[Date Prev][Date Next][Thread Prev][Thread Next][Date Index][Thread Index]

HVL for Xe-133 -Reply



Peter Fundarek wrote:

>Would anyone out there in radsafer land have information on 
>the HVL in lead for Xe-133?  I have seen two values, 0.2 mm 
>Pb and 0.8 mm Pb.  Based on its decay energies (similar to 
>I-125) I would have expected something thinner at about the 
>same as I-125, 0.02 mm Pb.  

You've already got a couple of excellent accurate answers and one
even provided a measured value! Let me add a couple of comments you
might find of interest, and they will be ballistic deficit, er, I
mean deficient.  

A source that emits x-rays and gamma rays of different energies
cannot be said to have a single HVL (without a little qualification).
As the low energies are preferentially removed early on in the
shield, the radiation hardens and the effective energy increases. I
believ it is common to give the HVL that exists at depths where the
low energy stuff has been cleaned out. I think I have seen this
referred to as a deep penetrating half thickness. In other words, we
can limit our consideration to the high energy  radiation as long as
its intensity is half way decent.

With Xe-133  (5.2 day) there is, as was pointed out, an 81 keV gamma
in addition to the 30 keV x-rays. I-125 doesn't have anything near 81
keV so you shouldn't expect the same HVL for xenon and iodine.

With a little hand waving, the HVL can be calculated as follows if
you are so inclined:

HVL  =  0.693/u where u is the linear attenuation coefficient.

Doing this for I-125 (ca 30 keV) we get 0.693/337 = 0.002 cm (0.02mm)
For Xe-133 (ca 80 keV) we get 0.693/26.4  =  0.026 cm (0.26mm)

There happens to be another Xe-133 with a 2.19 day half life, ie a
metastable state and a gamma at 233 keV. Doing the same thing here we
get

HVL = 0.693/11.2  = 0.06 cm (0.6 mm) Could this be a possible
explanation for the 0.8mm value you found? 

These calculated values would apply to good geometry situations.
Under broad beam bad geometry, the HVL would be expected to increase
because gammas scattered out can now also be scattered back.

Perhaps you could buy those of us who answered a beer at the
Brunswick Tavern should we happen to be in the neighborhood?

Best of luck

Paul Frame
Professional Training Programs
Oak Ridge Institute for Science and Education
framep@orau.gov