On Thu, Dec 1, 2011 at 1:01 PM, Berke Durak <[email protected]> wrote:
> 2011/12/1 Joshua Cude <[email protected]>: > > Putting in a gamma source is easy. Putting one in that could account for > the > > 10 kW of power? Not so easy. This would correspond to megacuries, or at > > least hundreds of kCi. That's 1000 times more radioactive than the > sources > > in devices used for radiotherapy, and unshielded, they are extremely > > dangerous. > > Curies don't mean much by themselves. > > Your figure is assuming 6.5 MeV gammas. > > Do we know the spectrum of the gammas inside the reactor? > > There was some talk of 500 keV gammas. > > In that case you get > > 10e3 [J/s] / ( 1.6e-19 [eV/J] x 500e3 [eV/photon] ) = 1.25e17 [photon/s] > > This is > > 1.25e17 [photon/s] / 3.7e10 [photon/s/curie] = 3.4e6 [curie] > So far, so good. I said megacuries. So, that fits. > > Assuming an area of 25 cm^2, that's a flux of 5e19 photon/m2/s. > > Then 2.5 cm of lead is approximately necessary and sufficient to > shield that flux to less than > 1 photon/m2/s. > > Whoa. That doesn't fit my experience in undergraduate physics, where we used a few microcurie Cs-137 source (gammas 660 keV), and counted the gammas through several cm of lead. It's a bit more energy, but a trillion times lower activity. 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. 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. And Villa cut a hole in the lead. And still measured nothing above background. Photons well below 200 keV would be largely stopped, but that doesn't fit the conceivable reactions. So, maybe they're inconceivable.
