It all depends on your initial dose rate, and what dose rate you are prepared to accept. For example if you have an initial dose rate of 1 Gy per minute, and would accept 7.5 micro Gy per hour, you would need 23 HVLs. However HVL is usually measured for narrow beams, and situations like this will often be in broad beam geometry, where scatter cannot be ignored. Several textbooks on radiation protection list TVL (tenth value layers) for various shielding materials, which are more appropriate for this sort of calculation. In the example above, 7 TVLs might be appropriate. However none of these are enough to "stop" the radiation, as attenuation is an exponential process.
Dear Moradi,
I am providing below the HVL calculation for all energies for typical absorber materials used for gamma ray attenuation
Refer to Handbook of Chemisty and Physics (Google search), latest Edition, 2016. Here you find table of mass attenuation coefficients of important elements for energies from 10 keV to 1 GeV. Taking lead (Z=82) as the attenuating element, we find:
µ/ρ for 1MeV = 7.1 x 10-2 cm2/gm;
µ/ρ for 1GeV = 11.5 x 10-2 cm2/gm;
Now HVL = 0.693/µ, where µ is the linear attenuation coefficient and µ/ρ is the mass attenuation coefficient.
Hence HVL in lead (ρ, density of lead = 11.34) for 1MeV is (0.693)/( 7.1 x 10-2 x 11.34) = 0.861 cm of Pb or 8.61 mm;
And HVL in lead (ρ, density of lead = 11.34) for 1GeV is (0.693)/( 11.5 x 10-2 x 11.34) = 0.531 cm of Pb or 5.31mm.
The TVL (tenth value layer) values can be calculated using the relation, TVL = 3.32 HVL
As for your other question: how many HVL values are required, it depends on the radiation dose equivalent rate level at a particular point of the beam and the transmitted level desired after attenuation by the absorber (Remember, the radiation attenuation is exponential and no amount of absorber thickness will stop the beam completely). One HVL will reduce the dose rate level by one-half; one TVL will reduce the level by one-tenth.
We will take an illustrative example. A 1 MeV source (close to a Co-60 source, with an average energy of 1.25 MeV) is emitting gamma radiation whose dose equivalent rate is 1Sv/hr (roughly an air kerma rate of 100 rads/hr) at a specified distance from the source. A barrier of lead is required so that the dose equivalent rate on the other side of the barrier is safe for an occupational worker (20 mSv/yr or 10 µSv/hr, assuming the total working hours/year is 50 weeks/yr and 40 hrs/week). How many TVLs and HVLs are required?
1 TVL reduces the dose equivalent rate by a factor of 10
Therefore, 5 TVLs reduce the dose equivalent rate by a factor of 100000, i.e., the dose rate reduces to 10 µSv/hr
So the thickness of lead barrier required = 5 x 8.61 x 3.32 = 14.3 cm (about 17 HVLs)
The above computation assumes narrow beam geometry, for which the mass attenuation coefficients are given in the Handbook mentioned above. In practice, there will be a lot of scatter from the absorber,etc. for a broad beam for which build-up corrections are required.
Note: I did not elaborate on the mathematics of the above computations because they are available in any book on radiation protection.
I hope the above will be of help. Please give your feedback.
Dear Mr. Thomas and Mr. Ananthanarayanan,
I quite understood that Dose Rate is the determining factor.
Thank you very very much for your answers.
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