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]\,\!\, e^{-n \sigma t} =\gt e^{.218} = .80 =\gt 80%[/math] of the photons get through the Carbon
What about Aluminum?
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^{-n \sigma t}=e^{.24}=.79=\gt 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.