On Thu, Dec 1, 2011 at 3:30 PM, Joshua Cude <[email protected]> wrote: > So far, so good. I said megacuries. So, that fits.
Great. > Looking it up on the NIST web site, I get a linear attenuation > coefficient of 1.8 cm^-1 for 500 keV gammas, which corresponds to an > attenuation at 2.5 cm of exp(-1.8*2.5) = 1% or so. Have I made a > mistake in there somewhere? Because that corresponds to a lethal > 10^15 gammas/s total into 4pi solid angle. No, you're right. My mistake, 2.5 cm won't stop that. I used Geant4 to fire some photons at slabs of lead of different thicknesses in Geant4 and got these results : 10000 x 500 keV gammas into Pb absorber thickness -> energy deposited in absorber per photon 0.1 cm -> 62 keV 0.5 cm -> 257 keV 1.0 cm -> 378 keV 2.0 cm -> 474 keV 3.0 cm -> 496 keV 5.0 cm -> 496 keV + 0.463 keV This fits with the formula within a factor of two. > Even with 10 cm of lead, as Villa mentioned, attenuation is 1e-8, > leaving some 10^9 gammas/s total. Still absolutely unmistakeable, > even with a small detector sampling some tiny part of the solid > angle. Right. But what thickness of lead do we need to reduce the photon flux to something comparable with a background radiation level of 10 mSv / year? Isn't that 10e-3 J/kg/year or 317 pJ/kg/s? So we need I/I0 of 317e-12/10e3 = 31.7e-15 or less. That's a thickness of - ln(I/I0)/lambda = 17 cm. > Photons well below 200 keV would be largely stopped Yes, but check out the curve at http://physics.nist.gov/PhysRefData/XrayMassCoef/ElemTab/z82.html Indeed, the absorption of photons in lead depends strongly on the energy even at 500 keV. I believe the lead thicknesses required to reduce the flux to background levels for a bunch of energies are as follows: 500 keV -> mu rho = 1.614e-1 x 11.34 = 1.83 -> 17 cm 400 keV -> mu rho = 2.323e-1 x 11.34 = 2.63 -> 11.8 cm 300 keV -> mu rho = 4.031e-1 x 11.34 = 4.57 -> 6.80 cm 200 keV -> mu rho = 9.985e-1 x 11.34 = 11.32 -> 2.74 cm 100 keV -> mu rho = 2.323e-1 x 11.34 = 62.93 -> 0.49 cm These are not unreasonable thicknesses, especially because: (a) Rossi doesn't want to give the exact energy (obviously). 500 keV sounds like a round number, anything between 200 and 500 keV could be informally called that. (b) We don't know much about the geometry of the device. (c) We don't know if the gammas are emitted isotropically. If they are emitted along a beam, a tube 17 cm long instead of a sphere 17 cm thick may be enough... but then, you'll need to make sure that the lead can take the heat. > but that doesn't fit the conceivable reactions. So, maybe they're > inconceivable. Not much point in speculating about what is conceivable and what is not, since if all this is for real, we're talking new physics here. -- Berke Durak
