511 keV photon attenuation in tungsten: Difference between revisions

Latest revision as of 20:44, 20 September 2012

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

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

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

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

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

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

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 [http://physics.nist.gov/PhysRefData/XrayMassCoef/tab3.html Photon attenuation in elemental matters]
-For 0.5 MeV photon, <math>\mu / \rho = 0.1378~(cm^2/g)</math>, <math>\mu_{en} / \rho = 0.0744~(cm^2/g)</math>
+https://wiki.iac.isu.edu/index.php/TF_SPIM_e-gamma#Mass_Attenuation_Coefficient
+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>