Shielding Equation Derivations For X And Gamma Rays
- 1. Derivation of the shielding formulas (M J Rhoades) Gammas and X-rays We know the three major interactions that occur with gamma and X-rays in materials. This matter could be our bodies so we had better consider ways to shield against it. I hate to think of all those electrons and positrons flying around in the cells of my body. So, let's talk about the attenuation of X and gamma rays. To start, let's consider a piece of lead shielding between a point source and you. Gammas start out with an initial intensity (this can be dose rate, count rate, energy, and so on.), and hit the shield. Gammas can, (depending on their energy), either goes on through, create a photo electron, Compton scatter, or create an electron-positron pair. If we described the probability that any one of the interactions above would occur, we would want to find the cross section of absorption for each of the processes. The symbol for cross section of absorption is σ. We would also want to total the different cross sections and have a σtotal . The equation for this would then be as follows: σtotal = σpe + σpp + Zσcs Where: σtotal = the total cross section of absorption for the energy of our gamma. σpe = the cross section for the photoelectric effect in cm2/atom. σpp = cross section for pair production in cm2/atom. Zσcs = cross section for Compton scattering x the number protons in cm2/atom Example: Pb208* σtotal = (6.1766 x 10−23 cm2/atom) + (1.3 x 10−25 cm2/atom) + (1.548 x 10-23cm2/atom) σtotal = 2.1787 x 10-23cm2/atom * From NIST data @ 1.3 MeV gamma. Now, back to our lead shield, we have the data for the cross section for absorption of the lead atoms for the three interactions. The next thing we need to know is how many atoms are packed into that shield, for this we must know the atom density. The equation for this is as follows: N= Where: N = atom density in atoms/cm3 ρ = density in g/cm3 NAv = Avogadro's number ( 6.02 x 1023 atoms/mole ) NUPT 221-01/(25629) Page 1
- 2. M = gram atomic weight in gram* atoms (11.35/ 3 )(6.02 10 23 / ) For Lead 208 we have: N = 208/ N = 3.285 x 1022 atoms/cm3 We now have the information we need to determine the gamma attenuation for our 1.3 Mev gamma, the 1/2 thickness formula, the 1/10 thickness formula, the linear attenuation coefficient, the mass attenuation coefficient and more. A gamma point source emits gammas at 1.3 MeV gammas at our lead shield at some intensity, say: gammas / m2/sec. Obviously, IO If γ Pb Nσ γ Detector dt the gammas pass into the shield, some are absorbed by the lead atoms. The amount of the absorption is a function of Nσtotal and the thickness of the shield. Let me go over that again, from the equations we talked about above, the intensity is reduced by the N, (atom density) x σtotal (the total cross section for absorption) and the thickness. Now, think about if we added a little NUPT 221-01/(25629) Page 2
- 3. bit of thickness, (dt, change in thickness to the shield) yes that's right, the amount of gamma making it through would drop a little. Or, if we were to put this in terms of an equation, it would look like the following: dI = -NσtotalI dt Where: dI = the change in the intensity. N = the atom density. σtotal = the total absorption cross section. I = the original intensity of gamma entering the shield. dt = the change in thickness. The negative sign comes from the fact that the change is negative, or reduced by the factors. The equation in this form really does not help us very much. Differential equations in this form just tell a story and are not good for plugging and chugging out answers. First, I want to let you in on little secret before we go any further. (Most of you may have all ready known, but I did not) Nσtotal = µ (The linear attenuation coefficient) I just wanted to show you where "µ" came from. And, starting from scratch, as I did, is all ways a good idea in my opinion. Ok, back to our equation. We are going to have to take definite integrals on both sides of the equation. It has to be definite integrals because we do not want to deal with any constants. I am going to start this on the next page so the equations are not split up. ( Don't have to keep thumbing back and forth.) NUPT 221-01/(25629) Page 3
- 4. Nσtotal = µ As stated above dI = -µI dt Substituting µ for Nσtotal. = -µ dt Dividing both sides by I = 0 −µ Taking definite integrals both sides, defined areas Final intensity = If thickness Rule of log dx/x ln = -µ 0 Rule: constants can move through the integral sign Initial intensity = Io 0 thickness ln (If) - ln (Io) = - µ( t-0t) Integral complete right side ln = -µ t Rule of logs on left side, = −µ Removing logs both sides = e -µ t Multiply through by Io, complete. This equation is also the basis for the half and 1/10 layer calculations as well as working with dose rates. NUPT 221-01/(25629) Page 4
- 5. While we have this formula fresh in our minds, we may as well get our 1/2 layer and 1/10 layer formulas derived, and finish up with our Pb 208 example. First the 1/2 layer formula: If = Io e-µ t We are after the thickness of Pb which will reduce the gamma ray intensity by 1/2. So let the ratio of If to Io be equal to 1/2. 1 -u t =e Dividing both sides by I and putting in our desired ratio. 2 this just means that Io is being cut in half. 1 = e-µ t By taking the natural log both sides 2 1 ln 2 = ln e-µ t - .693 = - µ t Result after taking the logs both sides. −.693 = t1/2 Renaming the t to reflect the half value thickness. −µ Dividing by minus µ .693 = t1/2 multiplying the left side by -1/-1, complete µ For the 1/10 layer formula just use 1/10 as the ratio for If / Io 1 -µ t =e 10 -2.3o3 = -µ t1/ 10 2.303 = t1/ 10 µ Remember that: Nσtotal = µ, let's go ahead and multiply my figures out for Pb208 with our 1.33 MeV gamma from cobalt 60. NUPT 221-01/(25629) Page 5
- 6. N = 3.285 x 1022 atoms/cm3 σtotal = 2.1787 x 10-23 cm2 / atom Then: Nσtotal = (3.285 x 1022 atoms/cm3)(2.1787 x 10-23 cm2/atom ) = .7157 cm-1 = µ So, for our shield using Pb 208, and our 1.33 MeV gamma calculations. .693 = t1/2 µ .693 = .968 cm = 9.68 mm of Pb 208 .7157 −1 This agrees very well for the lead 1/2 value listed for Pb206 in Martin and Harbison P.69.( I am sure that the t1/2 layer values listed in table 8.1 are on the conservative side), Other equations related to the intensity formula, I gave you are as follows: Dt = D0 e-µ t Dose rate instead of intensity If = I0 e (-µ/ρ) ρt This formula is for using the "Mass attenuation coefficient." Most tables list the mass attenuation coefficient instead of µ "the linear attenuation" but, all you have to do is multiply by the density to get µ, so no problem. µ So the mass attenuation coefficient is just = Ok , I kept my promise as stated earlier in the text that you can solve a lot of problems by under sanding the base formula we worked through above. (Note: All of the base information for working with these formulas, for all the nuclides, can be found by going to www. nist.gov/. NUPT 221-01/(25629) Page 6