Using Carbon or Aluminum to block photons: Difference between revisions
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nσt = <math>1.2 \cdot 10^{23} \frac{atoms}{cm^{3}} \times .4 \cdot 10^{-24} cm^{2} \times 4.56 cm = .218</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> | <math>\,\!\, e^{-n \sigma t} => e^{.218} = .80 => 80%</math> of the photons get through the Carbon | ||
== What about Aluminum? == | == What about Aluminum? == | ||
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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> | 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^{- | <math>\,\!\, 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. | Since more gets through Carbon, we're going to forget about Aluminum and focus solely on using Carbon. | ||
Revision as of 10:27, 5 June 2008
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
What about Aluminum?
20 MeV for Aluminum
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
10 MeV hitting 3.9 cm of Aluminum
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.