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

20 MeV for Carbon

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

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

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

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

[math]\displaystyle{ \,\!\, e^{-n \sigma t} => e^{.218} = .80 => 80% }[/math] of the photons get through the Carbon

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

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

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

Therefore, the thickness of our Aluminum is 3.9 cm

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

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

[math]\displaystyle{ \,\!\, e^{-n \sigma t}=e^{.24}=.79=>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.