Difference between revisions of "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.
 
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}}
+
= 20 MeV for Carbon =
−
density of Carbon = ~2.3 \frac{g}{cm^{3}}
+
 
−
thickness = \frac{range}{density} = \frac{10.49}{2.3} = 4.56 cm^{2}
+
 
 +
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>

Revision as of 16:44, 4 June 2008

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]