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

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[http://physics.nist.gov/PhysRefData/XrayMassCoef/tab3.html Photon attenuation in elemental matters]
 
[http://physics.nist.gov/PhysRefData/XrayMassCoef/tab3.html Photon attenuation in elemental matters]
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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>
 
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>
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so, <math>\mu = 2.65265/cm = 0.265265/mm</math>
 
so, <math>\mu = 2.65265/cm = 0.265265/mm</math>
  
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:<math> I = I_0 e^{-\mu x}</math> = intensity of light
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<math> I = I_0 e^{-\mu x}</math> = intensity of light
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:if <math>I= \frac{I_0}{2}</math>
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if x=1mm, <math>\frac{1}{I_0}=0.767</math>

Latest revision as of 20:44, 20 September 2012

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]

[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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