1. Mar 9, 2009

### NJV

What thickness of lead is needed to neutralize brief exposure to gamma radiation, or at least reduce it to below 5 gray?

2. Mar 9, 2009

3. Mar 9, 2009

4. Mar 9, 2009

### DeShark

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.

5. Mar 9, 2009

### NJV

Thank you for the information. The equation proves useful.

6. Mar 10, 2009

### Bob S

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.

7. Mar 11, 2009

### Bob S

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).