Attenuation of the (homogeneous) X-ray
beam.
Absorber of
thickness x
Incident
monochromatic
beam of intensity
Io .
Attenuated beam
of intensity I
If you are unsure of the meaning of attenuation - now is the time to ask
We shall assume there is
no geometrical reason for
the beam to decrease
intensity i.e. a collimated
beam.
Absorber of
thickness x
Incident
monochromatic
beam of
intensity Io.
I
Attenuated
beam of
intensity I

Io e
 x
i.e. the intensity decreases exponentially with thickness.
What are the units of  ?
m-1
 is called the linear attenuation coefficient
Half Value Thickness or HVT or x1/2
The thickness of absorber required to reduce the intensity
to half its original value.
i.e. I = Io /2 when x = x1/2
Using the formula for attenuation we get
Io
2

Io e
 x1 / 2
Then taking natural logs
ln( 2)
So
1
2

e
 x1 / 2
ln(1 / 2)   x1/ 2

x1/ 2
Question
The HVT of 25keV X-rays in an absorber is 2.5mm. If
the initial intensity of the beam is 4 x 102 Wm-2, what is
its intensity after passing through a) 5mm and b) 7mm of
absorber? c) Sketch a curve to show the variation of
intensity through 10mm of the absorber.
Solution
a) 5mm is 2 HVT’s so intensity is halved, twice.
1st half = 2 x 102 Wm-2
2nd half = 1 x 102 Wm-2
Question
The HVT of 25keV X-rays in an absorber is 2.5mm. If the
initial intensity of the beam is 4 x 102 Wm-2, what is its
intensity after passing through a) 5mm and b) 7mm of
absorber? c) Sketch a curve to show the variation of
intensity through 10mm of the absorber.
b) 7mm is not a whole number of HVT’s so we can’t use
that method. We need I  I e  x calculating 
o
from .
ln( 2)  x1/ 2
ln( 2)   2.5x103
So
and  = 2.8x102 m-1
I
 4 x102 exp( 2.8x102 x7 x103 )
then
so I = 5.6 x 101 Wm-2
Question c) Sketch a curve to show the variation of
intensity through 10mm of the absorber.
Intensity /
100 Wm-2
4
3
2
1
5
(0,0)
Depth / mm
10