^{3}}[/math]

σ = [math].4 \cdot 10^{-24} cm^{2}[/math]

nσt = [math]1.2 \cdot 10^{23} \frac{atoms}{cm^{3}} \times .4 \cdot 10^{-24} cm^{2} \times 4.56 cm = .218[/math]

[math]exp^{-nσt} =\gt exp^{.218} = .80 =\gt 80%[/math] of the photons get through the Carbon

20 MeV for Aluminum

range is [math]10.54 \frac{g}{cm^{3}}[/math]

density of Carbon = [math]2.7 \frac{g}{cm^{3}}[/math]

thickness = [math]\frac{range}{density} = \frac{10.54}{2.7} = 3.9 cm[/math]

Therefore, the thickness of our Aluminum is 3.9 cm

10 MeV hitting 3.9 cm of Aluminum

n = [math]2.7 \frac{g}{cm^{3}} \times \frac{6.022 \cdot 10^{23} atoms}{27 g} = 6.022 \cdot 10^{22} \frac{atoms}{cm^{3}}[/math]

σ = [math]1.039 \cdot 10^{-24} cm^{2}[/math]

nσt = [math]6.022 \cdot 10^{22} \frac{atoms}{cm^{3}} \times 1.039 \cdot 10^{-24} cm^{2} \times 3.9 cm = .24[/math]

[math]\,\!\, e^{-nσt}=e^{.24}=.79[/math] of the photons get through the Aluminum

Since more gets through Carbon, we're going to forget about Aluminum and focus solely on using Carbon.

[math]\,\!\, e^{x \ln e}=e^{x \cdot 1}=e^x.[/math]