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The Mexican Cobalt Source

Dear Reader,

We now have details of the size of the cobalt-60 source stolen and then found in Mexico, it was 3000 Ci according to the IAEA. I was wondering why such a source should be allowed to be driven around without either a police or military escort. I think that if two policemen were to escort the truck with the source then the chances of it being stolen would be greatly reduced, but hindsight is a great thing. With hindsight everything is so clear, with hindsight I am sure that we would allow have more success in life. Just think you would be more able to get your dream job, have success in your love life and avoid doing stupid and embarrassing things. But sadly like the rest of you I can not change the past or rewrite history !

But back to cobalt sources.

I recall in my youth under the supervision of Barry E. Tyler (genius physics teacher) doing the Searle’s bar experiment which is a simple system where heat flows along a thermally isolated bar.

If we have a solid bar with a thermal conductivity of k (W / m K) which has a cross sectional area of A (square meters) and a length of L (meters). Then if a difference of x kelvin (ΔT) exists between the two ends then the thermal power (rate of heat energy transfer through the bar) will be given by

P = k ΔT A / L

The temperature gradient along the bar will be linear.

If on the other hand we were to have a sphere of a solid with a spherical void in the middle containing a heat source then the maths will be a bit more complex.

The surface area of a sphere is given by

A = 4πr2

So it should be clear that the area over which the heat is being transmitted is a function of the distance from the centre of the sphere.

We can say for Searle’s bar that the thermal power is given by

P = A k (dT/dx)

(dT/dx) is the temperature gradient along the bar.

So we can apply our equation to the sphere (x is the distance from the centre of the sphere)

P = 4πx2 k (dT/dx)

Rearrange to get

P / 4πx2 k = (dT/dx)

It is important to note that at the centre of a perfect sphere we could have an infinite thermal gradient, so we should be careful to make sure that the heat source at the centre has a non zero radius.

Things will get more complex in a real shield around a cobalt-60 source as the gamma photons will travel some distance into the lead before their give up their energy to the lead.

If we make a lead shield which is a sphere which has a spherical void which has a radius of 2 cm in the centre and is 12 cm in radius, then we can do some calculations. If we assume that all the radiation is absorbed inside the source at the centre of the sphere and that the thermal output of the source is 250 W then we can make a graph of the temperature of the sphere as a function of the distance from the centre of the sphere.

This was easy but the more complex case of the photons traveling some distance into the lead does seem more fun, now the decay energy of cobalt-60 is 2.824 MeV, of this 96 keV on average is the energy of the beta particles. So we know that 3.4 % of the energy has to be released in the source. If we choose to use the first approximation for gamma rays that they obey the Beer-Lambert rule and that the linear attenuation coefficient in lead is 0.6 cm-1 then we can make a new calculation.

250 W of power from the decay of cobalt-60 in a sphere of lead with a 2 cm radius caviety in the middle

250 W of power from the decay of cobalt-60 in a sphere of lead with a 2 cm radius cavity in the middle


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