Difference between revisions of "511 keV photon attenuation in tungsten"

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so, <math>\mu = 2.65265/cm = 0.265265/mm</math>
 
so, <math>\mu = 2.65265/cm = 0.265265/mm</math>
  
−
:<math> I = I_0 e^{-\mu x}</math> = intensity of light
+
<math> I = I_0 e^{-\mu x}</math> = intensity of light
−
:if <math>I= \frac{I_0}{2}</math>
+
 
 +
if x=1mm, <math>\frac{1}{I_0}=0.767</math>

Revision as of 04:00, 20 September 2012

Photon attenuation in elemental matters

For 0.5 MeV photon, [math]\mu / \rho = 0.1378~(cm^2/g)[/math], or [math]\mu_{en} / \rho = 0.0744~(cm^2/g)[/math]

[math]\frac{\mu}{rho}=0.1378\frac{cm^2}{g}[/math]

Tungsten density [math]{rho}=19.25 g/cm^3[/math]

so, [math]\mu = 2.65265/cm = 0.265265/mm[/math]

[math] I = I_0 e^{-\mu x}[/math] = intensity of light

if x=1mm, [math]\frac{1}{I_0}=0.767[/math]

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