Using Carbon or Aluminum to block photons

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

$\,\!\, e^{-n \sigma t} =\gt e^{-.218} = .80 =\gt 80%$ 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 = $6.022 \cdot 10^{22} \frac{atoms}{cm^{3}} \times 1.039 \cdot 10^{-24} cm^{2} \times 3.9 cm = .24$

$\,\!\, e^{-n \sigma t}=e^{-.24}=.79=\gt 79%$ 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.

So, we're keeping the same information for the 20 MeV for Carbon and calculate the rest of the information for 6, 8, 12, 13, and 14 MeV.

6 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 = $4.924 \cdot 10^{-25} cm^{2}$

nσt = $1.2 \cdot 10^{23} \frac{atoms}{cm^{3}} \times 4.924 \cdot 10^{-25} cm^{2} \times 4.56 cm = .269$

$\,\!\, e^{-n \sigma t} =\gt e^{-.269} = .76 =\gt 76%$ of the photons get through the Carbon

8 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 = $4.297 \cdot 10^{-25} cm^{2}$

nσt = $1.2 \cdot 10^{23} \frac{atoms}{cm^{3}} \times 4.297 \cdot 10^{-25} cm^{2} \times 4.56 cm = .235$

$\,\!\, e^{-n \sigma t} =\gt e^{-.235} = .79 =\gt 79%$ of the photons get through the Carbon

12 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 = $3.646 \cdot 10^{-25} cm^{2}$

nσt = $1.2 \cdot 10^{23} \frac{atoms}{cm^{3}} \times 3.646 \cdot 10^{-25} cm^{2} \times 4.56 cm = .199$

$\,\!\, e^{-n \sigma t} =\gt e^{-.199} = .83 =\gt 83%$ of the photons get through the Carbon

13 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 = $3.545 \cdot 10^{-25} cm^{2}$

nσt = $1.2 \cdot 10^{23} \frac{atoms}{cm^{3}} \times 3.545 \cdot 10^{-25} cm^{2} \times 4.56 cm = .194$

$\,\!\, e^{-n \sigma t} =\gt e^{-.194} = .82 =\gt 82%$ of the photons get through the Carbon

14 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 = $3.461 \cdot 10^{-25} cm^{2}$

nσt = $1.2 \cdot 10^{23} \frac{atoms}{cm^{3}} \times 3.461 \cdot 10^{-25} cm^{2} \times 4.56 cm = .189$

$\,\!\, e^{-n \sigma t} =\gt e^{-.189} = .83 =\gt 83%$ of the photons get through the Carbon

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