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

Latest revision as of 20:44, 20 September 2012

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

$\frac{\mu}{rho}=0.1378\frac{cm^2}{g}$

Tungsten density ${rho}=19.25 g/cm^3$

so, $\mu = 2.65265/cm = 0.265265/mm$

$I = I_0 e^{-\mu x}$ = intensity of light

if x=1mm, $\frac{1}{I_0}=0.767$

@@ Line 1: / Line 1: @@
 [http://physics.nist.gov/PhysRefData/XrayMassCoef/tab3.html Photon attenuation in elemental matters]
+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>
@@ Line 9: / Line 11: @@
 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>