# Radiation distribution in subcritical core after meltdown?

A robot (or more probably remotely operated vehicle) was trying to explore around the core of one of the Fukushima reactors and detected extremely high radiation (530 Sievert/hour).

I saw an even higher estimate on Slashdot based on the rate at which the electronics burned out~ 600 Sievert.

I would like to know the distribution of radiation. I assume that in normal operation, a reactor has a lot of neutron flux, but after meltdown, that is lower. Is there any data, or anyone who can give a reasonable distribution for gamma, x-ray, alpha, beta, neutron, etc?

I am asking because I was considering what it would take to create a shielded robot that could survive such conditions so feel free to speculate on that as part of the answer. If all the electronics could be put in a small box, it could be sealed. A robot could be designed with no solid state electronics out of shielding.

I know lead is good because it is dense. Tungsten is denser, is it better? Berylium is used to reflect neutrons, is that of any use? I know different materials match different speed neutrons for capture. What would be the best shielding possible for a CPU to go into hellish conditions like Fukushima?

I found a list of the fission products and their radioactive output. It's not complete, but it is indicative. You can count on 7MeV gamma radiation when a U-235 atom is split by a neutron. There are neutrons but I don't yet have an energy distribution.

Starting with the assumption that the CPU (let's say a Raspberry Pi) and the camera must be shielded or they will die instantly. Same for the motor controllers. Anything outside can only be simple copper. Even that may have a problem, for example insulation, but it should survive much longer.

I found the intensity function for photon radiation, so that is just an exponent law. $I=I_0(e^{-\mu d})$ the constant in the exponent is proportional to the thickness of the matrial, the density, etc.

For neutrons it's much more complicated, and I don't have a bulk equation for neutrons coming through. My plan was for an outer shell of Beryllium or Beryllium-Aluminum allow if that significantly reflects a class of neutrons. Then Boron to absorb neutrons in some energy band (if it matches the energy of any that are common in the environment). Then possibly water if that's necessary. The water layer might be useful to convect heat out of the inner shield.

Then because the thing needs power, I was thinking of using Lead acid or Lead gel-cell batteries as combination mass and shielding.

Finally on the inside, Cd 113 to catch any thermal neutrons that make it through. This is the last layer to catch neutrons specifically, and the idea is to keep any beta or gamma radiation that results outside the final layer so it can be absorbed by the inner shield. The innermost shield is tungsten, or given that tungsten is so hard to work, possibly tungsten fill inside a lead matrix.

I think it is at least possible that neutrons reflect through a whole in the shielding, so the fiberoptics and wires that leave the housing will have to go through bends designed to absorb any radiation.

My question here is whether Beryllium and Boron help at all. Beryllium is used in bombs but that might be a very specific energy level. Boron is a neutron absorber, but I am not sure whether the neutrons in the environment match its absorbtion spectra. If they do work at all, they are at least light, and on the outside which is the biggest surface area, that's a big deal.

• There is a fundamental difficulty with putting a neutron absorber on the inside: neutron capture is accompanied by gamma emission. Commented Mar 25, 2017 at 16:25
• There is still an inner layer of Tungsten/lead. The alternative is putting Cd113 outside the batteries, but then it's much bigger, and cadmium is quite heavy. There is a question of how much thickness would be possible.
– Dov
Commented Mar 26, 2017 at 11:29