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

Revision as of 03:59, 20 September 2012

For 0.5 MeV photon, , or

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

so,

= intensity of light

if

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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>
+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>