511 keV photon attenuation in tungsten: Difference between revisions

Revision as of 03:59, 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 [math]\displaystyle{ I= \frac{I_0}{2} }[/math]

@@ Line 1: / Line 1: @@
 [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>
+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 <math>I= \frac{I_0}{2}</math>