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

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 = 0.693 so x ½ = 0.693 / m

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-1 and x ½ = 0.693/m so x ½ = 0.693 / (88 cm-1) = 0.0079 cm But for hard X-rays with energy 433 keV, m = 2.2 cm-1 so x ½ = 0.693 / (2.2 cm-1) = 0.31 cm

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: http://physics.nist.gov/PhysRefData/FFast/Text/cover.html which produced this graph for lead (Pb):

Tables of linear attenuation coefficient m Data for Z = 82, E = 2 - 433 keV E (keV ) µ Total (cm-1) 2.0004844E+00 1.3412E+04 2.0104868E+00 1.3272E+04 2.0205393E+00 1.3133E+04 2.0306420E+00 1.2996E+04 … 4.479101E+01 1.1677E+02 4.788159E+01 9.8248E+01 5.118542E+01 8.2776E+01 5.471721E+01 6.9836E+01 5.849270E+01 5.8933E+01 3.544049E+02 3.1633E+00 3.788588E+02 2.7826E+00 4.050001E+02 2.4588E+00 4.329451E+02 2.1827E+00 The NIST database produces this table of m for lead (Pb):

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