Optimizing Lead Shielding for Gamma Radiation Protection

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To effectively reduce brief exposure to gamma radiation below 5 gray, the thickness of lead required depends on the intensity and energy of the gamma rays. The intensity diminishes exponentially with lead thickness, governed by the attenuation coefficient, which varies with energy. For example, 2 inches of lead can reduce Cobalt 60 radiation (with gamma energies of 1.1 MeV and 1.3 MeV) by a factor of 10. The most penetrating gamma energies fall between 1 and 2 MeV, where specific attenuation coefficients can be referenced for accurate calculations. Understanding the source's energy and intensity is crucial for determining the necessary lead shielding.
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What thickness of lead is needed to neutralize brief exposure to gamma radiation, or at least reduce it to below 5 gray?
 
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You need to provide more info about the source or strength of the gamma rays.
 
NJV said:
What thickness of lead is needed to neutralize brief exposure to gamma radiation, or at least reduce it to below 5 gray?

As has been stated, you need to know the intensity and energy of the x-rays (or gamma rays). Essentially, the intensity of radiation as it passes through a thickness of lead diminishes exponentially with the thickness of lead.

i.e. I(x) = I(0) exp (- m.x) + c

where I(x) is the intensity as a function of x, the thickness of lead and m is what is known as the attenuation co-efficient of lead. c being the background radiation.

However, there's a problem, because m is a function of the energy of the source. You should be able to find an attenuation coefficient vs energy diagram for lead somewhere on the net.

If you have a specific source, then it will emit certain energies with a certain intensity. You'd have to find out how much each was attenuated to find the diminished activity and finally work out the number of grays in the usual manner.
 
Thank you for the information. The equation proves useful.
 
The most penetrating x-ray (or photon) energy is about 1 or 2 MeV. Below about 0.5 MeV, the photoelectric effect off of bound electrons is significant, and above 2 MeV, pair production (of an electron and positron) becomes significant. I seem to recall that 2 inches of lead reduces Cobalt 60 radiation by about a factor of 10.
 
Bob S said:
The most penetrating x-ray (or photon) energy is about 1 or 2 MeV. Below about 0.5 MeV, the photoelectric effect off of bound electrons is significant, and above 2 MeV, pair production (of an electron and positron) becomes significant. I seem to recall that 2 inches of lead reduces Cobalt 60 radiation by about a factor of 10.

Cobalt 60 radiation has two gammas, 1.1 MeV and 1.3 MeV.

The lead gamma absorption cross section for 1 to 2 MeV gammas is in the range of 20 barns. For 50 KeV, the photoelectric cross section is a few kilobarns (compared to a few barns for carbon).
 
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