Gamma Ray Shielding -- How much is enough?

  • #1
Salman Khan
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If Shielding is required for gamma ray flux of 1e23 (let say E=1 Mev). As Shielding cannot fully attenuate flux to zero, how much percentage is technically recommended for shielding of above flux?.
So far as I know shield thickness is considered to minimise dose/flux Upto 1/10 of its unshielded value. Please explain ?
 
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  • #2
In the US the amount of shielding required depends on the amount of radiation generated called the workload, the fraction of the time the main radiation beam hits the shield, the time the area shielded is occupied, and the amount of radiation permitted in the area determined by the regulations for radiation workers or the general public.

What is the nature of your radiation source?
 
  • #3
If I want shield a detector from scattered radiation, what criteria will I follow for this shielding??
 
  • #4
You want to shield a detector? I would think it would be obvious. Scatter produces unwanted signals which uses processing time and can obscure the signal of interest. You would want to have enough shielding to reduce the effect of the scatter to an acceptable level. What signal-to-noise level is acceptable? If the signal is strong then you do not need as much shielding. If the signal is weak the shielding is critical. The source of the radiation may have characteristics that require consideration too
Salman Khan said:
how much percentage is technically recommended for shielding of above flux
I do not know nor do I believe that there is a technical recommendation for shielding a detector. Normally if experimental time is not an issue one would estimate the requirement and then test it and adjust as necessary. Scatter will mostly be from Compton scattering

Salman Khan said:
So far as I know shield thickness is conside red to minimise dose/flux Upto 1/10 of its unshielded value. Please explain ?
This is known as the tenth-value thickness. There is also the half-value thickness with an attenuation of 1/2. If the source is a pure gamma emitter then the tenth value layer tends to remain constant. If the source is a bremsstrahlung source the tenth value layer increases with each tenth value layer applied as they remove lower energy photons that are present.
 
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  • #5
Salman Khan said:
1e23 (let say E=1 Mev)
That would be a substantial gamma flux. However, 1 MeV gammas can be quickly attenuated in high Z materials, e.g., depleted U (Z=92), natural Th (Z=90), Pb (Z=82), or W (Z=76). Water can be used but the thickness would be greater.

I did a quick calculation with the PENELOPE code, and I find that with Fe (Z=26), nearly full attenuation of 1 MeV gammas occurs in 12 cm (down to 0.005, or a reduction factor of about 200); at 15 cm the dose is down to about 0.001, or a reduction factor of about 1000. It depends on the source strength and the required dose at the surface of the shield.

Using U, the dose would be reduced by a factor of 200 to 0.005 at about 4 cm, or down to 0.001 at 5.2 cm. Pb and W would be somewhere in between.

If one were to use water as a shield, those dose at 20 cm would only be reduced to 0.33 of original, at 50 cm, the dose is reduced to a fraction about 0.035. The dose is further reduced to about 0.01 at 67 cm, and to about 0.005 at 76 cm, and then to about 0.001 at close to 100 cm, or 1 m. One will see some bremsstrahlung radiation over several cm.

I did some quick and simple calculations. For a more thorough analysis, one would have to use the exact geometry of source and shielding, source intensity and source gamma spectrum, or a bounding energy, then follow up with an experiment with dosimeters.

Besides a high Z material, one would use distance from the source to reduce dose.
 
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  • #6
I think we need more information about the experimental setup to rationally estimate what should be done.
 

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