# Californium(Cf)-252 shielding design..

by Nucengable
Tags: californiumcf252, design, shielding
 P: 42 Hello , If I wanted to design a radiation shield to protect from line source of Ca-252 with 5.3 Ci what is the first thing I should think about..?
P: 21,827
 Quote by Nucengable Hello , If I wanted to design a radiation shield to protect from line source of Cf-252 with 5.3 Ci what is the first thing I should think about..?
Is that strictly Cf-252, or Cf-252 + Be.

One wants to look at neutron and gamma emissions from Cf-252 and daughters, so one must consider the spectra and decay chain.

Cf-252 has a specific purpose, which is usually as a neutron source. It has a geometry designed for that purpose. The shield is provided as separate structure for storage and/or transportation.
 P: 42 its Cf-252 without Be in my case I don't want to use it as neutron source , no transportation .. so the design geometry is not that important to me , the purpose of the design to have less than 10 microSv/hr at the surface of the shield.. the priority is be simple and to use the "right" shielding materials with the right thickness.. but what about Alpha emissions from Cf-252 ? .. I though I can use these materials and in the same order Hydrogenous medium Gd-155 4 or maybe 5 mean free paths 2-3 mm Al or maybe lead
 P: 163 Californium(Cf)-252 shielding design.. You will likely want a two layered geometry. First you will want hydrogen rich materials to moderate the neutrons. This is usually water, oil, wax or plastics like polyethylene. You might also want to add something to it with a big absorption cross-section. This is usually boron. They sell borated polyethylene for this purpose. Depending on the budget you can even get enriched boron. Whenever a neutron is absorbed you will generate a photon. So on top of the gammas from the source you will also have gammas being produced in the shielding. The energy depends on the isotope absorbing the neutron, I think hydrogen was around 2.1 MeV (but you should really check). So you need to make the outside layer a gamma shield. Which is anything with a high electron density. Lead is the most common material used but you can use pretty much anything depending on your application.
P: 42
 Quote by Hologram0110 You will likely want a two layered geometry. First you will want hydrogen rich materials to moderate the neutrons. This is usually water, oil, wax or plastics like polyethylene. You might also want to add something to it with a big absorption cross-section. This is usually boron. They sell borated polyethylene for this purpose. Depending on the budget you can even get enriched boron. Whenever a neutron is absorbed you will generate a photon. So on top of the gammas from the source you will also have gammas being produced in the shielding. The energy depends on the isotope absorbing the neutron, I think hydrogen was around 2.1 MeV (but you should really check). So you need to make the outside layer a gamma shield. Which is anything with a high electron density. Lead is the most common material used but you can use pretty much anything depending on your application.
Thank you, that was very simple , NOW I just need to do the math.
 P: 163 No problem. Keep in mind that the gammas from the neutron absorption will occur in your neutron shield. If you are doing the calculation by hand the only way I can think of to do it is to assume that all the gammas make it to the surface of the neutron shield. You then add enough gamma shielding to lower that to the desired dose rate. This will be a conservative estimate because some of the gammas will be absorbed in the neutron shield. Without running a neutron transport code it will be hard to tell where they will be deposited.
P: 21,827
 Quote by Nucengable its Cf-252 without Be in my case I don't want to use it as neutron source , no transportation .. so the design geometry is not that important to me , the purpose of the design to have less than 10 microSv/hr at the surface of the shield.. the priority is be simple and to use the "right" shielding materials with the right thickness.. but what about Alpha emissions from Cf-252 ? .. I though I can use these materials and in the same order Hydrogenous medium Gd-155 4 or maybe 5 mean free paths 2-3 mm Al or maybe lead
Firstly, the Cf-252 is going to be encapsulated in some sealed stainless steel tube. Noone would use bare Cf-252 as a neutron source.

Secondly, alpha-particles are pretty stopped by paper or thin sheet metal, so those would be automatically shielded by the capsule wall. Alphas would also be stopped by self-shielding in the Cf material anyway. Betas are more penetrating, but basically if one shields for neutrons and γ's, the betas will be stopped as well. On the other hand, one will need to consider the thermal effects of alpha and betas on the capsule and surrounding shielding.

Shielding for gammas requires high-Z material, usually lead, but it could be DU as well. If one is shielding for neutrons, then one has to consider structural material with low activation potential, thermalizing fast neutrons, and capturing the thermal neutrons without unnecessarily increasing the gamma emissions (activation).

So the structural material must be strong, corrosion resistant, with relatively low activation from neutrons - which could be a low nickel SS. There should be some hydogenous material of which Hologram gave examples, but one could also use a metal hydride, e.g., LiH or ZrHx, and then a high-Z material to attenuate gammas. The shielding must also handle the thermal burden - and so much maintain strength as well as be thermodynamically/chemically stable.
 P: 42 Thank you very much "Hologram" ,"Astronuc " , you made it so clear ... may I come back to you if I had problems with my calculations ? I have this as project and I have to deliver it by a week its got 30% of the grade. I know its simple but I've been always interested in physics more than math. ... Thanks again
 Admin P: 21,827 Well - it's similar requirements to space reactor shielding. We'll be here. Physics and math go hand-in-hand. Physics is very quantitative.
 P: 42 yes actually it is .. Thank you very much Astronuc I really appreciated I wish there're more people like you in this world .. So by taking a thought about this I came up with this .. Firstly I have a cylindrical 5.3 Ci (Cf-252) neutron source with (height= 1inch and diameter=1mm) which its about (0.29 g) and I guess its too huge to use as a neutron source but this was the deal. it emits an alpha particle 97% of the time and fissions spontaneously about 3% of the time emitting approximately 3.6 neutrons per fission .. anyway Cf emits 2.314 x 10^6 neutrons per sec per μg then I found the intensity of 0.29 g of Cf with average neutron's energy 2.1 Mev range between 0.003-15.0 Mev ... First when I'm designing the shield should I consider the energy of neutron as an average of 2.1 Mev or maximum energy 15 Mev? ... For the neutron response function I used the tabulated data in "Radiation Shielding" book for "Shultis" when I looked up in the appendix I still confused should I use the value for the "effective dose equivalent" or the "dose equivalent".? .... but the major problem , HOW can I know the number of gammas produced per second in the borated polyethylene after absorbing the neutrons ? and in what energy? ... Thank you very much.
 Admin P: 21,827 One could bin the neutron energy spectrum by 1 MeV bin widths. Then one has to determine the scattering and absorption cross-sections in the shielding. The neutron flux/current is attenuated, and that attenuation curve provides source for (n,γ) reactions. Sounds like one is doing back of the envelope calcs as opposed to Monte Carlo transport with a code like MCNP.
 P: 42 Well now I'm officially lost !! I can think about one thing to do with the absorption cross section , absorption rate of neutrons is equals the production rate of gammas but still I would get lost with the units (cm^-3 s^-2) and with the (s^-2) all the design is to find the thickness how would I find the volume then. and even so I couldn't find any data for the borated polyethylene "cross section , build up factor" and the scattering cross section will provide me with information about how many neutrons will scatter to the appropriate energy to be absorbed but can't figure it out how that could help me, I still undergrad student in my 3rd year I guess there's still a lot to go through... but I will figure it out somehow. ... Thanks a lot
 P: 1 hello admin :) I have the same problem about californium source shield design . I dont know how to start my project :( can u help me ?
P: 628
 Hello , If I wanted to design a radiation shield to protect from line source of Ca-252 with 5.3Ci what is the first thing I should think about..?
On a personal note, the first thing I'd say you should think about is why someone would consider giving you authority on that much nuclear material yet you still need to ask such a question.

I don't understand how you (Nuc.. and Ash..) might come across, or be responsible for, 5.3Ci of Cf-252 and need to ask these questions. I thought/imagined such quantities were under strict control from ORNL and you should not be getting your hands on such quantities unless you were VERY well briefed in the first place.

Could you enlighten us on the use, and provenance, of this Cf-252? Is that much Cf available to production lines these days for some production-based purpose?

Cf-252 has a specific activity of 19.85TBq/g. 5.3Ci is 0.196TBq, so that'd be 10mg of material (where did you get 0.29g from?). In any case, it is a shed-load of neutrons.

I suspect you are confused. In the UK I seem to recall there is some test/industrial sample limit for some classifications at 200kBq, which is 5.4μCi. Maybe there is a similar limit elsewhere. Is this what you have/will have, 5.4μCi?
P: 628
The following indicates the maximum sources made are of 50mg 252-Cf, so I don't think you can have a 0.29g source as you suggest.

If you were to have a 5.3Ci source (10mg) then the info below shows it would cost some $32,000 commercially (2000 prices) to loan. For information, from a 2000 ORNL publication (http://www.ornl.gov/~webworks/cpr/pres/107270_.pdf);  Californium Industrial Loan Program: The DOE inventory of sealed 252-Cf sources is stored at the Radiochemical Engineering Development Center of Oak Ridge National Laboratory (ORNL). Californium-252 is produced, purified, and encapsulated at ORNL as a byproduct of DOE’s heavy element program. While source material is sold to commercial vendors at a current price of$60/μg of 252-Cf, plus encapsulation, packaging, and transportation charges, government researchers and contractors can obtain DOE sources on loan without charge for the radioisotope. If an appropriate source is available from the DOE source inventory at ORNL, the loanee pays only the technical service charges incurred for source preparation, shipment, and return. As part of the loan agreement, DOE requires source return to ORNL after use, eliminating source disposal concerns and costs to the user. One microgram of 252-Cf emits 2.314 × 10^6 fast neutrons/s with a 2.645-year half-life. Typical costs for loan and return of a <7μg source total ~$11,000. (This fee is waived under the University Loan Program for university research and teaching applications.) Similar costs for a source in the range of 7μg to 3 mg (neutron intensities ranging up to 7 × 10^9/s ) total ~$20,000, while sources in the 3- to 5-mg range total ~$28,000. Loan/return costs of pre-existing sources from inventory containing >5 mg total ~$32,000. Sources containing >8 mg typically require custom fabrication, but a source containing the maximum permitted 252-Cf content of 50 mg (neutron intensity ~10^11/s) can be obtained for ~\$51,000. None of these costs include transportation charges. Sources with neutron intensities <10^8/s can be obtained at lower cost from commercial vendors, but potential costs for end-of-use source disposal should be evaluated. Loan costs for high-intensity sources compare very favorably with procurement costs for electronic neutron generators and accelerators with comparable intensities. The choice of radioisotopic vs electronic neutron sources is dependent on the specific application and on a variety of practical factors.
P: 21,827
 Quote by cmb On a personal note, the first thing I'd say you should think about is why someone would consider giving you authority on that much nuclear material yet you still need to ask such a question. I don't understand how you (Nuc.. and Ash..) might come across, or be responsible for, 5.3Ci of Cf-252 and need to ask these questions. I thought/imagined such quantities were under strict control from ORNL and you should not be getting your hands on such quantities unless you were VERY well briefed in the first place. Could you enlighten us on the use, and provenance, of this Cf-252? Is that much Cf available to production lines these days for some production-based purpose? Cf-252 has a specific activity of 19.85TBq/g. 5.3Ci is 0.196TBq, so that'd be 10mg of material (where did you get 0.29g from?). In any case, it is a shed-load of neutrons. I suspect you are confused. In the UK I seem to recall there is some test/industrial sample limit for some classifications at 200kBq, which is 5.4μCi. Maybe there is a similar limit elsewhere. Is this what you have/will have, 5.4μCi?
I believe it is an exercise for a class project.

Cf-252 is used for primary neutron sources for startup of fresh cores in nuclear reactors.
PF Gold
P: 3,515
 Firstly I have a cylindrical 5.3 Ci (Cf-252) neutron source with (height= 1inch and diameter=1mm) which its about (0.29 g) and I guess its too huge to use as a neutron source but this was the deal. it emits an alpha particle 97% of the time and fissions spontaneously about 3% of the time emitting approximately 3.6 neutrons per fission

i hope these were just homework questions.

what earthly use could anyone have for such a thing?
i dont want one in my neighborhood.

is my arithmetic right here?
 Cf emits 2.314 x 10^6 neutrons per sec per μg
2.314E6n/ug times 2.9e5 ug equals
6.7E11 n/sec from a cylinder about 0.040 X 1 inch...
that's about 8E11 n/cm^2/sec, nearly same as the neutron flux in my school's research reactor at full power(1E12) ?

can you imagine if that thing got misplaced?
it'd be like "The Mouse that Roared" .

old jim
P: 628
 Quote by Astronuc I believe it is an exercise for a class project.
I sure hope so!

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