511 keV photon attenuation in tungsten

Photon attenuation in elemental matters

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]