|
Absorber
|
Density, p
|
Attenuation coefficient Mu/p [cm^2/g]
|
Mu
|
Half value layer (HVL), [cm]
|
|
Copper
|
8.960
|
0.455
|
4.077
|
0.1700
|
|
Tin
|
7.298
|
1.667
|
12.17
|
0.05694
|
|
Lead
|
11.360
|
5.461
|
62.04
|
0.01117
|
a) Estimate the thickness of each absorber required to attenuate a 100 keV photon beam by a 99.99 %.
Which absorber material would be ideal for building a wall to shield you from the photons? Why?
b) If the energy of the photon beam is increased to 1000 keV, the attenuation coefficient for lead decreases to 0.0701 cm^2/g. What is the HVL for lead at 1000 keV? How does this half value layer (HVL) thickness compare to the HVL value at 100 keV? The reciprocal of the linear attenuation coefficient is the mean free path length (lamda), which is the average distance a photon travels in the absorber before it interacts in a collision. Obtain the mean free path length for lead at the two photon energies and comment on the values.