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 $10.49 \frac{g}{cm^{3}}$

density of Carbon = $~2.3 \frac{g}{cm^{3}}$

thickness = $\frac{range}{density} = \frac{10.49}{2.3} = 4.56 cm$

Therefore, the thickness of our Carbon is 4.56 cm

10 MeV hitting 4.56 cm of Carbon

n = $2.3 \frac{g}{cm^{3}} \times \frac{6.022 \cdot 10^{23} atoms}{12 g} = 1.2 \cdot 10^{23} \frac{atoms}{cm^{3}}$

σ = $.4 \cdot 10^{-24} cm^{2}$

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

$exp^{-nσt} =\gt exp^{.218} = .80 =\gt 80%$ of the photons get through the Carbon

20 MeV for Aluminum

range is $10.54 \frac{g}{cm^{3}}$

density of Carbon = $2.7 \frac{g}{cm^{3}}$

thickness = $\frac{range}{density} = \frac{10.54}{2.7} = 3.9 cm$

Therefore, the thickness of our Aluminum is 3.9 cm

10 MeV hitting 3.9 cm of Aluminum

n = $2.7 \frac{g}{cm^{3}} \times \frac{6.022 \cdot 10^{23} atoms}{27 g} = 6.022 \cdot 10^{22} \frac{atoms}{cm^{3}}$

σ = $1.039 \cdot 10^{-24} cm^{2}$

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

$\,\!\, e^{-nσt}=e^{.24}=.79$ 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.

$\,\!\, e^{x \ln e}=e^{x \cdot 1}=e^x.$