Title: Loglog graph of the exponential expx
1Log-log graph of the exponential exp(-x)
exp(0) 1
exp(-0.693) 0.5 ½
exp(-1) 1/e 0.37
x
2Half-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
3Calculation 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
4Graphs 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/c
over.html
which produced this graph for lead (Pb)
5Tables of linear attenuation coefficient m
Data for Z 82, E 2 - 433 keV E (keV ) µ
Total (cm-1) 2.0004844E00 1.3412E04
2.0104868E00 1.3272E04 2.0205393E00
1.3133E04 2.0306420E00 1.2996E04
4.479101E01 1.1677E02 4.788159E01
9.8248E01 5.118542E01 8.2776E01 5.471721E01
6.9836E01 5.849270E01 5.8933E01
3.544049E02 3.1633E00 3.788588E02
2.7826E00 4.050001E02 2.4588E00 4.329451E02
2.1827E00
The NIST database produces this table of m for
lead (Pb)
6Half-thickness data from ORTEC-online. (link)
X
Gamma rays from Co-60
X
X