From Ohm

Attenuation of the radiation intensity $I$ of monochromatic (mono-energetic) rays while travelling through a homogeneous material with thickness d and attenuation coefficient $\mu$:

$I(d) = I_0 \cdot e^{−\mu \cdot d}$

What is the thickness of a material that absorbs exactly half of the rays intensity?

$I/I_0 = e^{−\mu \cdot d} = 0.5$
$D_\frac12 = \frac{ln 2}{\mu}$
$ln 2 \approx 0.7$