1. Log-log graph of the exponential exp(-x)+
exp(-0.693) = 0.5 = ½ + +
exp(-1) = 1/e = 0.37
exp(-x) x

2. Half-thickness for absorption of X-rays
For a particular thickness x ½ the intensity is decreased to ½ of its original magnitude. So if
I(x½) = Io exp(- m x ½) = ½ Io
we solve to find the half-thickness x ½.
exp(- m x ½) = ½ and m x ½ =
so x ½ = / m

3. Calculation of half-thickness
To calculate x ½ we need to know m.
As an example, for X-rays of energy 50 keV,
m = 88 cm and x ½ = /m so
x ½ = / (88 cm-1) = cm
But for hard X-rays with energy 433 keV,
m = 2.2 cm-1 so
x ½ = / (2.2 cm-1) = cm

4. Graphs of linear attenuation coefficient m
The linear attenuation coefficient m can be obtained from tables, or from automated databases such as the NIST database:
which produced
this graph for
lead (Pb):

5. Tables of linear attenuation coefficient m
Data for Z = 82, E = keV
E (keV ) µ Total (cm-1)
E E E E E E E E+04 …
E E E E E E E E E E+01
E E E E E E E E+00
The NIST database produces this table
of m for lead (Pb):

6. Half-thickness data from ORTEC-online. (link)X
Gamma rays from Co-60 X X